Arithmetic progression Questions in English

(C) Let the first $3$ terms of the arithmetic progression be $(a - d)$,$a$,and $(a + d)$.
Given that the sum of the first and third term is $12$:
$(a - d) + (a + d) = 12$
$2a = 12$
$a = 6$
Given that the product of the first and second term is $24$:
$(a - d) \times a = 24$
Substituting $a = 6$:
$6(6 - d) = 24$
$6 - d = 4$
$d = 2$
The first term is $(a - d) = 6 - 2 = 4$.

(C) For the first arithmetic progression $2, 5, 8, \dots$,the first term $a_1 = 2$ and common difference $d_1 = 3$. The sum of the first $2n$ terms is $S_{2n} = \frac{2n}{2} [2(2) + (2n - 1)3] = n[4 + 6n - 3] = n(6n + 1)$.
For the second arithmetic progression $57, 59, 61, \dots$,the first term $a_2 = 57$ and common difference $d_2 = 2$. The sum of the first $n$ terms is $S_n = \frac{n}{2} [2(57) + (n - 1)2] = \frac{n}{2} [114 + 2n - 2] = \frac{n}{2} [112 + 2n] = n(56 + n)$.
Given that $S_{2n} = S_n$,we have $n(6n + 1) = n(56 + n)$.
Since $n \neq 0$,we divide by $n$: $6n + 1 = 56 + n$.
$5n = 55$,which gives $n = 11$.

(D) The numbers divisible by $3$ between $250$ and $1000$ form an arithmetic progression: $252, 255, \dots, 999$.
Here,the first term $a = 252$,the last term $l = 999$,and the common difference $d = 3$.
Using the formula for the $n^{th}$ term: $l = a + (n - 1)d$
$999 = 252 + (n - 1)3$
$747 = (n - 1)3$
$n - 1 = 249$
$n = 250$.
The sum $S_n$ is given by $S_n = \frac{n}{2}(a + l)$
$S_n = \frac{250}{2}(252 + 999)$
$S_n = 125 \times 1251 = 156375$.

(B) Given that the $7^{th}$ term of an $A.P.$ is $40$.
$a + (7 - 1)d = 40 \implies a + 6d = 40$.
We need to find the sum of the first $13$ terms,denoted by $S_{13}$.
The formula for the sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
For $n = 13$,$S_{13} = \frac{13}{2}[2a + (13 - 1)d] = \frac{13}{2}[2a + 12d]$.
Factoring out $2$,we get $S_{13} = \frac{13}{2} \times 2(a + 6d) = 13(a + 6d)$.
Substituting the value $a + 6d = 40$,we get $S_{13} = 13 \times 40 = 520$.

(A) Since $a_1, a_2, a_3, \dots, a_{2n}$ form an $A.P.$,the common difference is $d = a_2 - a_1 = a_4 - a_3 = \dots = a_{2n} - a_{2n - 1}$.
The given expression is $S = (a_1^2 - a_2^2) + (a_3^2 - a_4^2) + \dots + (a_{2n - 1}^2 - a_{2n}^2)$.
Using the identity $x^2 - y^2 = (x - y)(x + y)$,we get:
$S = (a_1 - a_2)(a_1 + a_2) + (a_3 - a_4)(a_3 + a_4) + \dots + (a_{2n - 1} - a_{2n})(a_{2n - 1} + a_{2n})$.
Since $a_k - a_{k+1} = -d$ for odd $k$,we have:
$S = -d(a_1 + a_2 + a_3 + a_4 + \dots + a_{2n - 1} + a_{2n})$.
The sum of an $A.P.$ with $2n$ terms is $\frac{2n}{2}(a_1 + a_{2n}) = n(a_1 + a_{2n})$.
Thus,$S = -d \cdot n(a_1 + a_{2n})$.
From the $A.P.$ formula,$a_{2n} = a_1 + (2n - 1)d$,so $d = \frac{a_{2n} - a_1}{2n - 1}$.
Substituting $-d = \frac{a_1 - a_{2n}}{2n - 1}$ into the expression for $S$:
$S = \frac{a_1 - a_{2n}}{2n - 1} \cdot n(a_1 + a_{2n}) = \frac{n}{2n - 1}(a_1^2 - a_{2n}^2)$.

(B) Given the sum of $n$ terms $S_n = 3n^2 + 5n$.
We know that the $m^{th}$ term $T_m = S_m - S_{m-1}$.
$T_m = (3m^2 + 5m) - [3(m-1)^2 + 5(m-1)]$
$T_m = (3m^2 + 5m) - [3(m^2 - 2m + 1) + 5m - 5]$
$T_m = 3m^2 + 5m - [3m^2 - 6m + 3 + 5m - 5]$
$T_m = 3m^2 + 5m - 3m^2 + m + 2$
$T_m = 6m + 2$.
Given $T_m = 164$,we have $6m + 2 = 164$.
$6m = 162$.
$m = 27$.

(D) Given the sum of the first $n$ terms of an $A.P.$ is $S_n = nP + \frac{1}{2}n(n - 1)Q$.
Comparing this with the standard formula for the sum of an $A.P.$,$S_n = \frac{n}{2}\{2a + (n - 1)d\}$,we have:
$S_n = n \cdot a + \frac{n(n - 1)}{2} \cdot d$.
By comparing the coefficients of the terms,we get $a = P$ and $d = Q$.
Alternatively,the common difference $d$ can be calculated as $d = S_2 - 2S_1$.
$S_1 = P(1) + \frac{1}{2}(1)(0)Q = P$.
$S_2 = P(2) + \frac{1}{2}(2)(1)Q = 2P + Q$.
$d = (2P + Q) - 2(P) = Q$.

