Sequences and Series
The sum to $20$ terms of the series $2 \cdot 2^{2}-3^{2}+2 \cdot 4^{2}-5^{2}+2 \cdot 6^{2}-\ldots \ldots$ is equal to __________.
Explanation:
For $k \in \mathbb{N}$, if the sum of the series $1+\frac{4}{k}+\frac{8}{k^{2}}+\frac{13}{k^{3}}+\frac{19}{k^{4}}+\ldots$ is 10 , then the value of $k$ is _________.
Explanation:
$10 = 1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$
We isolate the 1 to get :
$9 = \frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$ .......(1)
Divide each term in the equation by $k$ :
$\frac{9}{k} = \frac{4}{k^2}+\frac{8}{k^3}+\frac{13}{k^4}+\ldots$ .......(2)
Subtracting the second equation from the first, we obtain :
$9\left(1-\frac{1}{k}\right) = \frac{4}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \ldots$
This is equivalent to :
$S = 9\left(1-\frac{1}{k}\right) = \frac{4}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \ldots$ .........(3)
Divide both sides by $k$ again, we get : $\frac{S}{k} = \frac{4}{k^2} + \frac{4}{k^3} + \frac{5}{k^4} + \ldots$ ..........(4)
Subtracting equation (4) from (3), we have :
$(1-\frac{1}{k})S = \frac{4}{k} + \frac{1}{k^3} + \frac{1}{k^4} + \ldots$
In this equation, you've treated the infinite series on the right-hand side as a geometric series with a ratio of $1/k$, so the sum of this series can be expressed as $\frac{1/k^3}{1 - 1/k}$.
Therefore, we get :
$9\left(1-\frac{1}{k}\right)^2 = \frac{4}{k} + \frac{1/k^3}{1 - 1/k}.$
Now, simplifying this equation, we get :
$9(k-1)^2 = 4k^2 - 4k + 1$
$ \Rightarrow $$9(k^2 - 2k + 1) = 4k^2 - 4k + 1$
$ \Rightarrow $$9k^2 - 18k + 9 = 4k^2 - 4k + 1$
$ \Rightarrow $$9k^2 - 4k^2 - 18k + 4k + 9 - 1 = 0$
$ \Rightarrow $$5k^2 - 14k + 8 = 0$
This is a standard form quadratic equation, $ax^2 + bx + c = 0$, where $a=5$, $b=-14$, and $c=8$.
The solutions can be found using the quadratic formula, $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Substituting the values for $a$, $b$, and $c$, we have :
$ \Rightarrow $$k = \frac{14 \pm \sqrt{(-14)^2 - 4\times5\times8}}{2\times5}$
$ \Rightarrow $$k = \frac{14 \pm \sqrt{196 - 160}}{10}$
$ \Rightarrow $$k = \frac{14 \pm \sqrt{36}}{10}$
$ \Rightarrow $$k = \frac{14 \pm 6}{10}$
So, the solutions are $k = 2$ or $k = 0.8$. However, since the question mentions $k \in \mathbb{N}$, i.e., $k$ is a natural number, the only valid solution is $k = 2$.
Let $S=109+\frac{108}{5}+\frac{107}{5^{2}}+\ldots .+\frac{2}{5^{107}}+\frac{1}{5^{108}}$. Then the value of $\left(16 S-(25)^{-54}\right)$ is equal to ___________.
