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Progression and Sequence Questions in Gujarati

Competitive Exam Quantitative Aptitude · Progression and Sequence · Progression and Sequence

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Showing 50 of 597 questions in Gujarati

301
MediumMCQ
એક સમગુણોત્તર શ્રેણીના પ્રથમ બે પદોનો સરવાળો $12$ છે. ત્રીજા અને ચોથા પદનો સરવાળો $48$ છે. જો સમગુણોત્તર શ્રેણીના પદો વારાફરતી ધન અને ઋણ હોય,તો પ્રથમ પદ શોધો.
A
$-4$
B
$-12$
C
$12$
D
$4$

Solution

(B) ધારો કે પ્રથમ પદ $a$ છે અને સામાન્ય ગુણોત્તર $r$ છે.
પ્રથમ બે પદો $a$ અને $ar$ છે. તેમનો સરવાળો $a + ar = a(1 + r) = 12$ છે ........ $(i)$
ત્રીજું અને ચોથું પદ $ar^2$ અને $ar^3$ છે. તેમનો સરવાળો $ar^2 + ar^3 = ar^2(1 + r) = 48$ છે ........ $(ii)$
સમીકરણ $(ii)$ ને સમીકરણ $(i)$ વડે ભાગતા:
$\frac{ar^2(1 + r)}{a(1 + r)} = \frac{48}{12}$
$r^2 = 4$
$r = \pm 2$
કારણ કે સમગુણોત્તર શ્રેણીના પદો વારાફરતી ધન અને ઋણ છે,તેથી સામાન્ય ગુણોત્તર $r$ ઋણ હોવો જોઈએ. તેથી,$r = -2$.
$r = -2$ ની કિંમત સમીકરણ $(i)$ માં મૂકતા:
$a(1 + (-2)) = 12$
$a(-1) = 12$
$a = -12$
આમ,પ્રથમ પદ $-12$ છે.
302
DifficultMCQ
શ્રેણી $1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ ના અનંત પદોનો સરવાળો કેટલો થાય?
A
$3$
B
$4$
C
$6$
D
$2$

Solution

(A) ધારો કે સરવાળો $S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$ છે.
આ એક અંકગણિતીય-ભૂમિતિ શ્રેણી છે જ્યાં અંશના પદો $1, 2, 6, 10, 14, \dots$ બીજા પદથી સમાંતર શ્રેણીમાં છે અને છેદ $3$ ની ઘાત છે.
$S = 1 + \frac{2}{3} + \frac{6}{3^2} + \frac{10}{3^3} + \frac{14}{3^4} + \dots$
$3$ વડે ભાગતા: $\frac{S}{3} = \frac{1}{3} + \frac{2}{3^2} + \frac{6}{3^3} + \frac{10}{3^4} + \dots$
બંને સમીકરણોની બાદબાકી કરતા:
$S - \frac{S}{3} = 1 + (\frac{2}{3} - \frac{1}{3}) + (\frac{6}{3^2} - \frac{2}{3^2}) + (\frac{10}{3^3} - \frac{6}{3^3}) + (\frac{14}{3^4} - \frac{10}{3^4}) + \dots$
$\frac{2S}{3} = 1 + \frac{1}{3} + \frac{4}{3^2} + \frac{4}{3^3} + \frac{4}{3^4} + \dots$
$\frac{2S}{3} = 1 + \frac{1}{3} + [\frac{4}{3^2} + \frac{4}{3^3} + \frac{4}{3^4} + \dots]$
કૌંસમાં રહેલી શ્રેણી એ અનંત ભૂમિતિ શ્રેણી છે જેમાં પ્રથમ પદ $a = \frac{4}{9}$ અને સામાન્ય ગુણોત્તર $r = \frac{1}{3}$ છે.
સરવાળો $= \frac{a}{1-r} = \frac{4/9}{1 - 1/3} = \frac{4/9}{2/3} = \frac{4}{9} \times \frac{3}{2} = \frac{2}{3}$.
તેથી,$\frac{2S}{3} = 1 + \frac{1}{3} + \frac{2}{3} = 1 + 1 = 2$.
$S = 2 \times \frac{3}{2} = 3$.
303
DifficultMCQ
એક વ્યક્તિએ $4500$ ચલણી નોટો ગણવાની છે. ધારો કે $a_n$ એ $n^{th}$ મિનિટમાં ગણેલી નોટોની સંખ્યા દર્શાવે છે. જો $a_1 = a_2 = \ldots = a_{10} = 150$ હોય અને $a_{10}, a_{11}, \ldots$ એ સામાન્ય તફાવત $-2$ ધરાવતી સમાંતર શ્રેણી $(A.P.)$ માં હોય,તો બધી નોટો ગણવા માટે તેને લાગતો સમય ............... મિનિટ છે.
A
$34$
B
$125$
C
$135$
D
$24$

Solution

(A) પ્રથમ $10$ મિનિટમાં ગણાયેલી નોટોની સંખ્યા $150 \times 10 = 1500$ છે.
બાકી રહેલી નોટોની સંખ્યા = $4500 - 1500 = 3000$.
ધારો કે પ્રથમ $10$ મિનિટ પછી લાગતો સમય $n$ મિનિટ છે. ગણાયેલી નોટોની શ્રેણી $a_{11} = 150 - 2 = 148$ થી શરૂ થતી સમાંતર શ્રેણી $(A.P.)$ છે,જેમાં સામાન્ય તફાવત $d = -2$ છે.
આ સમાંતર શ્રેણીના $n$ પદોનો સરવાળો $S_n = \frac{n}{2}[2a + (n-1)d]$ સૂત્ર દ્વારા મળે છે.
કિંમતો મૂકતા: $3000 = \frac{n}{2}[2(148) + (n-1)(-2)]$.
$3000 = \frac{n}{2}[296 - 2n + 2] = \frac{n}{2}[298 - 2n] = n(149 - n)$.
$3000 = 149n - n^2$,જે દ્વિઘાત સમીકરણ $n^2 - 149n + 3000 = 0$ આપે છે.
સમીકરણના અવયવ પાડતા: $(n - 24)(n - 125) = 0$.
આથી $n = 24$ અથવા $n = 125$ મળે છે.
નોટોની સંખ્યા ઋણ ન હોઈ શકે,તેથી $a_{10+n} = 148 + (n-1)(-2)$ તપાસતા,$n=125$ માટે પદ ઋણ થઈ જાય છે,જે શક્ય નથી. તેથી $n = 24$.
કુલ સમય = $10 + 24 = 34$ મિનિટ.
304
MediumMCQ
એક વ્યક્તિ તેની નોકરીના પ્રથમ ત્રણ મહિનામાં દરેક વખતે $200$ ની બચત કરે છે. ત્યારબાદના દરેક મહિનામાં તેની બચત અગાઉના મહિનાની બચત કરતા $40$ જેટલી વધે છે. નોકરીની શરૂઆતથી તેની કુલ બચત $11040$ કેટલા મહિના પછી થશે?
A
$19$
B
$20$
C
$21$
D
$18$

Solution

(C) પ્રથમ ત્રણ મહિનાની બચત $200, 200, 200$ છે. કુલ $= 600$.
ધારો કે કુલ મહિનાની સંખ્યા $n$ છે. ચોથા મહિનાથી શરૂ થતી બચત એક સમાંતર શ્રેણી $(AP)$ બનાવે છે,જેમાં પ્રથમ પદ $a = 240$ અને સામાન્ય તફાવત $d = 40$ છે.
આ સમાંતર શ્રેણીમાં પદોની સંખ્યા $(n - 3)$ છે.
બચતનો સરવાળો: $600 + \frac{n-3}{2} [2(240) + (n-3-1)40] = 11040$.
$600 + \frac{n-3}{2} [480 + (n-4)40] = 11040$.
$600 + (n-3) [240 + (n-4)20] = 11040$.
$600 + (n-3) [20n + 160] = 11040$.
$(n-3) [20n + 160] = 10440$.
$(n-3) [n + 8] = 522$.
$n^2 + 5n - 24 = 522$.
$n^2 + 5n - 546 = 0$.
દ્વિઘાત સમીકરણ ઉકેલતા: $(n + 26)(n - 21) = 0$.
$n > 0$ હોવાથી,$n = 21$ મહિના મળે છે.
305
DifficultMCQ
વિધાન $-1$: શ્રેણી $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ નો સરવાળો $8000$ છે.
વિધાન $-2$: કોઈપણ પ્રાકૃતિક સંખ્યા $n$ માટે,$\sum_{k=1}^{n} (k^3 - (k-1)^3) = n^3$.
A
વિધાન $-1$ ખોટું છે,વિધાન $-2$ સાચું છે.
B
વિધાન $-1$ સાચું છે,વિધાન $-2$ ખોટું છે.
C
વિધાન $-1$ સાચું છે,વિધાન $-2$ સાચું છે; વિધાન $-2$ એ વિધાન $-1$ ની સાચી સમજૂતી નથી.
D
વિધાન $-1$ સાચું છે,વિધાન $-2$ સાચું છે; વિધાન $-2$ એ વિધાન $-1$ ની સાચી સમજૂતી છે.

Solution

(D) વિધાન $-2$: $\sum_{k=1}^{n} (k^3 - (k-1)^3)$ એ ટેલિસ્કોપિંગ શ્રેણી છે. તેનું વિસ્તરણ કરતા: $(1^3 - 0^3) + (2^3 - 1^3) + (3^3 - 2^3) + \dots + (n^3 - (n-1)^3) = n^3$. તેથી,વિધાન $-2$ સાચું છે.
વિધાન $-1$: શ્રેણી $\sum_{k=1}^{20} ((k-1)^2 + (k-1)k + k^2)$ છે.
નોંધો કે $(k-1)^2 + (k-1)k + k^2 = \frac{k^3 - (k-1)^3}{k - (k-1)} = k^3 - (k-1)^3$.
$k=1$ માટે,$T_1 = 1^3 - 0^3 = 1$.
$k=2$ માટે,$T_2 = 2^3 - 1^3 = 7 = 1+2+4$.
$k=20$ માટે,$T_{20} = 20^3 - 19^3 = 8000 - 6859 = 1141 = 361 + 380 + 400$.
સરવાળો $\sum_{k=1}^{20} (k^3 - (k-1)^3) = 20^3 = 8000$ થાય છે. તેથી,વિધાન $-1$ સાચું છે.
વિધાન $-1$ એ વિધાન $-2$ માં આપેલ નિત્યસમ પરથી સીધું મેળવી શકાય છે,તેથી વિધાન $-2$ એ વિધાન $-1$ ની સાચી સમજૂતી છે.
306
MediumMCQ
જો $x, y, z$ એ $A.P.$ માં હોય અને $\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ પણ બીજા $A.P.$ માં હોય,તો:
A
$x = y = z$
B
$x = y = -z$
C
$x = 1, y = 2, z = 3$
D
$x = 2, y = 4, z = 6$

Solution

(A) આપેલ છે કે $x, y, z$ એ $A.P.$ માં છે,તેથી $2y = x + z$ ....$(1)$.
વળી,$\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ એ $A.P.$ માં છે,તેથી $2 \tan^{-1} y = \tan^{-1} x + \tan^{-1} z$.
સૂત્ર $\tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a+b}{1-ab} \right)$ નો ઉપયોગ કરતા,આપણને મળે $2 \tan^{-1} y = \tan^{-1} \left( \frac{x+z}{1-xz} \right)$.
બંને બાજુ $\tan$ લેતા: $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.
$(1)$ પરથી $x+z = 2y$ મૂકતા,$\frac{2y}{1-y^2} = \frac{2y}{1-xz}$.
આનો અર્થ એ થાય કે $2y = 0$ અથવા $\frac{1}{1-y^2} = \frac{1}{1-xz}$.
જો $2y = 0$,તો $y = 0$,જેનો અર્થ છે $x+z = 0$,એટલે કે $z = -x$. આ $A.P.$ ની શરત સંતોષે છે $(x, 0, -x)$.
જો $1-y^2 = 1-xz$,તો $y^2 = xz$. કારણ કે $x, y, z$ એ $A.P.$ અને $G.P.$ બંનેમાં છે,તેથી તેઓ સમાન હોવા જોઈએ,એટલે કે $x = y = z$.
307
MediumMCQ
શ્રેણી $0.7, 0.77, 0.777, \dots$ ના પ્રથમ $20$ પદોનો સરવાળો કેટલો થાય?
A
$\frac{7}{81}(179 - 10^{-20})$
B
$\frac{7}{9}(99 - 10^{-20})$
C
$\frac{7}{81}(179 + 10^{-20})$
D
$\frac{7}{9}(99 + 10^{-20})$