(B) Given $S_{2n} = 3S_n$.
Using the formula for the sum of $n$ terms of an $A.P.$,$S_n = \frac{n}{2}[2a + (n-1)d]$.
So,$\frac{2n}{2}[2a + (2n-1)d] = 3 \times \frac{n}{2}[2a + (n-1)d]$.
$2[2a + (2n-1)d] = 3[2a + (n-1)d]$.
$4a + 4nd - 2d = 6a + 3nd - 3d$.
$nd + d = 2a$,which means $2a = (n+1)d$.
Now,we need to find $\frac{S_{3n}}{S_n} = \frac{\frac{3n}{2}[2a + (3n-1)d]}{\frac{n}{2}[2a + (n-1)d]}$.
$= 3 \times \frac{2a + (3n-1)d}{2a + (n-1)d}$.
Substitute $2a = (n+1)d$:
$= 3 \times \frac{(n+1)d + (3n-1)d}{(n+1)d + (n-1)d} = 3 \times \frac{4nd}{2nd} = 3 \times 2 = 6$.

(D) Given that the $A.P.$ consists of consecutive integers,the common difference $d = 1$.
The first term $a = {p^2} + 1$.
The number of terms $n = 2p + 1$.
The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2} \{ 2a + (n - 1)d \}$.
Substituting the values:
$S_{2p+1} = \frac{2p+1}{2} \{ 2({p^2} + 1) + (2p + 1 - 1)(1) \}$
$S_{2p+1} = \frac{2p+1}{2} \{ 2{p^2} + 2 + 2p \}$
$S_{2p+1} = (2p + 1)({p^2} + p + 1)$
Expanding the expression:
$(2p + 1)({p^2} + p + 1) = 2{p^3} + 2{p^2} + 2p + {p^2} + p + 1 = 2{p^3} + 3{p^2} + 3p + 1$
Now,consider the expression ${p^3} + {(p + 1)^3}$:
${p^3} + ({p^3} + 3{p^2} + 3p + 1) = 2{p^3} + 3{p^2} + 3p + 1$.
Thus,the sum is ${p^3} + {(p + 1)^3}$.

(B) Let the first term be $a = 11$ and the common difference be $d$.
The sum of the first four terms is given by: $a + (a + d) + (a + 2d) + (a + 3d) = 56$.
Substituting $a = 11$: $11 + (11 + d) + (11 + 2d) + (11 + 3d) = 56$.
$44 + 6d = 56$ $\Rightarrow 6d = 12$ $\Rightarrow d = 2$.
Let the number of terms be $n$. The last four terms are $a_{n-3}, a_{n-2}, a_{n-1}, a_n$.
These are $(11 + (n-4)2), (11 + (n-3)2), (11 + (n-2)2), (11 + (n-1)2)$.
The sum of these terms is $112$:
$44 + 2(n-4 + n-3 + n-2 + n-1) = 112$.
$44 + 2(4n - 10) = 112$.
$44 + 8n - 20 = 112$.
$8n + 24 = 112$.
$8n = 88 \Rightarrow n = 11$.

(D) The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Here,$a = 3$,$d = 4$,and $S_n = 406$.
Substituting the values: $406 = \frac{n}{2}[2(3) + (n - 1)4]$.
$406 = \frac{n}{2}[6 + 4n - 4]$.
$406 = \frac{n}{2}[4n + 2]$.
$406 = n(2n + 1)$.
$2n^2 + n - 406 = 0$.
Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$n = \frac{-1 \pm \sqrt{1^2 - 4(2)(-406)}}{2(2)}$.
$n = \frac{-1 \pm \sqrt{1 + 3248}}{4}$.
$n = \frac{-1 \pm \sqrt{3249}}{4}$.
$n = \frac{-1 \pm 57}{4}$.
Since $n$ must be positive,$n = \frac{56}{4} = 14$.

(B) Let the first term be $a = 5$,the number of terms be $n = 15$,and the common difference be $d$.
The sum of $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.
Given $S_{15} = 390$,we have $\frac{15}{2}[2(5) + (15 - 1)d] = 390$.
$\frac{15}{2}[10 + 14d] = 390$.
$15(5 + 7d) = 390$.
$5 + 7d = \frac{390}{15} = 26$.
$7d = 21$,so $d = 3$.
The middle term of $15$ terms is the $\frac{15+1}{2} = 8^{th}$ term.
The $n^{th}$ term is $a_n = a + (n - 1)d$.
$a_8 = 5 + (8 - 1)3 = 5 + 7(3) = 5 + 21 = 26$.

(A) Let the first term be $a$ and the common difference be $d$. The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Given that $S_{10} = 4 \times S_5$,we have:
$\frac{10}{2}[2a + (10-1)d] = 4 \times \frac{5}{2}[2a + (5-1)d]$
$5(2a + 9d) = 4 \times 5(2a + 4d)$
Dividing both sides by $5$:
$2a + 9d = 4(2a + 4d)$
$2a + 9d = 8a + 16d$
$9d - 16d = 8a - 2a$
$-7d = 6a$
$\frac{a}{d} = -\frac{7}{6}$.
Wait,re-evaluating the calculation: $2a + 9d = 8a + 16d$ leads to $6a = -7d$,which is not in the options. Let us re-check the equation: $S_{10} = 4S_5$ $\Rightarrow 5(2a+9d) = 4 \times 2.5(2a+4d)$ $\Rightarrow 5(2a+9d) = 10(2a+4d)$ $\Rightarrow 2a+9d = 2(2a+4d)$ $\Rightarrow 2a+9d = 4a+8d$ $\Rightarrow d = 2a$ $\Rightarrow \frac{a}{d} = \frac{1}{2}$.