Explanation:
$\frac{S}{5}=\frac{109}{5}+\frac{108}{5^2}+\frac{107}{5^3}+\ldots .+\frac{2}{5^{108}}+\frac{1}{5^{109}}$ .............(ii)
On subtracting Eq. (ii) from Eq. (i), we get
$ \begin{aligned} \frac{4 S}{5} & =109-\frac{1}{5}-\frac{1}{5^2}-\ldots .-\frac{1}{5^{108}}-\frac{1}{5^{109}} \\\\ \frac{4 S}{5} & =109-\left[\frac{1}{5}+\frac{1}{5^2}+\ldots+\frac{1}{5^{108}}+\frac{1}{5^{109}}\right] \end{aligned} $
This form a GP with $r=\frac{1}{5}$
$ \begin{aligned} \frac{4 S}{5} & =109-\frac{1}{5}\left[\frac{1-\frac{1}{5^{109}}}{1-\frac{1}{5}}\right]\\\\ &=109-\frac{1}{4}\left[1-\frac{1}{5^{109}}\right] \\\\ & =109-\frac{1}{4}+\frac{1}{4} \times \frac{1}{5^{109}} \end{aligned} $
$ \therefore $ $ 4 S=\frac{5}{4}\left[435+\frac{1}{5^{109}}\right] $
$ \Rightarrow 16 S=2175+\frac{1}{5^{108}} $
$ 16 S-(25)^{-54}=2175 $
Suppose $a_{1}, a_{2}, 2, a_{3}, a_{4}$ be in an arithmetico-geometric progression. If the common ratio of the corresponding geometric progression is 2 and the sum of all 5 terms of the arithmetico-geometric progression is $\frac{49}{2}$, then $a_{4}$ is equal to __________.
Explanation:
$ \begin{aligned} & \frac{(a-2 d)}{4}, \frac{(c-d)}{2}, a, 2(a+d), 4(a+2 d) \\\\ & \text { or } a_1, a_2, 2, a_3, a_4 \text { (Given) } \\\\ & \Rightarrow a=2 \end{aligned} $
also sum of thes A.G.P. is $\frac{49}{2}$
$ \begin{aligned} & \Rightarrow \frac{2-2 d}{4}+\frac{2-d}{2}+2+2(2+d)+4(2+2 d)=\frac{49}{2} \\\\ & \Rightarrow \frac{1}{4}[2-2 d+4-2 d+8+16+8 d+32+32 d]=\frac{49}{2} \\\\ & \Rightarrow 36 d+62=98 \end{aligned} $
$ \Rightarrow 36 d=36 \Rightarrow d=1 $
Hence, $a_4=4(a+2 d)=4(2+2 \times 1)=16$
The sum of all those terms, of the arithmetic progression 3, 8, 13, ...., 373, which are not divisible by 3, is equal to ____________.
Explanation:
Which are not divisible by 3 is
$ \begin{aligned} & =(3+8+13+18+\ldots+373) -(3+18+33+\ldots+363) \\\\ & =\frac{75}{2}(3+373)-\frac{25}{2}(3+363) \\\\ & =\frac{75}{2} \times 376-\frac{25}{2} \times 366 \\\\ & =75 \times 188-25 \times 183 \\\\ & =14100-4575 \\\\ & =9525 \end{aligned} $
Let $0 < z < y < x$ be three real numbers such that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an arithmetic progression and $x, \sqrt{2} y, z$ are in a geometric progression. If $x y+y z+z x=\frac{3}{\sqrt{2}} x y z$ , then $3(x+y+z)^{2}$ is equal to ____________.
Explanation:
$ \Rightarrow \frac{1}{x}+\frac{1}{z}=\frac{2}{y} $ ........... (i)
and $x, \sqrt{2} y, z$ are in G.P.
$ \Rightarrow 2 y^2=x z $ .......... (ii)
from (i), $\frac{2}{y}=\frac{x+z}{x z}=\frac{x+z}{2 y^2}$
$ \Rightarrow 4 y=x+z $
$ \begin{aligned} & \text { Also, } x y+y z+z x=\frac{3}{\sqrt{2}} x y z \\\\ & y(4 y)+x z=\frac{3}{\sqrt{2}}\left(2 y^2\right) y \\\\ & \Rightarrow 4 y^2+2 y^2=3 \sqrt{2} y^3 \\\\ & \Rightarrow 6 y^2=3 \sqrt{2} y^3 \Rightarrow y=\sqrt{2} \\\\ & \therefore 3(x+y+z)^2=3(5 y)^2=3(5 \sqrt{2})^2 \\\\ & =150 \end{aligned} $
If
$(20)^{19}+2(21)(20)^{18}+3(21)^{2}(20)^{17}+\ldots+20(21)^{19}=k(20)^{19}$,
then $k$ is equal to ___________.