Solution

(C) ધારો કે સરવાળો $S = 0.7 + 0.77 + 0.777 + \dots$ $20$ પદો સુધી છે.
$S = 7[0.1 + 0.11 + 0.111 + \dots]$ $20$ પદો સુધી.
$9$ વડે ગુણતા અને ભાગતા:
$S = \frac{7}{9}[0.9 + 0.99 + 0.999 + \dots]$ $20$ પદો સુધી.
$S = \frac{7}{9}[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + \dots]$ $20$ પદો સુધી.
$S = \frac{7}{9}[20 - (0.1 + 0.01 + 0.001 + \dots)]$ $20$ પદો સુધી.
કૌંસમાં રહેલ પદ એ સમગુણોત્તર શ્રેણી છે જ્યાં $a = 0.1$,$r = 0.1$,અને $n = 20$.
સરવાળો $= \frac{a(1 - r^n)}{1 - r} = \frac{0.1(1 - (0.1)^{20})}{1 - 0.1} = \frac{0.1(1 - 10^{-20})}{0.9} = \frac{1}{9}(1 - 10^{-20})$.
કિંમત મૂકતા:
$S = \frac{7}{9}[20 - \frac{1}{9}(1 - 10^{-20})] = \frac{7}{9}[\frac{180 - 1 + 10^{-20}}{9}] = \frac{7}{81}(179 + 10^{-20})$.
308
DifficultMCQ
જો $(10)^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + ... + 10(11)^9 = k(10)^9$ હોય,તો $k$ ની કિંમત શોધો.
A
$100$
B
$110$
C
$\frac{121}{10}$
D
$\frac{441}{100}$

Solution

(A) ધારો કે આપેલ શ્રેણી $S = 10^9 + 2(11)^1(10)^8 + 3(11)^2(10)^7 + \dots + 10(11)^9 = k(10)^9$ છે.
બંને બાજુ $10^9$ વડે ભાગતા,આપણને મળે:
$k = 1 + 2\left(\frac{11}{10}\right) + 3\left(\frac{11}{10}\right)^2 + \dots + 10\left(\frac{11}{10}\right)^9$ ......$(i)$
ધારો કે $x = \frac{11}{10}$. તો $k = 1 + 2x + 3x^2 + \dots + 10x^9$.
$x$ વડે ગુણતા:
$xk = x + 2x^2 + 3x^3 + \dots + 9x^9 + 10x^{10}$ ......$(ii)$
$(i)$ માંથી $(ii)$ બાદ કરતા:
$k(1 - x) = 1 + x + x^2 + \dots + x^9 - 10x^{10}$
$k(1 - x) = \frac{1(x^{10} - 1)}{x - 1} - 10x^{10}$
અહીં $x = \frac{11}{10}$ હોવાથી,$1 - x = -\frac{1}{10}$ અને $x - 1 = \frac{1}{10}$ થાય.
$k\left(-\frac{1}{10}\right) = \frac{(\frac{11}{10})^{10} - 1}{\frac{1}{10}} - 10\left(\frac{11}{10}\right)^{10}$
$k\left(-\frac{1}{10}\right) = 10\left(\frac{11}{10}\right)^{10} - 10 - 10\left(\frac{11}{10}\right)^{10}$
$k\left(-\frac{1}{10}\right) = -10$
$k = 100$.
309
DifficultMCQ
ત્રણ ધન સંખ્યાઓ વધતી જતી $G.P.$ બનાવે છે. જો આ $G.P.$ માં વચ્ચેના પદને બમણું કરવામાં આવે,તો નવી સંખ્યાઓ $A.P.$ માં હોય છે,તો $G.P.$ નો સામાન્ય ગુણોત્તર શોધો:
A
$2 - \sqrt{3}$
B
$2 + \sqrt{3}$
C
$\sqrt{2} + \sqrt{3}$
D
$3 + \sqrt{2}$

Solution

(B) ધારો કે $G.P.$ માં ત્રણ ધન સંખ્યાઓ $a, ar, ar^2$ છે,જ્યાં વધતી જતી $G.P.$ માટે $r > 1$ છે.
જો વચ્ચેના પદને બમણું કરવામાં આવે,તો સંખ્યાઓ $a, 2ar, ar^2$ બને છે.
આ સંખ્યાઓ $A.P.$ માં હોવાથી,વચ્ચેનું પદ બાકીની બે સંખ્યાઓનો સમાંતર મધ્યક છે:
$2(2ar) = a + ar^2$
$4ar = a(1 + r^2)$
$a$ ધન સંખ્યા હોવાથી,આપણે $a$ વડે ભાગી શકીએ:
$r^2 - 4r + 1 = 0$
દ્વિઘાત સૂત્ર $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ નો ઉપયોગ કરતા:
$r = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$
$G.P.$ વધતી જતી હોવાથી,સામાન્ય ગુણોત્તર $r$ એ $1$ કરતા મોટો હોવો જોઈએ.
તેથી,$r = 2 + \sqrt{3}$.
310
DifficultMCQ
શ્રેણી $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$ ના પ્રથમ $9$ પદોનો સરવાળો કેટલો થાય?
A
$192$
B
$71$
C
$96$
D
$142$

Solution

(C) શ્રેણીનું $n$-મું પદ $T_n = \frac{\sum_{k=1}^n k^3}{\sum_{k=1}^n (2k-1)}$ દ્વારા આપવામાં આવે છે.
આપણે જાણીએ છીએ કે $\sum_{k=1}^n k^3 = \left[\frac{n(n+1)}{2}\right]^2$ અને પ્રથમ $n$ એકી સંખ્યાઓનો સરવાળો $\sum_{k=1}^n (2k-1) = n^2$ થાય છે.
તેથી,$T_n = \frac{\left[\frac{n(n+1)}{2}\right]^2}{n^2} = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4}$.
આપણે પ્રથમ $9$ પદોનો સરવાળો શોધવાનો છે,$S_9 = \sum_{n=1}^9 T_n = \sum_{n=1}^9 \frac{(n+1)^2}{4}$.
$S_9 = \frac{1}{4} \sum_{n=1}^9 (n+1)^2 = \frac{1}{4} (2^2 + 3^2 + \dots + 10^2)$.
$1^2$ ઉમેરતા અને બાદ કરતા,આપણને મળે છે $S_9 = \frac{1}{4} [(\sum_{k=1}^{10} k^2) - 1^2]$.
$m=10$ માટે સૂત્ર $\sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6}$ નો ઉપયોગ કરતા:
$S_9 = \frac{1}{4} [\frac{10(11)(21)}{6} - 1] = \frac{1}{4} [385 - 1] = \frac{384}{4} = 96$.
311
DifficultMCQ
જો $m$ એ બે ભિન્ન વાસ્તવિક સંખ્યાઓ $l$ અને $n$ $(l, n > 1)$ નો સમાંતર મધ્યક $(A.M.)$ હોય અને $G_1, G_2$ તથા $G_3$ એ $l$ અને $n$ વચ્ચેના ત્રણ સમગુણોત્તર મધ્યકો હોય,તો $G_1^4 + 2G_2^4 + G_3^4$ ની કિંમત શોધો:
A
$4l^2m^2n^2$
B
$4l^2mn$
C
$4lm^2n$
D
$4lmn^2$

Solution

(C) આપેલ છે કે $m$ એ $l$ અને $n$ નો સમાંતર મધ્યક છે,તેથી $m = \frac{l+n}{2}$,જેનો અર્થ છે કે $2m = l+n$.
$G_1, G_2, G_3$ એ $l$ અને $n$ વચ્ચેના ત્રણ સમગુણોત્તર મધ્યકો હોવાથી,શ્રેણી $l, G_1, G_2, G_3, n$ એ સમગુણોત્તર શ્રેણી $(GP)$ માં છે.
ધારો કે સામાન્ય ગુણોત્તર $r$ છે. તેથી $G_1 = lr, G_2 = lr^2, G_3 = lr^3$ અને $n = lr^4$,જેનો અર્થ છે કે $r^4 = \frac{n}{l}$.
હવે,પદાવલિ $E = G_1^4 + 2G_2^4 + G_3^4$ ની ગણતરી કરીએ.
$E = (lr)^4 + 2(lr^2)^4 + (lr^3)^4 = l^4r^4 + 2l^4r^8 + l^4r^{12}$.
$E = l^4r^4(1 + 2r^4 + r^8) = l^4r^4(1 + r^4)^2$.
$r^4 = \frac{n}{l}$ મૂકતા:
$E = l^4 \left(\frac{n}{l}\right) \left(1 + \frac{n}{l}\right)^2 = l^3n \left(\frac{l+n}{l}\right)^2$.
$E = l^3n \frac{(l+n)^2}{l^2} = ln(l+n)^2$.
$l+n = 2m$ હોવાથી,$E = ln(2m)^2 = 4lm^2n$.
312
DifficultMCQ
જો એક અચળ ન હોય તેવી $A.P.$ ના $2^{nd}, 5^{th}, \text{અને } 9^{th}$ પદો $G.P.$ માં હોય, તો આ $G.P.$ નો સામાન્ય ગુણોત્તર શોધો:
A
$1$
B
$\frac{7}{4}$
C
$\frac{8}{5}$
D
$\frac{4}{3}$

Solution

(D) ધારો કે $A.P.$ નું પ્રથમ પદ $a$ અને સામાન્ય તફાવત $d$ છે.
$2^{nd}$ પદ $a + d$ છે.
$5^{th}$ પદ $a + 4d$ છે.
$9^{th}$ પદ $a + 8d$ છે.
આ પદો $G.P.$ માં હોવાથી, ધારો કે સામાન્ય ગુણોત્તર $r$ છે.
તેથી, $(a + 4d) = (a + d)r$ અને $(a + 8d) = (a + d)r^2$.
પ્રથમ સમીકરણ પરથી, $a + 4d = ar + dr \implies a(1-r) = d(r-4) \implies a = \frac{d(r-4)}{1-r}$.
આ કિંમત બીજા સમીકરણમાં મૂકતા: $\frac{d(r-4)}{1-r} + 8d = r^2 \left( \frac{d(r-4)}{1-r} + d \right)$.
$d$ વડે ભાગતા (કારણ કે અચળ ન હોય તેવી $A.P.$ માટે $d \neq 0$):
$\frac{r-4 + 8 - 8r}{1-r} = r^2 \left( \frac{r-4 + 1-r}{1-r} \right)$.
$\frac{4-7r}{1-r} = r^2 \left( \frac{-3}{1-r} \right)$.
$4 - 7r = -3r^2 \implies 3r^2 - 7r + 4 = 0$.
$(3r - 4)(r - 1) = 0$.
$A.P.$ અચળ ન હોવાથી, $d \neq 0$, જેનો અર્થ છે કે $r \neq 1$.
તેથી, $r = \frac{4}{3}$.
313
DifficultMCQ
જો શ્રેણી ${\left( {1\frac{3}{5}} \right)^2} + {\left( {2\frac{2}{5}} \right)^2} + {\left( {3\frac{1}{5}} \right)^2} + {4^2} + \dots$ ના પ્રથમ દસ પદોનો સરવાળો $\frac{16}{5}m$ હોય,તો $m$ ની કિંમત શોધો:
A
$100$
B
$99$
C
$102$
D
$101$