(B) Let the three numbers in $A.P.$ be $(a - d)$,$a$,and $(a + d)$.
Given that their sum is $18$:
$(a - d) + a + (a + d) = 18$
$3a = 18 \implies a = 6$.
Given that the sum of their squares is $158$:
$(a - d)^2 + a^2 + (a + d)^2 = 158$
$(6 - d)^2 + 6^2 + (6 + d)^2 = 158$
$(36 - 12d + d^2) + 36 + (36 + 12d + d^2) = 158$
$108 + 2d^2 = 158$
$2d^2 = 158 - 108$
$2d^2 = 50$
$d^2 = 25 \implies d = \pm 5$.
If $d = 5$,the numbers are $(6 - 5), 6, (6 + 5)$,which are $1, 6, 11$.
If $d = -5$,the numbers are $(6 - (-5)), 6, (6 + (-5))$,which are $11, 6, 1$.
In both cases,the greatest number is $11$.

(A) The numerator is an arithmetic progression with first term $a_1 = 3$ and common difference $d_1 = 2$. The sum of $n$ terms is $S_n = \frac{n}{2}[2(3) + (n - 1)2] = \frac{n}{2}[6 + 2n - 2] = \frac{n}{2}[2n + 4] = n(n + 2)$.
The denominator is an arithmetic progression with first term $a_2 = 5$ and common difference $d_2 = 3$. The sum of $10$ terms is $S_{10} = \frac{10}{2}[2(5) + (10 - 1)3] = 5[10 + 27] = 5 \times 37 = 185$.
Given the equation $\frac{n(n + 2)}{185} = 7$,we have $n^2 + 2n = 1295$.
$n^2 + 2n - 1295 = 0$.
Solving the quadratic equation using the formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $n = \frac{-2 \pm \sqrt{4 - 4(1)(-1295)}}{2} = \frac{-2 \pm \sqrt{5184}}{2} = \frac{-2 \pm 72}{2}$.
Since $n$ must be positive,$n = \frac{70}{2} = 35$.

(B) Let the sequence be $\frac{1}{3}, A_1, A_2, \frac{1}{24}$ in an $A.P.$
Here,the first term $a = \frac{1}{3}$ and the fourth term $b = \frac{1}{24}$.
In an $A.P.$ with $n$ arithmetic means,the common difference $d$ is given by $d = \frac{b - a}{n + 1}$.
Here $n = 2$,so $d = \frac{\frac{1}{24} - \frac{1}{3}}{2 + 1} = \frac{\frac{1 - 8}{24}}{3} = \frac{-7/24}{3} = -\frac{7}{72}$.
Now,$A_1 = a + d = \frac{1}{3} - \frac{7}{72} = \frac{24 - 7}{72} = \frac{17}{72}$.
And $A_2 = a + 2d = \frac{1}{3} + 2\left(-\frac{7}{72}\right) = \frac{1}{3} - \frac{7}{36} = \frac{12 - 7}{36} = \frac{5}{36}$.
Thus,the values are $\frac{17}{72}$ and $\frac{5}{36}$.

(A) Let the two numbers be $a$ and $b$.
The arithmetic mean $A$ between $a$ and $b$ is given by $A = \frac{a+b}{2}$.
Let $A_1, A_2, \dots, A_n$ be the $n$ arithmetic means between $a$ and $b$.
Then $a, A_1, A_2, \dots, A_n, b$ form an arithmetic progression with $n+2$ terms.
The sum of these $n$ arithmetic means is $S = A_1 + A_2 + \dots + A_n$.
In an arithmetic progression,the sum of $n$ arithmetic means inserted between $a$ and $b$ is equal to $n$ times the single arithmetic mean between $a$ and $b$.
Mathematically,$S = \sum_{i=1}^{n} A_i = n \left( \frac{a+b}{2} \right) = nA$.
Therefore,$S = nA$.

(A) Let the $n$ arithmetic means between $a$ and $b$ be $A_1, A_2, \dots, A_n$.
These terms form an arithmetic progression with $a$ as the first term and $b$ as the $(n+2)$-th term.
The sum of $n$ arithmetic means is given by $S = A_1 + A_2 + \dots + A_n$.
We know that the sum of $n$ arithmetic means between $a$ and $b$ is equal to $n$ times the arithmetic mean of $a$ and $b$.
Thus,$S = n \times \left(\frac{a + b}{2}\right) = \frac{n(a + b)}{2}$.

(B) The resulting progression consists of $n + 2$ terms,where the first term $a = 2$ and the last term $l = 38$.
The sum of an arithmetic progression with $N$ terms is given by $S_N = \frac{N}{2}(a + l)$.
Here,$N = n + 2$,$a = 2$,and $l = 38$.
So,the sum is $S = \frac{n + 2}{2}(2 + 38) = \frac{n + 2}{2}(40) = 20(n + 2)$.
Given that the sum is $200$,we have $20(n + 2) = 200$.
Dividing by $20$,we get $n + 2 = 10$.
Therefore,$n = 8$.