Explanation:
Now,
$\begin{aligned} k\left(\frac{21}{20}\right)=\left(\frac{21}{20}\right) & +2\left(\frac{21}{20}\right)^2+3\left(\frac{21}{20}\right)^3 +\ldots+20\left(\frac{21}{20}\right)^{20}\end{aligned}$ ........(ii)
On subtracting Equation (ii) from Equation (i), we get
$ \begin{aligned} & k\left(\frac{-1}{20}\right)=1+\frac{21}{20}+\left(\frac{21}{20}\right)^2+\ldots \ldots+\left(\frac{21}{20}\right)^{19}-20\left(\frac{21}{20}\right)^{20} \\\\ & \Rightarrow k\left(\frac{-1}{20}\right)=\frac{1\left(\left(\frac{21}{20}\right)^{20}-1\right)}{\frac{21}{20}-1}-20\left(\frac{21}{20}\right)^{20} \\\\ & \Rightarrow k\left(\frac{-1}{20}\right)=20\left(\left(\frac{21}{20}\right)^{20}-1\right)-20\left(\frac{21}{20}\right)^{20} \\\\ & =20\left(\frac{21}{20}\right)^{20}-20-20\left(\frac{21}{20}\right)^{20} \\\\ & \Rightarrow k\left(\frac{-1}{20}\right)=-20 \\\\ & \Rightarrow k=400 \end{aligned} $
The sum of the common terms of the following three arithmetic progressions.
$3,7,11,15, \ldots ., 399$,
$2,5,8,11, \ldots ., 359$ and
$2,7,12,17, \ldots ., 197$,
is equal to _____________.
Explanation:
Common terms are 47, 107, 167
Sum $=321$
Let $a_{1}=8, a_{2}, a_{3}, \ldots, a_{n}$ be an A.P. If the sum of its first four terms is 50 and the sum of its last four terms is 170 , then the product of its middle two terms is ___________.
Explanation:
Explanation:
Separating odd placed and even placed terms we get
$ \begin{aligned} & \mathrm{S}=\left(1.1^2+3.5^2+\ldots .15 .(29)^2\right)-\left(2.3^2+4.7^2\right. \\ & +\ldots .+14 .(27)^2 \end{aligned} $
$ \begin{aligned} & =\sum_{r=1}^{8}(2 r-1)(4 r-3)^{2}-\sum_{r=1}^{7} 2 r(4 r-1)^{2} \\\\ & =\sum_{r=1}^{8} (32 r^{3}-64 r^{2}+42 r-9)-2\sum_{r=1}^{7} 16 r^{3}-8 r^{2}+r \\\\ & =32 \times 36^{2}-64 \times 204+1512-72 \\\\ & -2\left(16 \times 28^{2}-1120+28\right) \\\\ & =6592 \end{aligned} $
Let $a_{1}, a_{2}, \ldots, a_{n}$ be in A.P. If $a_{5}=2 a_{7}$ and $a_{11}=18$, then
$12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}}+\frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}}+\ldots+\frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to ____________.
Explanation:
$ \begin{aligned} & a+10 d=18 \\\\ & a_{5}=2 a_{7} \\\\ & a+4 d=2(a+6 d) \\\\ & a=-8 d \end{aligned} $
(i) and (ii) $\Rightarrow a=-72, d=9$.
On rationalising the denominator, given expression
$=12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{11}}}{-d}+\frac{\sqrt{a_{11}}-\sqrt{a_{12}}}{-d}+\ldots+\frac{\sqrt{a_{17}}-\sqrt{a_{18}}}{-d}\right]$
$=12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{18}}}{-d}\right]$
$=12\left[\frac{\sqrt{a_{11}-d}-\sqrt{a_{11}+7 d}}{-d}\right]$
$=12\left[\frac{\sqrt{18-9}-\sqrt{18+63}}{-9}\right]$ $=12 \times \frac{2}{3}=8$
$ \begin{aligned} & S_1=3+7+11+15+19+\ldots . . \\\\ & S_2=1+6+11+16+21+\ldots . . \end{aligned} $
is :
Explanation:
First common term is 11
Common difference of series of common terms is LCM (4, 5) = 20
$a_8=a+7d$
$=11+7\times20=151$
Let $\sum_\limits{n=0}^{\infty} \frac{\mathrm{n}^{3}((2 \mathrm{n}) !)+(2 \mathrm{n}-1)(\mathrm{n} !)}{(\mathrm{n} !)((2 \mathrm{n}) !)}=\mathrm{ae}+\frac{\mathrm{b}}{\mathrm{e}}+\mathrm{c}$, where $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbb{Z}$ and $e=\sum_\limits{\mathrm{n}=0}^{\infty} \frac{1}{\mathrm{n} !}$ Then $\mathrm{a}^{2}-\mathrm{b}+\mathrm{c}$ is equal to ____________.
