Mix Examples - Sequences and Series Questions in English

(A) Let the five numbers in $AP$ be $(a-2d), (a-d), a, (a+d), (a+2d)$,where $d \neq 0$.
Given that the $1^{st}$,$3^{rd}$,and $4^{th}$ terms are in $GP$.
Therefore,$(a-d)^2 = (a-2d)(a+d)$.
Expanding both sides:
$a^2 - 2ad + d^2 = a^2 - ad - 2d^2$.
Simplifying the equation:
$-2ad + d^2 = -ad - 2d^2$.
$3d^2 - ad = 0$.
$d(3d - a) = 0$.
Since $d \neq 0$,we have $3d - a = 0$,which implies $a = 3d$.
Substituting $a = 3d$ into the terms:
$(3d-2d), (3d-d), 3d, (3d+d), (3d+2d) \Rightarrow d, 2d, 3d, 4d, 5d$.
Wait,let us re-evaluate the condition: $1^{st}$,$3^{rd}$,$4^{th}$ terms are in $GP$.
Terms are $T_1 = a-2d, T_3 = a, T_4 = a+d$.
Condition: $T_3^2 = T_1 \times T_4$.
$a^2 = (a-2d)(a+d) = a^2 + ad - 2ad - 2d^2 = a^2 - ad - 2d^2$.
$0 = -ad - 2d^2 \Rightarrow ad = -2d^2$.
Since $d \neq 0$,$a = -2d$.
The terms are: $(-2d-2d), (-2d-d), -2d, (-2d+d), (-2d+2d) \Rightarrow -4d, -3d, -2d, -d, 0$.
Thus,the $5^{th}$ term is always $0$.

(A) We are given $I(n) = n^n$ and $J(n) = 1 \times 3 \times 5 \times \ldots \times (2n - 1)$.
Using the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for the $n$ positive integers $1, 3, 5, \ldots, (2n - 1)$:
$\frac{1 + 3 + 5 + \ldots + (2n - 1)}{n} > \sqrt[n]{1 \times 3 \times 5 \times \ldots \times (2n - 1)}$
The sum of the first $n$ odd integers is $n^2$,so $\frac{n^2}{n} > (J(n))^{1/n}$.
This simplifies to $n > (J(n))^{1/n}$.
Raising both sides to the power of $n$,we get $n^n > J(n)$.
Thus,$I(n) > J(n)$.

(C) First,find $a = \min \{x^{2} + 2x + 3\}$. The minimum value of a quadratic $Ax^{2} + Bx + C$ is given by $\frac{4AC - B^{2}}{4A}$. Here $A=1, B=2, C=3$,so $a = \frac{4(1)(3) - (2)^{2}}{4(1)} = \frac{12 - 4}{4} = 2$.
Next,find $b = \lim_{\theta \rightarrow 0} \frac{1 - \cos \theta}{\theta^{2}}$. Using the identity $1 - \cos \theta = 2 \sin^{2}(\theta/2)$,we get $b = \lim_{\theta \rightarrow 0} \frac{2 \sin^{2}(\theta/2)}{\theta^{2}} = \lim_{\theta \rightarrow 0} \frac{2 \sin^{2}(\theta/2)}{4(\theta/2)^{2}} = \frac{2}{4}(1)^{2} = \frac{1}{2}$.
Now,evaluate the sum $S = \sum_{r=0}^{n} a^{r} b^{n-r} = \sum_{r=0}^{n} 2^{r} (\frac{1}{2})^{n-r} = \sum_{r=0}^{n} 2^{r} \cdot 2^{r-n} = \sum_{r=0}^{n} 2^{2r-n} = 2^{-n} \sum_{r=0}^{n} (2^{2})^{r} = 2^{-n} \sum_{r=0}^{n} 4^{r}$.
This is a geometric series with $n+1$ terms,first term $1$,and common ratio $4$. The sum is $2^{-n} \left[ \frac{1(4^{n+1} - 1)}{4 - 1} \right] = 2^{-n} \left[ \frac{4^{n+1} - 1}{3} \right] = \frac{4^{n+1} - 1}{3 \cdot 2^{n}}$.

(B) Given the recurrence relation: $a_{n+1} = \frac{1}{2}a_{n} + \frac{n^{2}-2n-1}{n^{2}(n+1)^{2}}$.
This can be rewritten as $a_{n+1} - \frac{1}{2}a_{n} = \frac{2}{(n+1)^{2}} - \frac{1}{n^{2}}$.
Let $b_{n} = a_{n} - \frac{2}{n^{2}}$. Then $a_{n} = b_{n} + \frac{2}{n^{2}}$.
Substituting this into the recurrence:
$b_{n+1} + \frac{2}{(n+1)^{2}} = \frac{1}{2}(b_{n} + \frac{2}{n^{2}}) + \frac{n^{2}-2n-1}{n^{2}(n+1)^{2}}$.
$b_{n+1} = \frac{1}{2}b_{n} + \frac{1}{n^{2}} - \frac{2}{(n+1)^{2}} + \frac{n^{2}-2n-1}{n^{2}(n+1)^{2}}$.
Simplifying the right side:
$b_{n+1} = \frac{1}{2}b_{n} + \frac{(n+1)^{2} - 2n^{2} + n^{2}-2n-1}{n^{2}(n+1)^{2}} = \frac{1}{2}b_{n} + \frac{n^{2}+2n+1-2n^{2}+n^{2}-2n-1}{n^{2}(n+1)^{2}} = \frac{1}{2}b_{n}$.
Since $b_{1} = a_{1} - \frac{2}{1^{2}} = 1 - 2 = -1$,we have $b_{n} = b_{1} \cdot (\frac{1}{2})^{n-1} = -(\frac{1}{2})^{n-1}$.
Thus,$\sum_{n=1}^{\infty} b_{n} = \sum_{n=1}^{\infty} -(\frac{1}{2})^{n-1} = -\frac{1}{1 - 1/2} = -2$.
Therefore,$|\sum_{n=1}^{\infty} b_{n}| = |-2| = 2$.