Solution

(D) આપેલ શ્રેણી ${\left( \frac{8}{5} \right)^2} + {\left( \frac{12}{5} \right)^2} + {\left( \frac{16}{5} \right)^2} + {\left( \frac{20}{5} \right)^2} + \dots$ છે,જેમાં $10$ પદો છે.
આને $\frac{1}{25} [8^2 + 12^2 + 16^2 + 20^2 + \dots + (4(n+1))^2]$ તરીકે લખી શકાય,જ્યાં $n=1$ થી $10$ છે.
$n$-મું પદ $T_n = {\left( \frac{4(n+1)}{5} \right)^2} = \frac{16}{25} (n^2 + 2n + 1)$ છે.
$10$ પદોનો સરવાળો $S_{10} = \frac{16}{25} \sum_{n=1}^{10} (n^2 + 2n + 1)$ થશે.
સૂત્રો $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum n = \frac{n(n+1)}{2}$,અને $\sum 1 = n$ નો ઉપયોગ કરતા:
$S_{10} = \frac{16}{25} [\frac{10(11)(21)}{6} + 2 \cdot \frac{10(11)}{2} + 10] = \frac{16}{25} [385 + 110 + 10] = \frac{16}{25} [505]$.
આપેલ છે કે $S_{10} = \frac{16}{5} m$,તેથી $\frac{16}{25} \times 505 = \frac{16}{5} m$.
બંને બાજુ $\frac{16}{5}$ વડે ભાગતા,$m = \frac{505}{5} = 101$ મળે છે.
314
DifficultMCQ
કોઈપણ ત્રણ ધન વાસ્તવિક સંખ્યાઓ $a, b, c$ માટે,જો $9(25a^2 + b^2) + 25(c^2 - 3ac) = 15b(3a + c)$ હોય,તો:
A
$a, b, c$ એ $G.P.$ માં છે.
B
$b, c, a$ એ $G.P.$ માં છે.
C
$b, c, a$ એ $A.P.$ માં છે.
D
$a, b, c$ એ $A.P.$ માં છે.

Solution

(C) આપેલ સમીકરણ: $9(25a^2 + b^2) + 25(c^2 - 3ac) = 15b(3a + c)$
પદોનું વિસ્તરણ કરતા: $225a^2 + 9b^2 + 25c^2 - 75ac = 45ab + 15bc$
પદોને ગોઠવતા: $225a^2 + 9b^2 + 25c^2 - 45ab - 15bc - 75ac = 0$
$2$ વડે ગુણતા: $450a^2 + 18b^2 + 50c^2 - 90ab - 30bc - 150ac = 0$
આને આ રીતે લખી શકાય: $(15a - 3b)^2 + (3b - 5c)^2 + (5c - 15a)^2 = 0$
વર્ગોનો સરવાળો શૂન્ય હોવા માટે,દરેક પદ શૂન્ય હોવું જોઈએ:
$15a - 3b = 0 \Rightarrow 3b = 15a \Rightarrow b = 5a$
$3b - 5c = 0 \Rightarrow 3b = 5c$
$5c - 15a = 0 \Rightarrow 5c = 15a \Rightarrow c = 3a$
હવે,$b, c, a$ માટે $A.P.$ ની શરત ચકાસો:
$2c = b + a \Rightarrow 2(3a) = 5a + a \Rightarrow 6a = 6a$
આમ,$2c = a + b$ હોવાથી,સંખ્યાઓ $b, c, a$ એ $A.P.$ માં છે.
315
DifficultMCQ
ધારો કે $a, b, c \in R$. જો $f(x) = ax^2 + bx + c$ એવું હોય કે $a + b + c = 3$ અને $f(x + y) = f(x) + f(y) + xy$ તમામ $x, y \in R$ માટે,તો $\sum_{n=1}^{10} f(n)$ ની કિંમત શોધો:
A
$255$
B
$330$
C
$165$
D
$190$

Solution

(B) આપેલ છે $f(x) = ax^2 + bx + c$ અને $a + b + c = 3$.
$f(1) = a + b + c$ હોવાથી,$f(1) = 3$ મળે.
આપેલ છે $f(x + y) = f(x) + f(y) + xy$.
$x=0, y=0$ મુકતા,$f(0) = f(0) + f(0) + 0 \Rightarrow f(0) = 0$. તેથી $c = 0$.
હવે,$f(x) = ax^2 + bx$.
$f(1) = a + b = 3$.
$f(x+y) = a(x+y)^2 + b(x+y) = ax^2 + ay^2 + 2axy + bx + by$ નો ઉપયોગ કરતા.
વળી $f(x) + f(y) + xy = ax^2 + bx + ay^2 + by + xy$.
$xy$ ના સહગુણકો સરખાવતા,$2a = 1 \Rightarrow a = 1/2$.
$a + b = 3$ હોવાથી,$b = 3 - 1/2 = 5/2$.
તેથી,$f(n) = \frac{1}{2}n^2 + \frac{5}{2}n = \frac{n^2 + 5n}{2}$.
આપણે $\sum_{n=1}^{10} f(n) = \sum_{n=1}^{10} \frac{n^2 + 5n}{2} = \frac{1}{2} [\sum_{n=1}^{10} n^2 + 5 \sum_{n=1}^{10} n]$ શોધવાનું છે.
સરવાળાના સૂત્રોનો ઉપયોગ કરતા: $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ અને $\sum n = \frac{n(n+1)}{2}$.
$n=10$ માટે: $\sum n^2 = \frac{10 \times 11 \times 21}{6} = 385$ અને $\sum n = \frac{10 \times 11}{2} = 55$.
સરવાળો $= \frac{1}{2} [385 + 5(55)] = \frac{1}{2} [385 + 275] = \frac{660}{2} = 330$.
316
DifficultMCQ
ધારો કે $A$ એ શ્રેણી $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots$ ના પ્રથમ $20$ પદોનો સરવાળો છે અને $B$ એ પ્રથમ $40$ પદોનો સરવાળો છે. જો $B - 2A = 100\lambda$ હોય,તો $\lambda$ ની કિંમત શોધો:
A
$248$
B
$464$
C
$496$
D
$232$

Solution

(A) શ્રેણી $a_n = n^2$ જો $n$ એકી હોય,અને $a_n = 2n^2$ જો $n$ બેકી હોય.
$A = \sum_{n=1}^{20} a_n = (1^2 + 3^2 + \dots + 19^2) + 2(2^2 + 4^2 + \dots + 20^2)$.
$B = \sum_{n=1}^{40} a_n = (1^2 + 3^2 + \dots + 39^2) + 2(2^2 + 4^2 + \dots + 40^2)$.
$B - 2A = \sum_{n=1}^{40} a_n - 2\sum_{n=1}^{20} a_n$.
આને $\sum_{n=21}^{40} a_n - \sum_{n=1}^{20} a_n$ તરીકે લખી શકાય.
એકી $n$ માટે,$a_n = n^2$. બેકી $n$ માટે,$a_n = 2n^2$.
$B - 2A = \sum_{k=11}^{20} (2k-1)^2 + \sum_{k=11}^{20} 2(2k)^2 - \sum_{k=1}^{10} (2k-1)^2 - \sum_{k=1}^{10} 2(2k)^2$.
$x^2 - y^2 = (x-y)(x+y)$ ગુણધર્મનો ઉપયોગ કરતા,$B - 2A = \sum_{k=1}^{10} [(2k+19)^2 - (2k-1)^2] + \sum_{k=1}^{10} 2[(2k+20)^2 - (2k)^2]$.
$= \sum_{k=1}^{10} (20)(4k+18) + \sum_{k=1}^{10} 2(20)(4k+20) = 20 \sum_{k=1}^{10} (4k+18 + 8k+40) = 20 \sum_{k=1}^{10} (12k+58)$.
$= 20 [12 \cdot \frac{10 \cdot 11}{2} + 580] = 20 [660 + 580] = 20 [1240] = 24800$.
$B - 2A = 100\lambda$ હોવાથી,$100\lambda = 24800$,તેથી $\lambda = 248$.
317
DifficultMCQ
ધારો કે $a_1, a_2, \dots, a_{49}$ એ $A.P.$ માં છે,જેથી $\sum_{k=0}^{12} a_{4k+1} = 416$ અને $a_9 + a_{43} = 66$ થાય. જો $a_1^2 + a_2^2 + \dots + a_{17}^2 = 140m$ હોય,તો $m = \dots$
A
$68$
B
$34$
C
$33$
D
$66$

Solution

(B) આપેલ છે કે $a_1, a_2, \dots, a_{49}$ એ સામાન્ય તફાવત $d$ સાથે $A.P.$ માં છે.
સરવાળો $\sum_{k=0}^{12} a_{4k+1} = a_1 + a_5 + a_9 + \dots + a_{49} = 416$.
આ $13$ પદો ધરાવતી $A.P.$ છે,જેમાં પ્રથમ પદ $a_1$ અને સામાન્ય તફાવત $4d$ છે.
સરવાળો $= \frac{13}{2} [2a_1 + (13-1)4d] = 416 \Rightarrow \frac{13}{2} [2a_1 + 48d] = 416 \Rightarrow 13(a_1 + 24d) = 416 \Rightarrow a_1 + 24d = 32 \dots (1)$.
વળી,$a_9 + a_{43} = (a_1 + 8d) + (a_1 + 42d) = 66 \Rightarrow 2a_1 + 50d = 66 \Rightarrow a_1 + 25d = 33 \dots (2)$.
સમીકરણ $(2)$ માંથી $(1)$ બાદ કરતા,$d = 1$ મળે છે. $d=1$ ને $(1)$ માં મૂકતા,$a_1 + 24 = 32 \Rightarrow a_1 = 8$.
હવે,$\sum_{r=1}^{17} a_r^2 = \sum_{r=1}^{17} [8 + (r-1)1]^2 = \sum_{r=1}^{17} (r+7)^2 = \sum_{r=1}^{17} (r^2 + 14r + 49) = 140m$.
સરવાળાના સૂત્રોનો ઉપયોગ કરતા: $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$,$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$.
$n=17$ માટે: $\frac{17 \times 18 \times 35}{6} + 14 \times \frac{17 \times 18}{2} + 49 \times 17 = 1785 + 2142 + 833 = 4760$.
$140m = 4760 \Rightarrow m = \frac{4760}{140} = 34$.
318
MediumMCQ
જો $a, b, c$ એ સમાંતર શ્રેણી ($A$.$P$.) માં હોય,તો સુરેખા $ax + by + c = 0$ હંમેશા કયા બિંદુમાંથી પસાર થશે?
A
$(-1, -2)$
B
$(1, -2)$
C
$(-1, 2)$
D
$(1, 2)$

Solution

(B) આપેલ છે કે $a, b, c$ એ સમાંતર શ્રેણી ($A$.$P$.) માં છે.
સમાંતર શ્રેણીના ગુણધર્મ મુજબ,$2b = a + c$ થાય,જેને $a - 2b + c = 0$ તરીકે લખી શકાય.
સુરેખાનું સમીકરણ $ax + by + c = 0$ છે.
આ સમીકરણને $a(1) + b(-2) + c = 0$ સાથે સરખાવતા,આપણે જોઈ શકીએ છીએ કે રેખા બિંદુ $(x, y) = (1, -2)$ માટે સમીકરણનું સમાધાન કરે છે.
તેથી,સુરેખા $ax + by + c = 0$ હંમેશા બિંદુ $(1, -2)$ માંથી પસાર થશે.
319
DifficultMCQ
જો $\frac{S_n}{S_m} = \frac{n^4}{m^4}$ હોય (જ્યાં $S_k$ એ સમાંતર શ્રેણી $a_1, a_2, \dots, \infty$ ના પ્રથમ $k$ પદોનો સરવાળો છે),તો $m$ અને $n$ ના સ્વરૂપમાં $\frac{a_{m+1}}{a_{n+1}}$ ની કિંમત શું થશે?
A
$\frac{(2m+1)^3}{(2n+1)^3}$
B
$\frac{(2n+1)^3}{(2m+1)^3}$
C
$\frac{(2m-1)^3}{(2n-1)^3}$
D
$\frac{(2m+1)^3}{(2n-1)^3}$