Explanation:
$\sum\limits_{n = 0}^\infty {{{{n^3}(2n!) + (2n - 1)(n!)} \over {n!\,.\,(2n)!}}} $
$ = \sum\limits_{n = 0}^\infty {{{{n^3}} \over {n!}} + {{2n - 1} \over {2n!}}} $
$ = \sum\limits_{n = 0}^\infty {{3 \over {(n - 2)!}} + {1 \over {(n - 3)!}} + {1 \over {(n - 1)!}} + {1 \over {(2n - 1)!}} - {1 \over {(2n)!}}} $
$ = 3e + e + e - {1 \over e}$
$ = 5e - {1 \over e}$
$\therefore$ $a = 5,b = - 1,c = 0$
$\therefore$ ${a^2} - b + c = 26$
Let $a_1=b_1=1$ and ${a_n} = {a_{n - 1}} + (n - 1),{b_n} = {b_{n - 1}} + {a_{n - 1}},\forall n \ge 2$. If $S = \sum\limits_{n = 1}^{10} {{{{b_n}} \over {{2^n}}}} $ and $T = \sum\limits_{n = 1}^8 {{n \over {{2^{n - 1}}}}} $, then ${2^7}(2S - T)$ is equal to ____________.
Explanation:
$\because$ ${a_n} = {a_{n - 1}} + (n - 1)$ and ${a_1} = {b_1} = 1$
${b_n} = {b_{n - 1}} + {a_{n - 1}}$
$\therefore$ ${b_{n + 1}} = 2{b_n} - {b_{n - 1}} + n - 1$
| $n$ | $b_n$ | $b_n-n$ |
|---|---|---|
| 1 | 1 | 0 |
| 2 | 2 | 0 |
| 3 | 4 | 1 |
| 4 | 8 | 4 |
| 5 | 15 | 10 |
| 6 | 26 | 20 |
| 7 | 42 | 35 |
| 8 | 64 | 56 |
| 9 | 93 | 84 |
| 10 | 130 | 120 |
$\therefore$ $2S - T = \left( {\sum\limits_{n = 1}^8 {{{{b_n} - n} \over {{2^{n - 1}}}}} } \right) + {{{b_9}} \over {{2^8}}} + {{{b_{10}}} \over {{2^9}}}$
$ = {{461} \over {128}}$
$\therefore$ ${2^7}(2S - T) = 461$
Let $\{ {a_k}\} $ and $\{ {b_k}\} ,k \in N$, be two G.P.s with common ratios ${r_1}$ and ${r_2}$ respectively such that ${a_1} = {b_1} = 4$ and ${r_1} < {r_2}$. Let ${c_k} = {a_k} + {b_k},k \in N$. If ${c_2} = 5$ and ${c_3} = {{13} \over 4}$ then $\sum\limits_{k = 1}^\infty {{c_k} - (12{a_6} + 8{b_4})} $ is equal to __________.