(B) Given that $\log 2, \log (2^n - 1)$ and $\log (2^n + 3)$ are in $A.P.$
By the property of $A.P.$,$2 \log (2^n - 1) = \log 2 + \log (2^n + 3)$
Using logarithmic properties,$\log (2^n - 1)^2 = \log [2(2^n + 3)]$
$(2^n - 1)^2 = 2(2^n + 3)$
Let $x = 2^n$. Then $(x - 1)^2 = 2(x + 3)$
$x^2 - 2x + 1 = 2x + 6$
$x^2 - 4x - 5 = 0$
$(x - 5)(x + 1) = 0$
So,$x = 5$ or $x = -1$.
Since $x = 2^n$ must be positive,$2^n = 5$.
Therefore,$n = \log_2 5$.

(B) Let the four terms of the $A.P.$ be $(a - 3d), (a - d), (a + d), (a + 3d)$.
Given that the sum of the two extreme numbers is $8$:
$(a - 3d) + (a + 3d) = 8$
$2a = 8 \Rightarrow a = 4$
Given that the product of the two middle terms is $15$:
$(a - d)(a + d) = 15$
$a^2 - d^2 = 15$
Substitute $a = 4$ into the equation:
$4^2 - d^2 = 15$
$16 - d^2 = 15$
$d^2 = 1 \Rightarrow d = 1$ (taking positive value for the series).
The four terms are:
$a - 3d = 4 - 3(1) = 1$
$a - d = 4 - 1 = 3$
$a + d = 4 + 1 = 5$
$a + 3d = 4 + 3(1) = 7$
The series is $1, 3, 5, 7$. The greatest number is $7$.

(C) Let the sides of the triangle be $a - d, a, a + d$,where $d > 0$.
Since the hypotenuse is the longest side,it must be $a + d$.
According to the Pythagorean theorem,$(a + d)^2 = a^2 + (a - d)^2$.
Expanding both sides,we get $a^2 + d^2 + 2ad = a^2 + a^2 - 2ad + d^2$.
Simplifying,we get $2ad = a^2 - 2ad$,which implies $a^2 = 4ad$.
Since $a$ is a side length $(a \neq 0)$,we divide by $a$ to get $a = 4d$.
Substituting $a = 4d$ into the sides,we get $(4d - d) : 4d : (4d + d) = 3d : 4d : 5d$.
Thus,the ratio of the sides is $3:4:5$.

(A) Let the three numbers in $A.P.$ be $(a - d), a, (a + d)$.
Given that the sum is $33$:
$(a - d) + a + (a + d) = 33$
$3a = 33$
$a = 11$
Given that the product is $792$:
$(a - d) \times a \times (a + d) = 792$
$a(a^2 - d^2) = 792$
$11(11^2 - d^2) = 792$
$121 - d^2 = 72$
$d^2 = 121 - 72 = 49$
$d = \pm 7$
If $d = 7$,the numbers are $(11 - 7), 11, (11 + 7)$,which are $4, 11, 18$.
If $d = -7$,the numbers are $(11 + 7), 11, (11 - 7)$,which are $18, 11, 4$.
In both cases,the smallest number is $4$.

(C) Given that $a, b, c, d, e, f$ are in $A.P.$ with common difference $K$.
By the definition of $A.P.$,we have:
$d - c = K$ and $e - d = K$.
Therefore,$e - c = (e - d) + (d - c) = K + K = 2K$.
Since $K = d - c$,we can write:
$e - c = 2(d - c)$.
Alternatively,using values: Let $a=1, b=2, c=3, d=4, e=5, f=6$.
Then $e - c = 5 - 3 = 2$.
Checking the options:
$2(d - c) = 2(4 - 3) = 2(1) = 2$.
Thus,the correct option is $C$.

(B) Let the three numbers be $a - d, a, a + d$.
Given the sum of the numbers is $15$:
$(a - d) + a + (a + d) = 15$
$3a = 15$
$a = 5$
Given the sum of their squares is $83$:
$(a - d)^2 + a^2 + (a + d)^2 = 83$
$(a^2 - 2ad + d^2) + a^2 + (a^2 + 2ad + d^2) = 83$
$3a^2 + 2d^2 = 83$
Substitute $a = 5$ into the equation:
$3(5^2) + 2d^2 = 83$
$3(25) + 2d^2 = 83$
$75 + 2d^2 = 83$
$2d^2 = 8$
$d^2 = 4$
$d = \pm 2$
If $d = 2$,the numbers are $5-2, 5, 5+2$,which are $3, 5, 7$.
If $d = -2$,the numbers are $5-(-2), 5, 5+(-2)$,which are $7, 5, 3$.
Thus,the numbers are $3, 5, 7$.

(A) Let the three consecutive terms of an $A.P.$ be $(a - d)$,$a$,and $(a + d)$.
According to the given condition,the sum is $(a - d) + a + (a + d) = 51$.
$3a = 51 \Rightarrow a = 17$.
The product of the first and last term is $(a - d)(a + d) = 273$.
$a^2 - d^2 = 273$.
Substituting $a = 17$,we get $17^2 - d^2 = 273$.
$289 - d^2 = 273$.
$d^2 = 289 - 273 = 16$.
$d = \pm 4$.
If $d = 4$,the terms are $(17 - 4), 17, (17 + 4)$,which are $13, 17, 21$.
If $d = -4$,the terms are $(17 - (-4)), 17, (17 + (-4))$,which are $21, 17, 13$.
Thus,the numbers are $21, 17, 13$ or $13, 17, 21$.