Solution

(A) આપેલ છે કે $\frac{S_n}{S_m} = \frac{n^4}{m^4}$.
સૂત્ર $S_n = \frac{n}{2}[2a_1 + (n-1)d]$ નો ઉપયોગ કરતા,$\frac{\frac{n}{2}[2a_1 + (n-1)d]}{\frac{m}{2}[2a_1 + (m-1)d]} = \frac{n^4}{m^4}$ મળે.
સાદુરૂપ આપતા,$\frac{2a_1 + (n-1)d}{2a_1 + (m-1)d} = \frac{n^3}{m^3}$ મળે.
પદોનો ગુણોત્તર શોધવા માટે,$\frac{a_{m+1}}{a_{n+1}}$ માટે સાદુરૂપ આપતા પરિણામ $\frac{(2m+1)^3}{(2n+1)^3}$ મળે છે.
320
MediumMCQ
$150$ કામદારોને એક કામ ચોક્કસ દિવસોમાં પૂરું કરવા માટે રોકવામાં આવ્યા હતા. બીજા દિવસે $4$ કામદારો ઓછા થયા,ત્રીજા દિવસે બીજા $4$ કામદારો ઓછા થયા અને આ રીતે ક્રમ ચાલુ રહ્યો. હવે કામ પૂરું કરવામાં $8$ દિવસ વધુ લાગે છે. તો કામ કેટલા દિવસમાં પૂરું થયું હશે?
A
$15$
B
$20$
C
$25$
D
$30$

Solution

(C) ધારો કે મૂળ દિવસોની સંખ્યા $n$ છે. કુલ કામ $150n$ માનવ-દિવસ છે.
પ્રશ્ન મુજબ,કામદારોની સંખ્યા સમાંતર શ્રેણીમાં છે: $150, 146, 142, \dots$ જે $(n+8)$ દિવસ સુધી ચાલે છે.
આ સમાંતર શ્રેણીનો સરવાળો $S = \frac{N}{2} [2a + (N-1)d]$ સૂત્ર દ્વારા મળે છે,જ્યાં $N = n+8$,$a = 150$,અને $d = -4$.
$S = \frac{n+8}{2} [2(150) + (n+8-1)(-4)] = 150n$
$\frac{n+8}{2} [300 - 4n - 28] = 150n$
$(n+8)(272 - 4n) = 300n$
$(n+8)(68 - n) = 75n$
$68n - n^2 + 544 - 8n = 75n$
$-n^2 + 60n + 544 = 75n$
$n^2 + 15n - 544 = 0$
દ્વિઘાત સમીકરણ ઉકેલતા: $(n + 32)(n - 17) = 0$.
$n$ ધન હોવાથી,$n = 17$.
તેથી,કામ પૂરું કરવા માટે લાગતા કુલ દિવસો $n + 8 = 17 + 8 = 25$ દિવસ છે.
321
DifficultMCQ
આપેલ છે કે $n$ સમાંતર મધ્યકો ($A$.$M$.) બે સંખ્યાઓના સમૂહ $a, 2b$ અને $2a, b$ ની વચ્ચે મૂકવામાં આવે છે,જ્યાં $a, b \in R$. જો આ બંને સમૂહ વચ્ચેનો $m^{th}$ મધ્યક સમાન હોય,તો ગુણોત્તર $a:b$ શું થાય?
A
$n - m + 1 : m$
B
$n - m + 1 : n$
C
$n : n - m + 1$
D
$m : n - m + 1$

Solution

(D) બે સંખ્યાઓ $x$ અને $y$ ની વચ્ચે $n$ મધ્યકો મૂકવામાં આવે ત્યારે $m^{th}$ સમાંતર મધ્યક $(A.M.)$ નું સૂત્ર $A_m = x + \frac{m(y - x)}{n + 1}$ છે.
પ્રથમ સમૂહ $a, 2b$ માટે,$m^{th}$ મધ્યક $A_m = a + \frac{m(2b - a)}{n + 1}$ છે.
બીજા સમૂહ $2a, b$ માટે,$m^{th}$ મધ્યક $A'_m = 2a + \frac{m(b - 2a)}{n + 1}$ છે.
આપેલ છે કે $A_m = A'_m$,તેથી:
$a + \frac{m(2b - a)}{n + 1} = 2a + \frac{m(b - 2a)}{n + 1}$
બંને બાજુથી $a$ બાદ કરતા:
$\frac{m(2b - a)}{n + 1} = a + \frac{m(b - 2a)}{n + 1}$
$(n + 1)$ વડે ગુણતા:
$m(2b - a) = a(n + 1) + m(b - 2a)$
$2bm - am = an + a + bm - 2am$
$a$ અને $b$ વાળા પદોને ગોઠવતા:
$2bm - bm - am + 2am = a(n + 1)$
$bm + am = a(n + 1)$
$b(m) = a(n + 1 - m)$
તેથી,ગુણોત્તર $\frac{a}{b} = \frac{m}{n - m + 1}$ મળે છે.
322
MediumMCQ
જો ${x_1}, {x_2}, {x_3}$ તેમજ ${y_1}, {y_2}, {y_3}$ સમાન સામાન્ય ગુણોત્તર સાથે સમગુણોત્તર શ્રેણી ($G$.$P$.) માં હોય,તો બિંદુઓ $({x_1}, {y_1}), ({x_2}, {y_2})$ અને $({x_3}, {y_3})$:
A
એક સીધી રેખા પર આવેલા છે
B
એક ઉપવલય પર આવેલા છે
C
એક વર્તુળ પર આવેલા છે
D
ત્રિકોણના શિરોબિંદુઓ છે

Solution

(A) આપેલ છે કે ${x_1}, {x_2}, {x_3}$ અને ${y_1}, {y_2}, {y_3}$ સમાન સામાન્ય ગુણોત્તર $r$ સાથે સમગુણોત્તર શ્રેણીમાં છે.
તેથી,આપણે લખી શકીએ: ${x_2} = r{x_1}, {x_3} = {r^2}{x_1}$ અને ${y_2} = r{y_1}, {y_3} = {r^2}{y_1}$.
બિંદુઓ $({x_1}, {y_1}), ({x_2}, {y_2}), ({x_3}, {y_3})$ સમરેખ છે કે નહીં તે તપાસવા માટે,આપણે આ બિંદુઓ દ્વારા બનતા ત્રિકોણનું ક્ષેત્રફળ ગણીએ:
ક્ષેત્રફળ $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
કિંમતો મૂકતા:
ક્ષેત્રફળ $= \frac{1}{2} |x_1(ry_1 - r^2y_1) + rx_1(r^2y_1 - y_1) + r^2x_1(y_1 - ry_1)|$
ક્ષેત્રફળ $= \frac{1}{2} |x_1y_1(r - r^2) + x_1y_1(r^3 - r) + x_1y_1(r^2 - r^3)|$
ક્ષેત્રફળ $= \frac{1}{2} |x_1y_1(r - r^2 + r^3 - r + r^2 - r^3)| = 0$.
ત્રિકોણનું ક્ષેત્રફળ $0$ હોવાથી,બિંદુઓ સમરેખ છે અને એક સીધી રેખા પર આવેલા છે.
323
MediumMCQ
ધારો કે ${a_n}$ એ ધન સંખ્યાઓની સમગુણોત્તર શ્રેણીનું ${n^{th}}$ પદ છે. ધારો કે $\sum\limits_{n = 1}^{100} {{a_{2n}}} = \alpha $ અને $\sum\limits_{n = 1}^{100} {{a_{2n - 1}}} = \beta $,જેથી $\alpha \ne \beta $,તો સામાન્ય ગુણોત્તર શોધો.
A
$\frac{\alpha }{\beta }$
B
$\frac{\beta }{\alpha }$
C
$\sqrt {\frac{\alpha }{\beta }} $
D
$\sqrt {\frac{\beta }{\alpha }} $

Solution

(A) ધારો કે સમગુણોત્તર શ્રેણી $a, ar, ar^2, \dots$ છે,જ્યાં $a > 0$ અને $r > 0$ છે.
આપેલ છે કે $\alpha = \sum_{n=1}^{100} a_{2n} = a_2 + a_4 + \dots + a_{200}$.
આ એક સમગુણોત્તર શ્રેણી છે જેનું પ્રથમ પદ $a_2 = ar$ અને સામાન્ય ગુણોત્તર $r^2$ છે.
તેથી,$\alpha = ar + ar^3 + \dots + ar^{199} = ar(1 + r^2 + r^4 + \dots + r^{198})$.
આપેલ છે કે $\beta = \sum_{n=1}^{100} a_{2n-1} = a_1 + a_3 + \dots + a_{199}$.
આ એક સમગુણોત્તર શ્રેણી છે જેનું પ્રથમ પદ $a_1 = a$ અને સામાન્ય ગુણોત્તર $r^2$ છે.
તેથી,$\beta = a + ar^2 + \dots + ar^{198} = a(1 + r^2 + r^4 + \dots + r^{198})$.
$\alpha$ ને $\beta$ વડે ભાગતા:
$\frac{\alpha}{\beta} = \frac{ar(1 + r^2 + r^4 + \dots + r^{198})}{a(1 + r^2 + r^4 + \dots + r^{198})} = r$.
આમ,સામાન્ય ગુણોત્તર $\frac{\alpha}{\beta}$ છે.
324
MediumMCQ
એક સમગુણોત્તર શ્રેણીના ત્રણ ક્રમિક પદોનો સરવાળો $14$ છે. જો પ્રથમ અને બીજા પદમાં $1$ ઉમેરવામાં આવે અને ત્રીજા પદમાંથી $1$ બાદ કરવામાં આવે,તો મળતા નવા પદો સમાંતર શ્રેણીમાં હોય છે. તો મૂળ પદોમાં સૌથી નાનું પદ કયું છે?
A
$1$
B
$2$
C
$4$
D
$8$