Explanation:
$\{ {a_k}\} $ be a G.P. with ${a_1} = 4,r = {r_1}$
And
$\{ {b_k}\} $ be G.P. with ${b_1} = 4,r = {r_2}$ $({r_1} < {r_2})$
Now
${C_k} = {a_k} + {b_k}$
${c_1} = 4 + 4 = 8$ and ${c_2} = 5$
${a_2} + {b_2} = 5$
$\therefore$ ${r_1} + {r_2} = {5 \over 4}$
and ${c_3} = {{13} \over 4} \Rightarrow r_4^2 + r_2^2 = {{13} \over {16}}$
$\therefore$ ${{25} \over {16}} - 2{r_1}{r_2} = {{13} \over {16}} \Rightarrow 2{r_1}{r_2} = {3 \over 4}$
$\therefore$ ${r_2} - {r_1} = \sqrt {{{25} \over {16}} - {3 \over 2}} = {1 \over 4}$
$\therefore$ ${r_2} = {3 \over 4},{r_1} = {1 \over 2}$
$\therefore$ ${a_6} = 4 \times {1 \over {{2^5}}} = {1 \over 8},{b_4} = 4 \times {{27} \over {64}} = {{27} \over {16}}$
and $\sum\limits_{K = 1}^\infty {{C_K} = 4\left[ {{1 \over {1 - {1 \over 2}}} + {1 \over {1 - {3 \over 4}}}} \right] = 24} $
$\therefore$ $\sum\limits_{K = 1}^\infty {{C_K} - (12{a_6} + 8{b_4}) = 09} $
Let $a_1,a_2,a_3,...$ be a $GP$ of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then $a_1a_9+a_2a_4a_9+a_5+a_7$ is equal to __________.
Explanation:
$\therefore a_{1} r^{3} \times a_{1} r^{5}=9$
$a_{1}^{2} r^{8}=9 \Rightarrow a_{1} r^{4}=3$
And
$ \begin{aligned} & a_{1}\left(r^{4}+r^{6}\right)=24 \\\\ \Rightarrow & 3\left(1+r^{2}\right)=24 \\\\ \therefore & r^{2}=7 \text { and } a_{1}=\frac{3}{49} \end{aligned} $
Now
$ \begin{aligned} & a_{1} a_{9}+a_{2} a_{4} a_{9}+a_{5}+a_{7} \\\\ & =a_{1}^{2} r^{8}+a_{1}^{3} r^{12}+24 \\\\ & =24+\frac{9}{7^{4}} \times 7^{4}+\frac{27}{7^{6}} \cdot 7^{6}=60 \end{aligned} $
For the two positive numbers $a,b,$ if $a,b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{a},10$ and $\frac{1}{b}$ are in an arithmetic progression, then $16a+12b$ is equal to _________.
Explanation:
If ${{{1^3} + {2^3} + {3^3}\, + \,...\,up\,to\,n\,terms} \over {1\,.\,3 + 2\,.\,5 + 3\,.\,7\, + \,...\,up\,to\,n\,terms}} = {9 \over 5}$, then the value of $n$ is
Explanation:
Now
Let $S=1.3+2.5+3.7+\ldots$
$ \begin{aligned} & T_{n}=n \cdot(2 n+1) \\\\ & \therefore S=\frac{2 n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2} \\\\ & \Rightarrow \frac{\left(\frac{n(n+1)}{2}\right)^{2}}{n(n+1)\left[\frac{2 n+1}{3}+\frac{1}{2}\right]}=\frac{9}{5} \\\\ & \Rightarrow 5 n^{2}-19 n-30=0 \\\\ & \Rightarrow(5 n+6)(n-5)=0 \\\\ & \therefore n=5 \end{aligned} $
The 4$^\mathrm{th}$ term of GP is 500 and its common ratio is $\frac{1}{m},m\in\mathbb{N}$. Let $\mathrm{S_n}$ denote the sum of the first n terms of this GP. If $\mathrm{S_6 > S_5 + 1}$ and $\mathrm{S_7 < S_6 + \frac{1}{2}}$, then the number of possible values of m is ___________
Explanation:
$ a r^{3}=500 \Rightarrow a=\frac{500}{r^{3}} $
Now,
$ \begin{aligned} & S_{6} > S_{5}+1 \\\\ & \frac{a\left(1-r^{6}\right)}{1-r}-\frac{a\left(1-r^{5}\right)}{1-r} > 1 \\\\ & a r^{5} > 1 \\\\ & \text { Now, } r=\frac{1}{m} \text { and } a=\frac{500}{r^{3}} \\\\ & \Rightarrow \quad m^{2} < 500 \\\\ & \because m > 0 \Rightarrow m \in(0,10 \sqrt{5}) \\\\ & \quad S_{7} < S_{6}+\frac{1}{2} \\\\ & \quad \frac{a\left(1-r^{6}\right)}{1-r}<\frac{a\left(1-r^{6}\right)}{1-r}+\frac{1}{2} \\\\ & \quad a r^{6}<\frac{1}{2} \end{aligned} $