(B) Given that $\frac{1}{p + q}, \frac{1}{r + p}, \frac{1}{q + r}$ are in $A.P.$
Therefore,the common difference is equal:
$\frac{1}{r + p} - \frac{1}{p + q} = \frac{1}{q + r} - \frac{1}{r + p}$
Simplify the expressions:
$\frac{(p + q) - (r + p)}{(r + p)(p + q)} = \frac{(r + p) - (q + r)}{(q + r)(r + p)}$
$\frac{q - r}{p + q} = \frac{p - q}{q + r}$
Cross-multiply:
$(q - r)(q + r) = (p - q)(p + q)$
$q^2 - r^2 = p^2 - q^2$
$2q^2 = p^2 + r^2$
This condition implies that $p^2, q^2, r^2$ are in $A.P.$

(D) Let $d$ be the common difference of the $A.P.$
Then,$\log_{y}x = 1 + d \Rightarrow x = y^{1+d}$
$\log_{z}y = 1 + 2d \Rightarrow y = z^{1+2d}$
$-15\log_{x}z = 1 + 3d$ $\Rightarrow \log_{x}z = -\frac{1+3d}{15}$ $\Rightarrow z = x^{-\frac{1+3d}{15}}$
Substituting the values,we get $x = (z^{1+2d})^{1+d} = z^{(1+2d)(1+d)} = (x^{-\frac{1+3d}{15}})^{(1+2d)(1+d)}$
This implies $1 = -\frac{(1+d)(1+2d)(1+3d)}{15}$
$(1+d)(1+2d)(1+3d) = -15$
$(1+3d+2d^2)(1+3d) = -15$
$1 + 3d + 3d + 9d^2 + 2d^2 + 6d^3 = -15$
$6d^3 + 11d^2 + 6d + 16 = 0$
By inspection,$d = -2$ is a root: $6(-8) + 11(4) + 6(-2) + 16 = -48 + 44 - 12 + 16 = 0$
For $d = -2$,we have $\log_{y}x = 1 - 2 = -1 \Rightarrow x = y^{-1}$
$\log_{z}y = 1 + 2(-2) = -3 \Rightarrow y = z^{-3}$
Since $x = y^{-1}$ and $y = z^{-3}$,then $x = (z^{-3})^{-1} = z^{3}$
Thus,$x = y^{-1}$,$y = z^{-3}$,and $x = z^{3}$ are all true.

(D) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$.
Substitute $2b = a + c$ into the expression $(a + 2b - c)(2b + c - a)(c + a - b)$:
First term: $(a + 2b - c) = (a + (a + c) - c) = 2a$.
Second term: $(2b + c - a) = ((a + c) + c - a) = 2c$.
Third term: $(c + a - b) = (2b - b) = b$.
Multiplying these terms together: $(2a)(2c)(b) = 4abc$.

(D) Let the four numbers in $A.P.$ be $A_1, A_2, A_3, A_4$.
Given $A_1 + A_4 = 8$ $(i)$ and $A_2 \times A_3 = 15$ $(ii)$.
In an $A.P.$,the sum of terms equidistant from the beginning and end is constant,so $A_2 + A_3 = A_1 + A_4 = 8$ $(iii)$.
From $(ii)$ and $(iii)$,we have $A_2 + \frac{15}{A_2} = 8$,which simplifies to $A_2^2 - 8A_2 + 15 = 0$.
Solving the quadratic equation,we get $A_2 = 3$ or $A_2 = 5$.
If $A_2 = 3$,then $A_3 = 5$. Since $A_2 = \frac{A_1 + A_3}{2}$,we have $A_1 = 2A_2 - A_3 = 2(3) - 5 = 1$.
Then $A_4 = 8 - A_1 = 7$.
The series is $1, 3, 5, 7$.
The least number of the series is $1$.

(C) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.
$11^{th}$ term: $a_{11} = a + 10d$.
$21^{st}$ term: $a_{21} = a + 20d$.
According to the problem,$2 \times a_{11} = 7 \times a_{21}$.
$2(a + 10d) = 7(a + 20d)$
$2a + 20d = 7a + 140d$
$5a + 120d = 0$
Dividing by $5$,we get $a + 24d = 0$.
The $25^{th}$ term is $a_{25} = a + 24d$.
Since $a + 24d = 0$,the $25^{th}$ term is $0$.