Solution

(B) ધારો કે સમગુણોત્તર શ્રેણીના ત્રણ ક્રમિક પદો $a/r, a, ar$ છે.
આપેલ છે કે તેમનો સરવાળો $14$ છે,તેથી $a/r + a + ar = 14$,જે સૂચવે છે કે $a(1/r + 1 + r) = 14$ ..... $(i)$
પ્રશ્ન મુજબ,જો પ્રથમ અને બીજા પદમાં $1$ ઉમેરવામાં આવે અને ત્રીજા પદમાંથી $1$ બાદ કરવામાં આવે,તો નવા પદો $a/r + 1, a + 1, ar - 1$ મળે છે.
આ પદો સમાંતર શ્રેણીમાં હોવાથી,$2(a + 1) = (a/r + 1) + (ar - 1)$.
$2a + 2 = a/r + ar$.
$2a + 2 = a(1/r + r)$.
સમીકરણ $(i)$ પરથી,$a(1/r + r) = 14 - a$. આ કિંમતને ઉપરના સમીકરણમાં મૂકતા:
$2a + 2 = 14 - a$.
$3a = 12$,તેથી $a = 4$.
$a = 4$ ને સમીકરણ $(i)$ માં મૂકતા: $4(1/r + 1 + r) = 14$.
$1/r + 1 + r = 3.5$.
$1/r + r = 2.5$.
$r^2 - 2.5r + 1 = 0$.
$2r^2 - 5r + 2 = 0$.
$(2r - 1)(r - 2) = 0$.
તેથી,$r = 2$ અથવા $r = 1/2$.
જો $r = 2$ હોય,તો પદો $4/2, 4, 4(2)$ એટલે કે $2, 4, 8$ મળે છે.
જો $r = 1/2$ હોય,તો પદો $4/(1/2), 4, 4(1/2)$ એટલે કે $8, 4, 2$ મળે છે.
બંને કિસ્સામાં પદો $2, 4, 8$ છે. તેથી સૌથી નાનું પદ $2$ છે.
325
MediumMCQ
ધારો કે $a$ અને $b$ એ $x^2 - 3x + p = 0$ ના બીજ છે અને $c$ અને $d$ એ $x^2 - 12x + q = 0$ ના બીજ છે,જ્યાં $a, b, c, d$ એક વધતી જતી સમગુણોત્તર શ્રેણી ($G$.$P$.) બનાવે છે. તો $(q + p) : (q - p)$ નો ગુણોત્તર કેટલો થાય?
A
$8 : 7$
B
$11 : 10$
C
$17 : 15$
D
આમાંથી કોઈ નહીં

Solution

(C) આપેલ છે કે $a$ અને $b$ એ $x^2 - 3x + p = 0$ ના બીજ છે,તેથી $a + b = 3$ અને $ab = p$.
આપેલ છે કે $c$ અને $d$ એ $x^2 - 12x + q = 0$ ના બીજ છે,તેથી $c + d = 12$ અને $cd = q$.
$a, b, c, d$ એ વધતી જતી સમગુણોત્તર શ્રેણીમાં હોવાથી,પદોને $a, ar, ar^2, ar^3$ તરીકે લો,જ્યાં $r > 1$.
તેથી $a + b = a(1 + r) = 3$ અને $ab = a^2r = p$.
તે જ રીતે $c + d = ar^2(1 + r) = 12$ અને $cd = a^2r^5 = q$.
સરવાળાના સમીકરણોનો ભાગાકાર કરતા: $\frac{ar^2(1 + r)}{a(1 + r)} = \frac{12}{3} \Rightarrow r^2 = 4$. શ્રેણી વધતી જતી હોવાથી,$r = 2$.
હવે,$r = 2$ ને સરવાળાના સમીકરણમાં મૂકતા: $a(1 + 2) = 3 \Rightarrow 3a = 3 \Rightarrow a = 1$.
તેથી $b = ar = 2$,$c = ar^2 = 4$,અને $d = ar^3 = 8$.
$p$ અને $q$ ની કિંમત શોધતા: $p = ab = 1 \times 2 = 2$ અને $q = cd = 4 \times 8 = 32$.
ગુણોત્તર $(q + p) : (q - p) = (32 + 2) : (32 - 2) = 34 : 30 = 17 : 15$.
326
MediumMCQ
નીચેની શ્રેણી $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots$ નો અનંત સુધીનો સરવાળો કેટલો થશે?
A
$3$
B
$4$
C
$7/2$
D
$9/2$

Solution

(C) આપેલી શ્રેણી $S = 2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots \infty$ છે.
આપણે આ શ્રેણીને $1/2$ અને $1/3$ ના ઘાતાંકવાળા પદોને અલગ જૂથમાં વહેંચીને ફરીથી લખી શકીએ છીએ:
$S = (1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots \infty) + (1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots \infty)$.
અહીં મૂળ શ્રેણીમાં શરૂઆતમાં $2$ હતા,જેને આપણે $1 + 1$ તરીકે લખીને બંને ભૂમિતિ શ્રેણીમાં વહેંચ્યા છે.
બંને શ્રેણીઓ અનંત ભૂમિતિ શ્રેણી છે,જેનો સરવાળાનું સૂત્ર $S = \frac{a}{1-r}$ છે.
પ્રથમ શ્રેણી માટે: $a = 1, r = 1/2$. સરવાળો $= \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$.
બીજી શ્રેણી માટે: $a = 1, r = 1/3$. સરવાળો $= \frac{1}{1 - 1/3} = \frac{1}{2/3} = 3/2$.
કુલ સરવાળો $= 2 + 3/2 = 7/2$.
327
MediumMCQ
જો $|\alpha| < 1$ અને $|\beta| < 1$ હોય,જ્યાં $s_1 = 1 - \alpha + \alpha^2 - \alpha^3 + \dots \infty$ અને $s_2 = 1 - \beta + \beta^2 - \beta^3 + \dots \infty$ હોય,તો $1 - \alpha\beta + \alpha^2\beta^2 - \alpha^3\beta^3 + \dots \infty$ ની કિંમત શું થાય?
A
$s_1s_2$
B
$\frac{s_1s_2}{1 + s_1s_2}$
C
$\frac{s_1s_2}{1 - s_1 - s_2 + 2s_1s_2}$
D
$\frac{1}{1 + s_1s_2}$

Solution

(C) આપેલ શ્રેણીઓ અનંત સમગુણોત્તર શ્રેણીઓ છે જેનો સામાન્ય ગુણોત્તર અનુક્રમે $-\alpha$ અને $-\beta$ છે.
$s_1 = \frac{1}{1 - (-\alpha)} = \frac{1}{1 + \alpha} \implies 1 + \alpha = \frac{1}{s_1} \implies \alpha = \frac{1}{s_1} - 1 = \frac{1 - s_1}{s_1}$.
તે જ રીતે,$s_2 = \frac{1}{1 + \beta} \implies \beta = \frac{1 - s_2}{s_2}$.
ધારો કે $S = 1 - \alpha\beta + \alpha^2\beta^2 - \alpha^3\beta^3 + \dots \infty$. આ એક અનંત સમગુણોત્તર શ્રેણી છે જેનો સામાન્ય ગુણોત્તર $-\alpha\beta$ છે.
$S = \frac{1}{1 - (-\alpha\beta)} = \frac{1}{1 + \alpha\beta}$.
$\alpha$ અને $\beta$ ની કિંમતો મૂકતા:
$S = \frac{1}{1 + (\frac{1 - s_1}{s_1})(\frac{1 - s_2}{s_2})} = \frac{1}{1 + \frac{(1 - s_1)(1 - s_2)}{s_1s_2}}$.
$S = \frac{s_1s_2}{s_1s_2 + (1 - s_1 - s_2 + s_1s_2)} = \frac{s_1s_2}{2s_1s_2 - s_1 - s_2 + 1}$.
328
DifficultMCQ
જો $1$ અને $\frac{1}{31}$ ની વચ્ચે $n$ હાર્મોનિક મધ્યકો હોય અને $7^{th}$ અને $(n - 1)^{th}$ હાર્મોનિક મધ્યકોનો ગુણોત્તર $9:5$ હોય,તો $n$ ની કિંમત શોધો.
A
$12$
B
$13$
C
$14$
D
$15$

Solution

(C) ધારો કે $1$ અને $\frac{1}{31}$ ની વચ્ચે $n$ હાર્મોનિક મધ્યકો $H_1, H_2, \dots, H_n$ છે.
તેથી $1, H_1, H_2, \dots, H_n, \frac{1}{31}$ એ હાર્મોનિક શ્રેણી $(HP)$ માં છે.
તેથી $1, \frac{1}{H_1}, \frac{1}{H_2}, \dots, \frac{1}{H_n}, 31$ એ સમાંતર શ્રેણી $(AP)$ માં છે.
ધારો કે આ $AP$ નો સામાન્ય તફાવત $d$ છે. પ્રથમ પદ $a = 1$ અને $(n+2)^{th}$ પદ $31$ છે.
$a + (n + 2 - 1)d = 31 \implies 1 + (n + 1)d = 31 \implies (n + 1)d = 30 \implies d = \frac{30}{n + 1}$.
$k^{th}$ હાર્મોનિક મધ્યક $H_k = \frac{1}{a + kd}$ દ્વારા મળે છે.
આપેલ છે કે $\frac{H_7}{H_{n-1}} = \frac{9}{5} \implies \frac{a + (n - 1)d}{a + 7d} = \frac{9}{5}$.
$a = 1$ મૂકતા: $\frac{1 + (n - 1)d}{1 + 7d} = \frac{9}{5} \implies 5 + 5(n - 1)d = 9 + 63d$.
$5(n - 1)d - 63d = 4 \implies d(5n - 5 - 63) = 4 \implies d(5n - 68) = 4$.
$d = \frac{30}{n + 1}$ મૂકતા: $\frac{30(5n - 68)}{n + 1} = 4 \implies 15(5n - 68) = 2(n + 1)$.
$75n - 1020 = 2n + 2 \implies 73n = 1022 \implies n = 14$.
329
MediumMCQ
જો સમીકરણ $ax^2 + bx + c = 0$ ના બીજનો સરવાળો તેમના વર્ગોના વ્યસ્તના સરવાળા જેટલો હોય,તો $bc^2, ca^2, ab^2$ શેમાં હશે?
A
$A.P.$
B
$G.P.$
C
$H.P.$
D
આમાંથી કોઈ નહીં

Solution

(A) ધારો કે સમીકરણ $ax^2 + bx + c = 0$ ના બીજ $\alpha$ અને $\beta$ છે.
દ્વિઘાત સમીકરણના ગુણધર્મો પરથી,$\alpha + \beta = -\frac{b}{a}$ અને $\alpha\beta = \frac{c}{a}$ મળે છે.
તેમના વર્ગોના વ્યસ્તનો સરવાળો $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}$ છે.
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ હોવાથી,કિંમતો મૂકતા:
$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{(-\frac{b}{a})^2 - 2(\frac{c}{a})}{(\frac{c}{a})^2} = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c^2}{a^2}} = \frac{b^2 - 2ac}{c^2}$ મળે.
આપેલ શરત મુજબ,$\alpha + \beta = \frac{1}{\alpha^2} + \frac{1}{\beta^2}$,તેથી $-\frac{b}{a} = \frac{b^2 - 2ac}{c^2}$.
ચોકડી ગુણાકાર કરતા $-bc^2 = a(b^2 - 2ac) = ab^2 - 2a^2c$ મળે.
પદોને ગોઠવતા,$2a^2c = ab^2 + bc^2$ મળે.
આથી,$bc^2, ca^2, ab^2$ એ $A.P.$ (સમાંતર શ્રેણી) માં છે.
330
MediumMCQ
જો $a, b, c, d$ અને $p$ ભિન્ન વાસ્તવિક સંખ્યાઓ એવી રીતે હોય કે $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le 0$,તો $a, b, c, d$ શેમાં છે?
A
$A.P.$
B
$G.P.$
C
$H.P.$
D
$ab = cd$

Solution

(B) આપેલ અસમતા: $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le 0$ ... $(i)$
ડાબી બાજુના પદને નીચે મુજબ ફરીથી લખી શકાય:
$(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bcp + c^2) + (c^2p^2 - 2cdp + d^2) \le 0$
આનું સાદું રૂપ:
$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \le 0$ ... $(ii)$
વાસ્તવિક સંખ્યાઓના વર્ગોનો સરવાળો હંમેશા અ-ઋણ હોવાથી,સરવાળો $0$ કે તેથી ઓછો થવા માટે દરેક પદ $0$ હોવું જોઈએ:
$(ap - b)^2 = 0 \Rightarrow ap = b \Rightarrow p = b/a$
$(bp - c)^2 = 0 \Rightarrow bp = c \Rightarrow p = c/b$
$(cp - d)^2 = 0 \Rightarrow cp = d \Rightarrow p = d/c$
આમ,$b/a = c/b = d/c = p$. આ દર્શાવે છે કે શ્રેણી $a, b, c, d$ નો સામાન્ય ગુણોત્તર $p$ અચળ છે,જેનો અર્થ છે કે $a, b, c, d$ એ $G.P.$ (ગુણોત્તર શ્રેણી) માં છે.
331
DifficultMCQ
જો $\alpha, \beta, \gamma$ એ અનુક્રમે $ca, ab$; $ab, bc$; અને $bc, ca$ વચ્ચેના સમગુણોત્તર મધ્યકો હોય,જ્યાં $a, b, c$ એ $A.P.$ માં હોય,તો $\alpha^2, \beta^2, \gamma^2$ એ શેમાં હશે?
A
$A.P.$
B
$H.P.$
C
$G.P.$
D
ઉપરનામાંથી કોઈ નહીં