(A) Let the $A$.$P$. be $a, a+d, \dots$ and the $G$.$P$. be $g, gr, \dots$.
Given that the sum of the first ten terms of the $A$.$P$. is $160$,we have $S_{10} = \frac{10}{2}(2a + 9d) = 160$,which simplifies to $2a + 9d = 32$.
Given that the sum of the first two terms of the $G$.$P$. is $8$,we have $g + gr = 8$,or $g(1+r) = 8$.
We are given $a = r$ and $g = d$. Substituting these into the equations:
$2r + 9g = 32$ and $g(1+r) = 8$.
From the second equation,$g = \frac{8}{1+r}$. Substituting this into the first equation:
$2r + 9(\frac{8}{1+r}) = 32 \Rightarrow 2r(1+r) + 72 = 32(1+r)$.
$2r^2 + 2r + 72 = 32 + 32r \Rightarrow 2r^2 - 30r + 40 = 0 \Rightarrow r^2 - 15r + 20 = 0$.
Let the roots be $r_1$ and $r_2$. Then $r_1 + r_2 = 15$ and $r_1r_2 = 20$.
The first term of the $G$.$P$. is $g = \frac{8}{1+r}$. The sum of all possible values of $g$ is:
$g_1 + g_2 = \frac{8}{1+r_1} + \frac{8}{1+r_2} = 8 \left( \frac{1+r_2 + 1+r_1}{(1+r_1)(1+r_2)} \right) = 8 \left( \frac{2 + (r_1+r_2)}{1 + (r_1+r_2) + r_1r_2} \right)$.
Substituting the values: $8 \left( \frac{2 + 15}{1 + 15 + 20} \right) = 8 \left( \frac{17}{36} \right) = \frac{34}{9}$.

(C) Given the sum of the first $n$ terms $S_n = 3n^2 + 5n$.
The $n^{th}$ term $T_n$ is given by $T_n = S_n - S_{n-1}$.
$T_n = (3n^2 + 5n) - (3(n-1)^2 + 5(n-1)) = 3n^2 + 5n - (3(n^2 - 2n + 1) + 5n - 5) = 3n^2 + 5n - (3n^2 - 6n + 3 + 5n - 5) = 6n + 2$.
The first $10$ terms are $8, 14, 20, \dots, 62$.
We need to find the sum of squares: $\sum_{n=1}^{10} (6n + 2)^2$.
$= \sum_{n=1}^{10} (36n^2 + 24n + 4) = 36 \sum_{n=1}^{10} n^2 + 24 \sum_{n=1}^{10} n + \sum_{n=1}^{10} 4$.
Using standard summation formulas: $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$.
$= 36 \left( \frac{10 \cdot 11 \cdot 21}{6} \right) + 24 \left( \frac{10 \cdot 11}{2} \right) + 4(10)$.
$= 6(10 \cdot 11 \cdot 21) + 12(110) + 40$.
$= 6(2310) + 1320 + 40 = 13860 + 1360 = 15220$.

(D) Let the $A$.$P$. be $a_n = a + (n-1)d$ and the $G$.$P$. be $g_n = ar^{n-1}$.
Given $a_1 = g_1 = a$.
From $a_2 + g_2 = 1$,we have $(a+d) + ar = 1 \implies d = 1 - a - ar$.
From $a_3 + g_3 = 4$,we have $(a+2d) + ar^2 = 4$.
Substitute $d$ into the second equation: $a + 2(1 - a - ar) + ar^2 = 4$.
$a + 2 - 2a - 2ar + ar^2 = 4 \implies ar^2 - 2ar - a = 2 \implies a(r^2 - 2r - 1) = 2$.
Since $a = 1/(1+r)$,we substitute: $\frac{r^2 - 2r - 1}{r+1} = 2$.
$r^2 - 2r - 1 = 2r + 2 \implies r^2 - 4r - 3 = 0$. This does not yield integer solutions. Re-evaluating the system for standard integer constraints: if $a = -1/2$ and $r = 3$,then $a_1 = -1/2, a_2 = -1/2 + d, g_2 = -3/2$. $a_2 + g_2 = -2 + d = 1 \implies d = 3$. Then $a_3 = -1/2 + 6 = 5.5$ and $g_3 = -1/2(9) = -4.5$. $a_3 + g_3 = 5.5 - 4.5 = 1$ (Incorrect).
Correcting the system: Let $a=1, r=2$. $a_1=1, g_1=1$. $a_2+g_2 = (1+d)+2 = 1 \implies d = -2$. $a_3+g_3 = (1-4)+4 = 1$ (Incorrect).
Using $a=-1, r=3$: $a_1=-1, g_1=-1$. $a_2+g_2 = (-1+d)-3 = 1 \implies d=5$. $a_3+g_3 = (-1+10)-9 = 0$ (Incorrect).
Given the options,for $a=-1/2, r=3$,$a_{10} = -0.5 + 9(3) = 26.5$ and $g_5 = -0.5(81) = -40.5$. Sum $= -14$. Testing $a=1, r=3$: $a_1=1, g_1=1$. $a_2+g_2 = 1+d+3 = 1 \implies d=-3$. $a_3+g_3 = 1-6+9 = 4$. This matches!
So $a_n = 1 + (n-1)(-3) = 4 - 3n$ and $g_n = 1(3)^{n-1}$.
$a_{10} = 4 - 30 = -26$. $g_5 = 3^4 = 81$.
$a_{10} + g_5 = -26 + 81 = 55$.