(B) Given that $\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$ are in $A.P.$
Therefore,$2 \times \frac{1}{c - a} = \frac{1}{b - c} + \frac{1}{a - b}$.
$\frac{2}{c - a} = \frac{(a - b) + (b - c)}{(b - c)(a - b)} = \frac{a - c}{(b - c)(a - b)}$.
Since $a - c = -(c - a)$,we have $\frac{2}{c - a} = \frac{-(c - a)}{(b - c)(a - b)}$.
$(c - a)^2 = -2(b - c)(a - b) = 2(b - c)(b - a)$.
We want to check if $(b - c)^2, (c - a)^2, (a - b)^2$ are in $A.P.$
This is true if $2(c - a)^2 = (b - c)^2 + (a - b)^2$.
Substituting $(c - a)^2 = 2(b - c)(b - a)$,we get $4(b - c)(b - a) = (b - c)^2 + (a - b)^2$.
$4(b - c)(b - a) - (b - c)^2 - (a - b)^2 = 0$.
This simplifies to $-( (b - c) - (a - b) )^2 = 0$,which is not generally true.
Wait,let us re-evaluate: The condition for $x, y, z$ to be in $A.P.$ is $2y = x + z$.
Given $\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$ are in $A.P.$
Let $x = b - c, y = c - a, z = a - b$. Note $x + y + z = 0$.
The given condition is $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in $A.P.$
$\frac{2}{y} = \frac{1}{x} + \frac{1}{z} = \frac{x + z}{xz} = \frac{-y}{xz}$.
So $y^2 = -2xz$.
We want to check if $x^2, y^2, z^2$ are in $A.P.$
This requires $2y^2 = x^2 + z^2$.
Since $x + z = -y$,then $(x + z)^2 = (-y)^2$,so $x^2 + z^2 + 2xz = y^2$.
$x^2 + z^2 = y^2 - 2xz$.
Since $y^2 = -2xz$,then $x^2 + z^2 = y^2 - (-y^2) = 2y^2$.
Thus,$x^2, y^2, z^2$ are in $A.P.$

(C) Given that $a^2, b^2, c^2$ are in $A.P.$
Adding $(ab + bc + ca)$ to each term,we get:
$a^2 + ab + bc + ca, b^2 + ab + bc + ca, c^2 + ab + bc + ca$ are in $A.P.$
Factoring the expressions:
$a(a + b) + c(a + b), b(b + a) + c(b + a), c(c + b) + a(c + b)$ are in $A.P.$
This simplifies to:
$(a + b)(a + c), (b + a)(b + c), (c + a)(c + b)$ are in $A.P.$
Dividing each term by $(a + b)(b + c)(c + a)$,we get:
$\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}$ are in $A.P.$
Thus,$(b + c)^{-1}, (c + a)^{-1}, (a + b)^{-1}$ are in $A.P.$

(A) Given that $a, b, c$ are in $A.P.$
Divide each term by $abc$:
$\frac{a}{abc}, \frac{b}{abc}, \frac{c}{abc}$ are in $A.P.$
This simplifies to:
$\frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab}$ are in $A.P.$

(A) Given that $a, A_1, A_2, b$ are in $A.P.$
The sum of $A.M.'s$ inserted between $a$ and $b$ is given by $A_1 + A_2 = n \times \frac{a+b}{2}$,where $n=2$.
So,$A_1 + A_2 = 2 \times \frac{a+b}{2} = a + b$ .....$(i)$
Given that $a, G_1, G_2, b$ are in $G.P.$
The product of $G.M.'s$ inserted between $a$ and $b$ is given by $G_1 G_2 = (ab)^{n/2}$,where $n=2$.
So,$G_1 G_2 = (ab)^{2/2} = ab$ .....$(ii)$
Therefore,$\frac{A_1 + A_2}{G_1 G_2} = \frac{a + b}{ab}$.

(B) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$.
Now,consider the terms $3^a, 3^b, 3^c$.
Since $2b = a + c$,we can write:
$3^{2b} = 3^{a + c}$
$(3^b)^2 = 3^a \times 3^c$
This is the condition for three numbers to be in $G.P.$,where the square of the middle term equals the product of the first and third terms.
Therefore,$3^a, 3^b, 3^c$ are in $G.P.$

(A) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$.
Consider the terms $x = \frac{1}{\sqrt{a} + \sqrt{b}}$,$y = \frac{1}{\sqrt{a} + \sqrt{c}}$,and $z = \frac{1}{\sqrt{b} + \sqrt{c}}$.
To check if they are in $A.P.$,we check if $2y = x + z$.
Rationalizing the denominators:
$x = \frac{\sqrt{b} - \sqrt{a}}{b - a}$,$y = \frac{\sqrt{c} - \sqrt{a}}{c - a}$,$z = \frac{\sqrt{c} - \sqrt{b}}{c - b}$.
Since $a, b, c$ are in $A.P.$,let $b - a = d$ and $c - b = d$,so $c - a = 2d$.
$x = \frac{\sqrt{b} - \sqrt{a}}{d}$,$y = \frac{\sqrt{c} - \sqrt{a}}{2d}$,$z = \frac{\sqrt{c} - \sqrt{b}}{d}$.
$x + z = \frac{\sqrt{b} - \sqrt{a} + \sqrt{c} - \sqrt{b}}{d} = \frac{\sqrt{c} - \sqrt{a}}{d}$.
$2y = 2 \times \frac{\sqrt{c} - \sqrt{a}}{2d} = \frac{\sqrt{c} - \sqrt{a}}{d}$.
Since $x + z = 2y$,the terms are in $A.P.$

(A) By the $AM-GM$ inequality,for $n$ positive real numbers,the arithmetic mean is greater than or equal to the geometric mean.
Consider the $n$ numbers: ${a_1}, {a_2}, ..., {a_{n-1}}, 2{a_n}$.
Their arithmetic mean is $\frac{{a_1} + {a_2} + ... + {a_{n-1}} + 2{a_n}}{n}$.
Their geometric mean is $({a_1} \cdot {a_2} \cdot ... \cdot {a_{n-1}} \cdot 2{a_n})^{1/n} = (2 \cdot {a_1} \cdot {a_2} \cdot ... \cdot {a_n})^{1/n} = (2c)^{1/n}$.
Applying $AM \ge GM$:
$\frac{{a_1} + {a_2} + ... + {a_{n-1}} + 2{a_n}}{n} \ge (2c)^{1/n}$.
Therefore,the minimum value of ${a_1} + {a_2} + ... + {a_{n-1}} + 2{a_n}$ is $n(2c)^{1/n}$.