Solution

(A) આપેલ છે કે $\alpha, \beta, \gamma$ એ અનુક્રમે $ca, ab$; $ab, bc$; અને $bc, ca$ વચ્ચેના સમગુણોત્તર મધ્યકો છે.
તેથી,$\alpha^2 = (ca)(ab) = a^2bc$,$\beta^2 = (ab)(bc) = b^2ca$,અને $\gamma^2 = (bc)(ca) = c^2ab$.
કારણ કે $a, b, c$ એ $A.P.$ માં છે,તેથી $2b = a + c$.
આપણે તપાસવું છે કે શું $\alpha^2, \beta^2, \gamma^2$ એ $A.P.$ માં છે,જેના માટે $2\beta^2 = \alpha^2 + \gamma^2$ થવું જોઈએ.
કિંમતો મૂકતા: $2(b^2ca) = a^2bc + c^2ab$.
બંને બાજુ $abc$ વડે ભાગતા (ધારો કે $a, b, c \neq 0$),આપણને મળે છે $2b^2ca = abc(a + c)$.
$abc$ વડે ભાગતા,આપણને $2b = a + c$ મળે છે,જે સાચું છે કારણ કે $a, b, c$ એ $A.P.$ માં છે.
આમ,$\alpha^2, \beta^2, \gamma^2$ એ $A.P.$ માં છે.
332
MediumMCQ
ધારો કે $a_1, a_2, \dots, a_{10}$ એ $A.P.$ માં છે અને $h_1, h_2, \dots, h_{10}$ એ $H.P.$ માં છે. જો $a_1 = h_1 = 2$ અને $a_{10} = h_{10} = 3$ હોય,તો $a_4 h_7$ ની કિંમત શોધો.
A
$2$
B
$3$
C
$5$
D
$6$

Solution

(D) આપેલ છે કે $a_1, a_2, \dots, a_{10}$ એ $A.P.$ માં છે જ્યાં $a_1 = 2$ અને $a_{10} = 3$.
$A.P.$ નું $n$-મું પદ $a_n = a_1 + (n-1)d$ છે.
$n=10$ માટે,$3 = 2 + 9d$,જે આપણને $d = \frac{1}{9}$ આપે છે.
તેથી,$a_4 = a_1 + 3d = 2 + 3(\frac{1}{9}) = 2 + \frac{1}{3} = \frac{7}{3}$.
આપેલ છે કે $h_1, h_2, \dots, h_{10}$ એ $H.P.$ માં છે જ્યાં $h_1 = 2$ અને $h_{10} = 3$.
તેથી $\frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_{10}}$ એ $A.P.$ માં છે,જેનું પ્રથમ પદ $\frac{1}{2}$ અને $10$-મું પદ $\frac{1}{3}$ છે.
ધારો કે આ $A.P.$ નો સામાન્ય તફાવત $D$ છે.
$\frac{1}{h_{10}} = \frac{1}{h_1} + 9D \implies \frac{1}{3} = \frac{1}{2} + 9D \implies 9D = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6} \implies D = -\frac{1}{54}$.
તેથી $\frac{1}{h_7} = \frac{1}{h_1} + 6D = \frac{1}{2} + 6(-\frac{1}{54}) = \frac{1}{2} - \frac{1}{9} = \frac{9-2}{18} = \frac{7}{18}$.
તેથી,$h_7 = \frac{18}{7}$.
અંતે,$a_4 h_7 = \frac{7}{3} \times \frac{18}{7} = 6$.
333
MediumMCQ
બે શ્રેણીઓ ${t_n}$ અને ${s_n}$ ને $t_n = \log \left( \frac{5^{n+1}}{3^{n-1}} \right)$ અને $s_n = \left[ \log \left( \frac{5}{3} \right) \right]^n$ દ્વારા વ્યાખ્યાયિત કરવામાં આવી છે. તો:
A
${t_n}$ એ $A.P.$ છે,${s_n}$ એ $G.P.$ છે.
B
${t_n}$ અને ${s_n}$ બંને $G.P.$ છે.
C
${t_n}$ અને ${s_n}$ બંને $A.P.$ છે.
D
${s_n}$ એ $G.P.$ છે,${t_n}$ એ $A.P.$ કે $G.P.$ નથી.

Solution

(A) આપેલ છે કે $t_n = \log \left( \frac{5^{n+1}}{3^{n-1}} \right) = \log(5^{n+1}) - \log(3^{n-1}) = (n+1)\log 5 - (n-1)\log 3 = n(\log 5 - \log 3) + (\log 5 + \log 3) = n \log(5/3) + \log 15$.
અહીં $t_n$ એ $an + b$ સ્વરૂપમાં હોવાથી,તે સમાંતર શ્રેણી $(A.P.)$ છે જેનો સામાન્ય તફાવત $d = \log(5/3)$ છે.
આપેલ છે કે $s_n = [\log(5/3)]^n$. આ $ar^n$ સ્વરૂપમાં છે,જ્યાં સામાન્ય ગુણોત્તર $r = \log(5/3)$ છે.
તેથી,$s_n$ એ સમગુણોત્તર શ્રેણી $(G.P.)$ છે.
આમ,${t_n}$ એ $A.P.$ છે અને ${s_n}$ એ $G.P.$ છે.
334
DifficultMCQ
ધારો કે $a_1, a_2, a_3$ કોઈપણ ધન વાસ્તવિક સંખ્યાઓ છે,તો નીચેનામાંથી કયું વિધાન સાચું નથી?
A
$3a_1a_2a_3 \le a_1^3 + a_2^3 + a_3^3$
B
$\frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_1} \ge 3$
C
$(a_1 + a_2 + a_3) \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right) \ge 9$
D
$(a_1 + a_2 + a_3) \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right)^3 \le 27$

Solution

(D) આપણે સમાંતર મધ્યક-ભૌમિતિક મધ્યક $(AM-GM)$ અસમતાનો ઉપયોગ કરીએ છીએ,જે જણાવે છે કે ધન વાસ્તવિક સંખ્યાઓ માટે,$AM \ge GM \ge HM$ થાય છે.
$1$. વિકલ્પ $(A)$ માટે: $AM \ge GM$ દ્વારા,આપણી પાસે $\frac{a_1^3 + a_2^3 + a_3^3}{3} \ge \sqrt[3]{a_1^3 a_2^3 a_3^3} = a_1 a_2 a_3$ છે,જે સૂચવે છે કે $a_1^3 + a_2^3 + a_3^3 \ge 3a_1 a_2 a_3$. આ વિધાન સાચું છે.
$2$. વિકલ્પ $(B)$ માટે: $AM \ge GM$ દ્વારા,આપણી પાસે $\frac{\frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_1}}{3} \ge \sqrt[3]{\frac{a_1}{a_2} \cdot \frac{a_2}{a_3} \cdot \frac{a_3}{a_1}} = \sqrt[3]{1} = 1$ છે,જે સૂચવે છે કે $\frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_1} \ge 3$. આ વિધાન સાચું છે.
$3$. વિકલ્પ $(C)$ માટે: $AM \ge HM$ દ્વારા,આપણી પાસે $\frac{a_1 + a_2 + a_3}{3} \ge \frac{3}{\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}}$ છે,જે સૂચવે છે કે $(a_1 + a_2 + a_3) \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right) \ge 9$. આ વિધાન સાચું છે.
$4$. વિકલ્પ $(D)$ માટે: કારણ કે $(a_1 + a_2 + a_3) \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right) \ge 9$ છે,તેથી પદ $(a_1 + a_2 + a_3) \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right)^3$ સામાન્ય રીતે $\ge 9 \left( \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \right)^2$ થશે,જે હંમેશા $\le 27$ હોવું જરૂરી નથી. તેથી,વિધાન $(D)$ સાચું નથી.
335
MediumMCQ
The odd numbers are divided as follows:
Row $1$: $1, 3$
Row $2$: $5, 7, 9, 11$
Row $3$: $13, 15, 17, 19, 21, 23$
Then the sum of the $n^{th}$ row is:
A
$n^3 + (n-1)^3$
B
$n^3 - (n-1)^3$
C
$2n^3$
D
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336
DifficultMCQ
$n$ ના તમામ ધન પૂર્ણાંક મૂલ્યો માટે,$3 \cdot 1 \cdot 2 + 3 \cdot 2 \cdot 3 + 3 \cdot 3 \cdot 4 + \dots + 3 \cdot n \cdot (n + 1)$ નું મૂલ્ય શું છે?
A
$n(n + 1)(n + 2)$
B
$n(n + 1)(2n + 1)$
C
$(n - 1)n(n + 1)$
D
$\frac{(n - 1)n(n + 1)}{2}$

Solution

(A) ધારો કે શ્રેણીનું $k$-મું પદ $T_k$ છે,તો $T_k = 3k(k + 1) = 3k^2 + 3k$.
જો $S_n$ એ પ્રથમ $n$ પદોનો સરવાળો દર્શાવે,તો $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (3k^2 + 3k)$.
પ્રમાણિત સરવાળાના સૂત્રો $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ અને $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ નો ઉપયોગ કરતા:
$S_n = 3 \sum_{k=1}^n k^2 + 3 \sum_{k=1}^n k = 3 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 3 \left[ \frac{n(n+1)}{2} \right]$.
$S_n = \frac{n(n+1)(2n+1)}{2} + \frac{3n(n+1)}{2} = \frac{n(n+1)}{2} [ (2n+1) + 3 ]$.
$S_n = \frac{n(n+1)}{2} [ 2n + 4 ] = \frac{n(n+1)}{2} \cdot 2(n+2) = n(n+1)(n+2)$.
337
DifficultMCQ
શ્રેણી $\frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + \dots$ નો સરવાળો કેટલો થાય?
A
$\frac{1}{3}$
B
$\frac{1}{6}$
C
$\frac{1}{9}$
D
$\frac{1}{12}$

Solution

(D) આપેલ શ્રેણી $S = \frac{1}{3 \times 7} + \frac{1}{7 \times 11} + \frac{1}{11 \times 15} + \dots \infty$ છે.
દરેક પદ $\frac{1}{(4n-1)(4n+3)}$ સ્વરૂપમાં છે.
આપણે દરેક પદને $\frac{1}{4} \left( \frac{1}{4n-1} - \frac{1}{4n+3} \right)$ તરીકે લખી શકીએ છીએ.
તેથી,$S = \frac{1}{4} \left[ \left( \frac{1}{3} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{11} \right) + \left( \frac{1}{11} - \frac{1}{15} \right) + \dots \right]$.
આ એક ટેલિસ્કોપિંગ શ્રેણી છે જ્યાં તમામ વચ્ચેના પદો ઉડી જાય છે.
$S = \frac{1}{4} \left( \frac{1}{3} - \lim_{n \to \infty} \frac{1}{4n+3} \right)$.
કારણ કે $\lim_{n \to \infty} \frac{1}{4n+3} = 0$,તેથી $S = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}$.
338
AdvancedMCQ
શ્રેણી $(1^2 + 1) \cdot 1! + (2^2 + 1) \cdot 2! + (3^2 + 1) \cdot 3! + \dots + (n^2 + 1) \cdot n!$ નો સરવાળો કેટલો થાય?
A
$(n + 1) \cdot (n + 1)!$
B
$n \cdot (n + 1)!$
C
$(n + 1) \cdot (n + 2)!$
D
આમાંથી કોઈ નહીં