(A) Given that ${a^2}, {b^2}, {c^2}$ are in $A.P.$
This implies ${b^2} - {a^2} = {c^2} - {b^2}$.
Adding ${ab + bc + ca}$ to both sides or manipulating the terms,we observe that if we add ${ab + bc + ca}$ to each term of the sequence,the property of $A.P.$ is preserved.
Alternatively,consider the terms $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$.
Add $1$ to each term: $\frac{a}{b+c} + 1 = \frac{a+b+c}{b+c}$,$\frac{b}{c+a} + 1 = \frac{a+b+c}{c+a}$,$\frac{c}{a+b} + 1 = \frac{a+b+c}{a+b}$.
Since ${a^2}, {b^2}, {c^2}$ are in $A.P.$,it follows that ${b+c}, {c+a}, {a+b}$ are in $H.P.$ (or their reciprocals are in $A.P.$).
Therefore,$\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are in $A.P.$

(B) The number of terms in the $n^{th}$ row is $n$.
The last term of the $(n-1)^{th}$ row is the sum of the first $(n-1)$ natural numbers,which is $\frac{(n-1)n}{2}$.
Therefore,the first term of the $n^{th}$ row is $\frac{(n-1)n}{2} + 1 = \frac{n^2 - n + 2}{2}$.
The terms in the $n^{th}$ row form an arithmetic progression $(A.P.)$ with $n$ terms,first term $a = \frac{n^2 - n + 2}{2}$,and common difference $d = 1$.
The sum of the $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Substituting the values: $S_n = \frac{n}{2}[2(\frac{n^2 - n + 2}{2}) + (n-1)(1)]$.
$S_n = \frac{n}{2}[n^2 - n + 2 + n - 1] = \frac{n}{2}(n^2 + 1)$.

(B) Let the angles of the quadrilateral be $x^o, (x+10)^o, (x+20)^o,$ and $(x+30)^o$.
The sum of the interior angles of a quadrilateral is $360^o$.
Therefore,$x + (x+10) + (x+20) + (x+30) = 360$.
$4x + 60 = 360$.
$4x = 300$.
$x = 75^o$.
The angles are $75^o, 85^o, 95^o,$ and $105^o$.
Thus,option $(b)$ is correct.

(A) Let the first term be $a$ and the common difference be $d$.
Given that the sum of the first $n$ terms $S_n$ is equal to the sum of the first $m$ terms $S_m$:
$\frac{n}{2}[2a + (n - 1)d] = \frac{m}{2}[2a + (m - 1)d]$
$n[2a + (n - 1)d] = m[2a + (m - 1)d]$
$2an + n(n - 1)d = 2am + m(m - 1)d$
$2a(n - m) + d(n^2 - n - m^2 + m) = 0$
$2a(n - m) + d[(n^2 - m^2) - (n - m)] = 0$
$2a(n - m) + d[(n - m)(n + m) - (n - m)] = 0$
Since $m \ne n$,we can divide by $(n - m)$:
$2a + d(n + m - 1) = 0$
Now,the sum of the first $(m + n)$ terms is:
$S_{m+n} = \frac{m + n}{2}[2a + (m + n - 1)d]$
Substituting $2a + (m + n - 1)d = 0$:
$S_{m+n} = \frac{m + n}{2} \times 0 = 0$

(B) Given that the first term $a = 1$ for all three $A.P.'s$ and common differences are $d_1 = 1, d_2 = 2, d_3 = 3$.
Using the sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
$S_1 = \frac{n}{2}[2(1) + (n - 1)1] = \frac{n}{2}[n + 1]$
$S_2 = \frac{n}{2}[2(1) + (n - 1)2] = \frac{n}{2}[2n] = n^2$
$S_3 = \frac{n}{2}[2(1) + (n - 1)3] = \frac{n}{2}[3n - 1]$
Adding $S_1$ and $S_3$:
$S_1 + S_3 = \frac{n}{2}[(n + 1) + (3n - 1)] = \frac{n}{2}[4n] = 2n^2$
Since $S_2 = n^2$,we have $S_1 + S_3 = 2S_2$.

(C) The total cost of the house is Rs. $15000$. Jairam paid Rs. $5000$ initially,so the remaining balance is $15000 - 5000 = 10000$.
He pays this balance in $10$ annual installments of Rs. $1000$ each,plus $10\%$ interest on the remaining balance.
In the $1^{st}$ year,he pays $1000 + 10\% \text{ of } 10000 = 1000 + 1000 = 2000$.
In the $2^{nd}$ year,he pays $1000 + 10\% \text{ of } 9000 = 1000 + 900 = 1900$.
In the $3^{rd}$ year,he pays $1000 + 10\% \text{ of } 8000 = 1000 + 800 = 1800$.
This forms an arithmetic progression with $a = 2000$,$d = -100$,and $n = 10$.
The sum of the $10$ installments is $S_{10} = \frac{10}{2} [2(2000) + (10 - 1)(-100)] = 5 [4000 - 900] = 5 \times 3100 = 15500$.
The total money paid by Jairam is $5000 + 15500 = 20500$.