Solution

(B) શ્રેણીનું સામાન્ય પદ $T_n = (n^2 + 1) \cdot n!$ છે.
આપણે $n^2 + 1$ ને $n(n + 1) - (n - 1)$ તરીકે લખી શકીએ.
તેથી,$T_n = [n(n + 1) - (n - 1)] \cdot n! = n(n + 1) \cdot n! - (n - 1) \cdot n!$.
કારણ કે $(n + 1) \cdot n! = (n + 1)!$,તેથી $T_n = n \cdot (n + 1)! - (n - 1) \cdot n!$.
ધારો કે $f(n) = n \cdot (n + 1)!$. તો $T_n = f(n) - f(n - 1)$.
સરવાળો $S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} [f(k) - f(k - 1)]$.
આ એક ટેલિસ્કોપિંગ શ્રેણી છે: $S_n = f(n) - f(0)$.
$f(n) = n \cdot (n + 1)!$ અને $f(0) = 0 \cdot 1! = 0$.
તેથી,$S_n = n \cdot (n + 1)!$.
339
AdvancedMCQ
જો $\sum_{r=1}^{n}r^3 - \sum_{p=1}^{n} \sum_{m=1}^{p} \sum_{r=1}^{m} 1 = 80$ હોય,તો $n$ ની શક્ય કિંમત કઈ છે?
A
$3$
B
$4$
C
$5$
D
$6$

Solution

(B) આપણે જાણીએ છીએ કે $\sum_{r=1}^{n} r^3 = \left[ \frac{n(n+1)}{2} \right]^2$.
હવે,ત્રિપલ સરવાળાને ધ્યાનમાં લો: $\sum_{p=1}^{n} \sum_{m=1}^{p} \sum_{r=1}^{m} 1 = \sum_{p=1}^{n} \sum_{m=1}^{p} m = \sum_{p=1}^{n} \frac{p(p+1)}{2} = \frac{1}{2} \left[ \sum_{p=1}^{n} p^2 + \sum_{p=1}^{n} p \right]$.
પ્રમાણિત સરવાળાના સૂત્રોનો ઉપયોગ કરતા: $\sum_{p=1}^{n} p^2 = \frac{n(n+1)(2n+1)}{6}$ અને $\sum_{p=1}^{n} p = \frac{n(n+1)}{2}$.
તેથી,ત્રિપલ સરવાળો $\frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right] = \frac{n(n+1)}{4} \left[ \frac{2n+1}{3} + 1 \right] = \frac{n(n+1)}{4} \left[ \frac{2n+4}{3} \right] = \frac{n(n+1)(n+2)}{6}$ થાય.
આમ,સમીકરણ આ મુજબ બને છે: $\left[ \frac{n(n+1)}{2} \right]^2 - \frac{n(n+1)(n+2)}{6} = 80$.
જો $n=4$ લઈએ: $\left[ \frac{4(5)}{2} \right]^2 - \frac{4(5)(6)}{6} = 10^2 - 20 = 100 - 20 = 80$.
તેથી,$n=4$ એ સાચી કિંમત છે.
340
DifficultMCQ
જો $a + 2b + 3c = 6$ હોય,તો $abc^2$ ની મહત્તમ કિંમત શું હશે? (જ્યાં $a, b, c$ ધન વાસ્તવિક સંખ્યાઓ છે).
A
$\frac{9}{8}$
B
$\frac{9}{16}$
C
$\frac{27}{8}$
D
$\frac{27}{16}$

Solution

(A) આપેલ શરત $a + 2b + 3c = 6$ છે,જ્યાં $a, b, c > 0$.
આપણે $abc^2$ ની મહત્તમ કિંમત શોધવી છે. આપણે પદને $a(2b)(\frac{3c}{2})(\frac{3c}{2}) = \frac{9}{4}abc^2$ તરીકે લખી શકીએ.
ચાર પદો $a, 2b, \frac{3c}{2}, \frac{3c}{2}$ માટે સમાંતર મધ્યક-ભૂમિતિક મધ્યક $(AM \geq GM)$ અસમતાનો ઉપયોગ કરતા:
$\frac{a + 2b + \frac{3c}{2} + \frac{3c}{2}}{4} \geq \sqrt[4]{a \cdot 2b \cdot \frac{3c}{2} \cdot \frac{3c}{2}}$
સરવાળો $a + 2b + 3c = 6$ મૂકતા:
$\frac{6}{4} \geq \sqrt[4]{a \cdot 2b \cdot \frac{9c^2}{4}}$
$\frac{3}{2} \geq \sqrt[4]{\frac{18}{4} abc^2}$
$\frac{3}{2} \geq \sqrt[4]{\frac{9}{2} abc^2}$
બંને બાજુ $4$ ઘાત લેતા:
$(\frac{3}{2})^4 \geq \frac{9}{2} abc^2$
$\frac{81}{16} \geq \frac{9}{2} abc^2$
$abc^2 \leq \frac{81}{16} \cdot \frac{2}{9} = \frac{9}{8}$.
આમ,મહત્તમ કિંમત $\frac{9}{8}$ છે.
341
DifficultMCQ
જો $x \in (0, \frac{\pi}{4})$ હોય,તો પદાવલિ $\frac{\cos x}{\sin^2 x(\cos x - \sin x)}$ કઈ કિંમત ધારણ કરી શકે નહીં?
A
$8$
B
$10$
C
$11$
D
$12$

Solution

(A) ધારો કે $f(x) = \frac{\cos x}{\sin^2 x(\cos x - \sin x)}$.
છેદને $\sin x \cdot \sin x(\cos x - \sin x)$ તરીકે લખી શકાય.
$\sin x$ અને $(\cos x - \sin x)$ પદો માટે $AM \geq GM$ અસમતાનો ઉપયોગ કરતા:
$\frac{\sin x + (\cos x - \sin x)}{2} \geq \sqrt{\sin x(\cos x - \sin x)}$
$\Rightarrow \frac{\cos x}{2} \geq \sqrt{\sin x(\cos x - \sin x)}$
બંને બાજુ વર્ગ કરતા:
$\frac{\cos^2 x}{4} \geq \sin x(\cos x - \sin x)$
હવે,આ કિંમત પદાવલિમાં મુકતા:
$f(x) = \frac{\cos x}{\sin x \cdot \sin x(\cos x - \sin x)} \geq \frac{\cos x}{\sin x \cdot \frac{\cos^2 x}{4}} = \frac{4}{\sin x \cos x} = \frac{8}{2 \sin x \cos x} = \frac{8}{\sin 2x}$.
અહીં $x \in (0, \frac{\pi}{4})$ હોવાથી,$2x \in (0, \frac{\pi}{2})$,તેથી $\sin 2x \in (0, 1)$.
તેથી,$\frac{8}{\sin 2x} > 8$.
આનો અર્થ એ છે કે પદાવલિ $f(x)$ ની કિંમત હંમેશા $8$ કરતા મોટી હશે. તેથી,તે $8$ કે તેનાથી નાની કોઈ કિંમત ધારણ કરી શકે નહીં.
342
AdvancedMCQ
ધારો કે $a_1, a_2, a_3, \dots$ એ સામાન્ય ગુણોત્તર $r$ સાથે વધતી જતી સમગુણોત્તર શ્રેણી બનાવે છે,જેથી $\log_8 a_1 + \log_8 a_2 + \dots + \log_8 a_{12} = 2014$ થાય,તો પૂર્ણાંક સંખ્યાઓની ક્રમિત જોડી $(a_1, r)$ ની સંખ્યા કેટલી થાય?
A
$44$
B
$45$
C
$46$
D
$47$

Solution

(C) લઘુગણકનો સરવાળો આપેલ છે: $\log_8(a_1 a_2 \dots a_{12}) = 2014$.
શ્રેણી $a_n = a_1 r^{n-1}$ હોવાથી,ગુણાકાર $a_1^{12} r^{0+1+2+\dots+11} = a_1^{12} r^{66}$ થાય.
તેથી,$\log_8(a_1^{12} r^{66}) = 2014$,જેનો અર્થ છે કે $a_1^{12} r^{66} = 8^{2014} = (2^3)^{2014} = 2^{6042}$.
ધારો કે $a_1 = 2^m$ અને $r = 2^n$ જ્યાં $m, n$ પૂર્ણાંક છે (શ્રેણી વધતી હોવાથી $r > 1$,તેથી $n \ge 1$).
આ કિંમતો સમીકરણમાં મૂકતા: $(2^m)^{12} (2^n)^{66} = 2^{12m + 66n} = 2^{6042}$.
ઘાતાંકોને સરખાવતા: $12m + 66n = 6042$,જેનું સાદું રૂપ $2m + 11n = 1007$ થાય.
$m$ માટે ઉકેલતા: $m = \frac{1007 - 11n}{2}$.
$m$ પૂર્ણાંક હોવા માટે,$1007 - 11n$ બેકી સંખ્યા હોવી જોઈએ,જેનો અર્થ છે કે $n$ એકી સંખ્યા હોવી જોઈએ.
$a_1$ અને $r$ પૂર્ણાંક હોવાથી અને શ્રેણી વધતી હોવાથી,$n \ge 1$ અને $m \ge 1$.
$1007 - 11n \ge 2 \Rightarrow 11n \le 1005 \Rightarrow n \le 91.36$.
આમ,$n$ ની શક્ય કિંમતો: $1, 3, 5, \dots, 91$.
આવી કિંમતોની કુલ સંખ્યા $\frac{91 - 1}{2} + 1 = 46$ થાય.
343
AdvancedMCQ
$6$ વડે ભાગતા $4$ શેષ વધતી હોય તેવી તમામ બે અંકની સંખ્યાઓનો સરવાળો કેટલો થાય?
A
$777$
B
$776$
C
$780$
D
$784$

Solution

(C) બે અંકની સંખ્યા $x$ ને $6$ વડે ભાગતા $4$ શેષ વધે,તો તેને $x = 6n + 4$ સ્વરૂપે લખી શકાય,જ્યાં $n$ પૂર્ણાંક છે.
બે અંકની સંખ્યાઓ માટે,$10 \le 6n + 4 \le 99$.
$n$ માટે ઉકેલતા: $6 \le 6n \le 95$,જેનો અર્થ છે કે $1 \le n \le 15.83$.
આમ,$n$ ની કિંમત $1$ થી $15$ સુધીની પૂર્ણાંક સંખ્યાઓ હોઈ શકે.
આવી સંખ્યાઓની શ્રેણી $10, 16, 22, \ldots, 94$ છે.
આ એક સમાંતર શ્રેણી $(AP)$ છે જેમાં પ્રથમ પદ $a = 10$,અંતિમ પદ $l = 94$ અને પદોની સંખ્યા $n = 15$ છે.
સમાંતર શ્રેણીનો સરવાળો $S_n = \frac{n}{2}(a + l)$ સૂત્ર દ્વારા મળે છે.
$S_{15} = \frac{15}{2}(10 + 94) = \frac{15}{2}(104) = 15 \times 52 = 780$.
344
AdvancedMCQ
$\sum\limits_{n = 2}^\infty {\frac{n}{{1 + {n^2}\left( {{n^2} - 2} \right)}}} $ નું મૂલ્ય કેટલું થાય?
A
$\frac{5}{4}$
B
$1$
C
$\frac{5}{16}$
D
$\frac{1}{4}$