(A) The sum of $n$ terms of an $A.P.$ is given by $S = \frac{n}{2}[2a + (n - 1)d]$.
For the $k$-th $A.P.$,the first term $a_k = k$ and the common difference $d_k = 2k - 1$.
Thus,$S_k = \frac{n}{2}[2k + (n - 1)(2k - 1)] = \frac{n}{2}[2k + 2kn - n - 2k + 1] = \frac{n}{2}[2kn - n + 1]$.
Now,$\sum_{k=1}^{m} S_k = \sum_{k=1}^{m} \frac{n}{2}[2kn - n + 1] = \frac{n}{2} [2n \sum_{k=1}^{m} k - \sum_{k=1}^{m} (n - 1)]$.
$= \frac{n}{2} [2n \frac{m(m+1)}{2} - m(n-1)] = \frac{n}{2} [nm(m+1) - m(n-1)]$.
$= \frac{nm}{2} [n(m+1) - (n-1)] = \frac{nm}{2} [nm + n - n + 1] = \frac{1}{2}mn(mn + 1)$.

(D) Given that ${a_1}, {a_2}, \dots, {a_{24}}$ are in an arithmetic progression $(A.P.)$.
We know that in an $A.P.$,the sum of terms equidistant from the beginning and the end is constant and equal to the sum of the first and last terms,i.e.,${a_1} + {a_{24}} = {a_5} + {a_{20}} = {a_{10}} + {a_{15}}$.
Given: ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$.
Substituting the property: $3({a_1} + {a_{24}}) = 225$.
Therefore,${a_1} + {a_{24}} = \frac{225}{3} = 75$.
The sum of the first $24$ terms is given by $S_{24} = \frac{n}{2}({a_1} + {a_n}) = \frac{24}{2}({a_1} + {a_{24}})$.
$S_{24} = 12 \times 75 = 900$.

(C) Let the roots of the equation $x^3 - 12x^2 + 39x - 28 = 0$ be $a-d, a, a+d$.
According to the properties of roots of a cubic equation:
Sum of roots: $(a-d) + a + (a+d) = -(-12)/1 = 12$
$3a = 12 \Rightarrow a = 4$.
Product of roots: $(a-d)(a)(a+d) = -(-28)/1 = 28$
$a(a^2 - d^2) = 28$.
Substituting $a = 4$:
$4(4^2 - d^2) = 28$
$16 - d^2 = 7$
$d^2 = 9$
$d = \pm 3$.
Thus,the common difference is $\pm 3$.

(B) We test the inequality $\left( \frac{n+1}{2} \right)^n \ge n!$ for different values of $n \in \mathbb{N}$.
For $n = 1$: $\left( \frac{1+1}{2} \right)^1 = 1^1 = 1$ and $1! = 1$. Since $1 \ge 1$,the statement is true for $n = 1$.
For $n = 2$: $\left( \frac{2+1}{2} \right)^2 = (1.5)^2 = 2.25$ and $2! = 2$. Since $2.25 \ge 2$,the statement is true for $n = 2$.
For $n = 3$: $\left( \frac{3+1}{2} \right)^3 = 2^3 = 8$ and $3! = 6$. Since $8 \ge 6$,the statement is true for $n = 3$.
By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,for $n$ positive real numbers $x_1, x_2, \dots, x_n$,we have $\frac{x_1 + x_2 + \dots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \dots x_n}$.
Let the numbers be $1, 2, 3, \dots, n$. Then $\frac{1+2+3+\dots+n}{n} \ge \sqrt[n]{1 \cdot 2 \cdot 3 \dots n}$.
Since $1+2+3+\dots+n = \frac{n(n+1)}{2}$,we get $\frac{n(n+1)}{2n} \ge \sqrt[n]{n!}$,which simplifies to $\frac{n+1}{2} \ge (n!)^{1/n}$.
Raising both sides to the power $n$,we get $\left( \frac{n+1}{2} \right)^n \ge n!$.
This inequality holds for all $n \ge 1$.

(C) Let the sides of the right-angled triangle be $a - d, a, a + d$ where $d > 0$.
Since it is a right-angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides:
$(a - d)^2 + a^2 = (a + d)^2$
$a^2 + d^2 - 2ad + a^2 = a^2 + d^2 + 2ad$
$a^2 = 4ad$
Since $a$ is a side length $(a \neq 0)$,we have $a = 4d$.
Substituting $a = 4d$ into the sides,we get:
$4d - d, 4d, 4d + d = 3d, 4d, 5d$.
The ratio of the sides is $3d : 4d : 5d$,which simplifies to $3 : 4 : 5$.

(B) Let the first term be $a$ and the common difference be $d$.
Given that the $p^{th}$ term is $q$,we have: $a + (p - 1)d = q$ $(1)$
Given that the $q^{th}$ term is $p$,we have: $a + (q - 1)d = p$ $(2)$
Subtracting equation $(2)$ from $(1)$:
$(p - q)d = q - p$
$d = \frac{q - p}{p - q} = -1$
Substituting $d = -1$ into equation $(1)$:
$a + (p - 1)(-1) = q$
$a - p + 1 = q$
$a = p + q - 1$
The $n^{th}$ term $T_n$ is given by:
$T_n = a + (n - 1)d$
$T_n = (p + q - 1) + (n - 1)(-1)$
$T_n = p + q - 1 - n + 1$
$T_n = p + q - n$