Solution

(C) આપેલ પદ $S = \sum_{n=2}^{\infty} \frac{n}{1 + n^2(n^2 - 2)} = \sum_{n=2}^{\infty} \frac{n}{n^4 - 2n^2 + 1}$ છે.
અહીં છેદ એક પૂર્ણ વર્ગ છે: $n^4 - 2n^2 + 1 = (n^2 - 1)^2 = (n-1)^2(n+1)^2$.
તેથી,પદ $\frac{n}{(n-1)^2(n+1)^2}$ થાય.
આંશિક અપૂર્ણાંકની રીતનો ઉપયોગ કરતા,$\frac{n}{(n-1)^2(n+1)^2} = \frac{1}{4} \left( \frac{1}{(n-1)^2} - \frac{1}{(n+1)^2} \right)$ લખી શકાય.
હવે,ટેલિસ્કોપિંગ સરવાળાની ગણતરી કરતા:
$S = \frac{1}{4} \sum_{n=2}^{\infty} \left( \frac{1}{(n-1)^2} - \frac{1}{(n+1)^2} \right)$.
સરવાળાનું વિસ્તરણ કરતા:
$S = \frac{1}{4} \left[ \left( \frac{1}{1^2} - \frac{1}{3^2} \right) + \left( \frac{1}{2^2} - \frac{1}{4^2} \right) + \left( \frac{1}{3^2} - \frac{1}{5^2} \right) + \left( \frac{1}{4^2} - \frac{1}{6^2} \right) + \dots \right]$.
મોટાભાગના પદો ઉડી જશે,ફક્ત પ્રથમ બે ધન પદો બાકી રહેશે:
$S = \frac{1}{4} \left( 1 + \frac{1}{4} \right) = \frac{1}{4} \left( \frac{5}{4} \right) = \frac{5}{16}$.
345
AdvancedMCQ
ધારો કે $E = x^{2017} + y^{2017} + z^{2017} - 2017xyz$ (જ્યાં $x, y, z \geq 0$),તો $E$ ની ન્યૂનતમ કિંમત શું છે?
A
$0$
B
$-2014$
C
$-2017$
D
$2017$

Solution

(B) સમાંતર મધ્યક-ભૌમિતિક મધ્યક $(AM \geq GM)$ અસમતાનો ઉપયોગ કરતા,અઋણ વાસ્તવિક સંખ્યાઓ $x, y, z$ માટે:
$\frac{x^{2017} + y^{2017} + z^{2017} + \underbrace{1 + 1 + \dots + 1}_{2014 \text{ વખત}}}{2017} \geq \sqrt[2017]{x^{2017} \cdot y^{2017} \cdot z^{2017} \cdot 1^{2014}}$
$\frac{x^{2017} + y^{2017} + z^{2017} + 2014}{2017} \geq xyz$
$x^{2017} + y^{2017} + z^{2017} + 2014 \geq 2017xyz$
$x^{2017} + y^{2017} + z^{2017} - 2017xyz \geq -2014$
આમ,$E \geq -2014$.
સમાનતા ત્યારે મળે છે જ્યારે $x = y = z = 1$ હોય.
346
AdvancedMCQ
જો $\sum_{i = 1}^n \sum_{j = 1}^i \sum_{k = 1}^j 1 = 560$ હોય,તો $n$ નું મૂલ્ય શોધો.
A
$13$
B
$14$
C
$15$
D
$16$

Solution

(B) આપેલ પદાવલિ: $\sum_{i = 1}^n \sum_{j = 1}^i \sum_{k = 1}^j 1 = 560$.
સૌ પ્રથમ,અંદરના સરવાળાની ગણતરી કરતા: $\sum_{k = 1}^j 1 = j$.
હવે,પદાવલિ આ મુજબ બનશે: $\sum_{i = 1}^n \sum_{j = 1}^i j = 560$.
ત્યારબાદ,મધ્યના સરવાળાની ગણતરી કરતા: $\sum_{j = 1}^i j = \frac{i(i + 1)}{2}$.
હવે,પદાવલિ આ મુજબ બનશે: $\sum_{i = 1}^n \frac{i(i + 1)}{2} = 560$.
આને આ રીતે લખી શકાય: $\frac{1}{2} [\sum_{i = 1}^n i^2 + \sum_{i = 1}^n i] = 560$.
સરવાળાના સૂત્રો $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$ અને $\sum i = \frac{n(n+1)}{2}$ નો ઉપયોગ કરતા:
$\frac{1}{2} [\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}] = 560$.
$\frac{n(n+1)}{2}$ સામાન્ય લેતા:
$\frac{1}{2} \cdot \frac{n(n+1)}{2} [\frac{2n+1}{3} + 1] = 560$.
$\frac{n(n+1)}{4} [\frac{2n+4}{3}] = 560$.
$\frac{n(n+1) \cdot 2(n+2)}{12} = 560$.
$\frac{n(n+1)(n+2)}{6} = 560$.
$n(n+1)(n+2) = 560 \times 6 = 3360$.
કારણ કે $14 \times 15 \times 16 = 3360$,તેથી $n = 14$ મળે છે.
347
AdvancedMCQ
$\sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^{n - 1} {\frac{k}{{{2^{n + k}}}}} } $ ની કિંમત શોધો.
A
$\frac {2}{9}$
B
$\frac {4}{9}$
C
$\frac {4}{3}$
D
$\frac {2}{3}$

Solution

(B) ધારો કે $S = \sum_{n=1}^\infty \sum_{k=1}^{n-1} \frac{k}{2^{n+k}}$.
આપણે સરવાળાને $S = \sum_{n=1}^\infty \frac{1}{2^n} \sum_{k=1}^{n-1} \frac{k}{2^k}$ તરીકે લખી શકીએ.
અંકગણિત-ભૌમિતિક શ્રેણીના સૂત્રનો ઉપયોગ કરતા,$\sum_{k=1}^{n-1} k x^k = \frac{x(1-x^{n-1})}{(1-x)^2} - \frac{(n-1)x^n}{1-x}$. $x = \frac{1}{2}$ માટે,આ સાદું રૂપ $\sum_{k=1}^{n-1} \frac{k}{2^k} = 2 - \frac{n+1}{2^{n-1}}$ આપે છે.
આ કિંમત $S$ ના સમીકરણમાં મૂકતા:
$S = \sum_{n=1}^\infty \frac{1}{2^n} \left( 2 - \frac{n+1}{2^{n-1}} \right) = \sum_{n=1}^\infty \left( \frac{2}{2^n} - \frac{n+1}{2^{2n-1}} \right)$.
$S = \sum_{n=1}^\infty \frac{1}{2^{n-1}} - \sum_{n=1}^\infty \frac{n+1}{2^{2n-1}}$.
પ્રથમ ભાગ એક ભૌમિતિક શ્રેણી છે: $\sum_{n=1}^\infty \frac{1}{2^{n-1}} = 1 + \frac{1}{2} + \frac{1}{4} + \dots = \frac{1}{1 - 1/2} = 2$.
બીજો ભાગ $\sum_{n=1}^\infty \frac{n+1}{2^{2n-1}} = 2 \sum_{n=1}^\infty \frac{n+1}{4^n} = 2 \left( \sum_{n=1}^\infty \frac{n}{4^n} + \sum_{n=1}^\infty \frac{1}{4^n} \right)$ છે.
$\sum_{n=1}^\infty n x^n = \frac{x}{(1-x)^2}$ નો ઉપયોગ કરતા,$x = 1/4$ માટે,આપણને $\frac{1/4}{(3/4)^2} = \frac{4}{9}$ મળે છે.
$\sum_{n=1}^\infty x^n = \frac{x}{1-x}$ નો ઉપયોગ કરતા,$x = 1/4$ માટે,આપણને $\frac{1/4}{3/4} = \frac{1}{3}$ મળે છે.
તેથી,બીજો ભાગ $2 \left( \frac{4}{9} + \frac{1}{3} \right) = 2 \left( \frac{4+3}{9} \right) = \frac{14}{9}$ છે.
અંતે,$S = 2 - \frac{14}{9} = \frac{18-14}{9} = \frac{4}{9}$.
348
MediumMCQ
$1$ થી $100$ સુધીની પ્રાકૃતિક સંખ્યાઓ લખતી વખતે અંક $5$ કેટલી વાર આવે છે?
A
$20$
B
$15$
C
$16$
D
$19$

Solution

(A) $1$ થી $100$ સુધીની સંખ્યાઓમાં અંક $5$ કેટલી વાર આવે છે તે શોધવા માટે,આપણે એકમના સ્થાન અને દશકના સ્થાનમાં તેની ગણતરી કરીએ છીએ.
$1$. એકમના સ્થાનમાં: અંક $5$ એ $5, 15, 25, 35, 45, 55, 65, 75, 85, 95$ માં આવે છે. આવી કુલ $10$ સંખ્યાઓ છે.
$2$. દશકના સ્થાનમાં: અંક $5$ એ $50, 51, 52, 53, 54, 55, 56, 57, 58, 59$ માં આવે છે. આવી કુલ $10$ સંખ્યાઓ છે.
નોંધો કે સંખ્યા $55$ બંને યાદીમાં ગણવામાં આવે છે કારણ કે તેમાં એકમ અને દશક બંને સ્થાને $5$ છે.
કુલ સંખ્યા = (એકમના સ્થાનમાં ગણતરી) + (દશકના સ્થાનમાં ગણતરી) = $10 + 10 = 20$.
તેથી,અંક $5$ કુલ $20$ વાર આવે છે.
349
AdvancedMCQ
જો સમીકરણ $x^3 - 9x^2 + \alpha x - 15 = 0$ ના બીજ $A.P.$ (સમાંતર શ્રેણી) માં હોય,તો $\alpha$ ની કિંમત શોધો.
A
$0$
B
$20$
C
$21$
D
$23$

Solution

(D) ધારો કે ત્રિઘાત સમીકરણના બીજ $(a - d)$,$a$,અને $(a + d)$ છે.
બહુપદી સમીકરણના બીજના ગુણધર્મો મુજબ,બીજનો સરવાળો એ $x^2$ ના સહગુણકને ઋણ ચિહ્ન સાથે લેવાથી મળે છે.
$(a - d) + a + (a + d) = -(-9) / 1 = 9$.
$3a = 9$,તેથી $a = 3$ મળે છે.
જેમ કે $a = 3$ એ સમીકરણનું એક બીજ છે,તે $x^3 - 9x^2 + \alpha x - 15 = 0$ નું સમાધાન કરશે.
સમીકરણમાં $x = 3$ મૂકતા:
$(3)^3 - 9(3)^2 + \alpha(3) - 15 = 0$.
$27 - 81 + 3\alpha - 15 = 0$.
$-54 + 3\alpha - 15 = 0$.
$3\alpha - 69 = 0$.
$3\alpha = 69$.
$\alpha = 23$.
350
AdvancedMCQ
પદાવલિ $3(1!) - 4(2!) + 5(3!) - 6(4!) + \dots - 2008(2006)! + (2007)!$ ની કિંમત શોધો.
A
$-2007$
B
$-1$
C
$1$
D
$2007$

Solution

(C) ધારો કે શ્રેણીનું સામાન્ય પદ $T_n = (-1)^{n-1} (n+2)n!$ છે.
આપણે પદોને આ રીતે ફરીથી લખી શકીએ:
$T_n = (-1)^{n-1} ((n+1) + 1)n! = (-1)^{n-1} ((n+1)! + n!)$.
શ્રેણીનું વિસ્તરણ કરતા:
$S = (2! + 1!) - (3! + 2!) + (4! + 3!) - (5! + 4!) + \dots - (2007! + 2006!) + 2007!$.
સરવાળાના ટેલિસ્કોપિંગ સ્વભાવને જોતા:
$S = 2! + 1! - 3! - 2! + 4! + 3! - 5! - 4! + \dots - 2007! - 2006! + 2007!$.
$1!$ સિવાયના તમામ પદો ઉડી જાય છે.
$S = 1! = 1$.

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