$\because r=\frac{1}{m}$ and $a=\frac{500}{r^{5}}$
$ \frac{1}{m^{3}}<\frac{1}{1000} $
$\Rightarrow m \in(10, \infty)$
Possible values of $m$ is $\{11,12,....22 \}$
$\because m \in N$
Total 12 values
$ \begin{aligned} &\text { Let }\left\{a_{n}\right\}_{n=0}^{\infty} \text { be a sequence such that } a_{0}=a_{1}=0 \text { and } \\\\ &a_{n+2}=3 a_{n+1}-2 a_{n}+1, \forall n \geq 0 . \end{aligned} $
Then $a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24}$ is equal to
Consider the sequence $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{1}=1, a_{2}=2$ and $a_{n+2}=\frac{2}{a_{n+1}}+a_{n}$ for $\mathrm{n}=1,2,3, \ldots .$ If $\left(\frac{\mathrm{a}_{1}+\frac{1}{\mathrm{a}_{2}}}{\mathrm{a}_{3}}\right) \cdot\left(\frac{\mathrm{a}_{2}+\frac{1}{\mathrm{a}_{3}}}{\mathrm{a}_{4}}\right) \cdot\left(\frac{\mathrm{a}_{3}+\frac{1}{\mathrm{a}_{4}}}{\mathrm{a}_{5}}\right) \ldots\left(\frac{\mathrm{a}_{30}+\frac{1}{\mathrm{a}_{31}}}{\mathrm{a}_{32}}\right)=2^{\alpha}\left({ }^{61} \mathrm{C}_{31}\right)$, then $\alpha$ is equal to :
Let the sum of an infinite G.P., whose first term is a and the common ratio is r, be 5 . Let the sum of its first five terms be $\frac{98}{25}$. Then the sum of the first 21 terms of an AP, whose first term is $10\mathrm{a r}, \mathrm{n}^{\text {th }}$ term is $\mathrm{a}_{\mathrm{n}}$ and the common difference is $10 \mathrm{ar}^{2}$, is equal to :
Suppose $a_{1}, a_{2}, \ldots, a_{n}$, .. be an arithmetic progression of natural numbers. If the ratio of the sum of first five terms to the sum of first nine terms of the progression is $5: 17$ and , $110 < {a_{15}} < 120$, then the sum of the first ten terms of the progression is equal to
Consider two G.Ps. 2, 22, 23, ..... and 4, 42, 43, .... of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is ${(2)^{{{225} \over 8}}}$, then $\sum\limits_{k = 1}^n {k(n - k)} $ is equal to :
The sum $\sum\limits_{n = 1}^{21} {{3 \over {(4n - 1)(4n + 3)}}} $ is equal to
The value of $1 + {1 \over {1 + 2}} + {1 \over {1 + 2 + 3}} + \,\,....\,\, + \,\,{1 \over {1 + 2 + 3 + \,\,.....\,\, + \,\,11}}$ is equal to:
The sum of the infinite series $1 + {5 \over 6} + {{12} \over {{6^2}}} + {{22} \over {{6^3}}} + {{35} \over {{6^4}}} + {{51} \over {{6^5}}} + {{70} \over {{6^6}}} + \,\,.....$ is equal to :
Let $\{ {a_n}\} _{n = 0}^\infty $ be a sequence such that ${a_0} = {a_1} = 0$ and ${a_{n + 2}} = 2{a_{n + 1}} - {a_n} + 1$ for all n $\ge$ 0. Then, $\sum\limits_{n = 2}^\infty {{{{a_n}} \over {{7^n}}}} $ is equal to:
If n arithmetic means are inserted between a and 100 such that the ratio of the first mean to the last mean is 1 : 7 and a + n = 33, then the value of n is :
Let A1, A2, A3, ....... be an increasing geometric progression of positive real numbers. If A1A3A5A7 = ${1 \over {1296}}$ and A2 + A4 = ${7 \over {36}}$, then the value of A6 + A8 + A10 is equal to
Let $S = 2 + {6 \over 7} + {{12} \over {{7^2}}} + {{20} \over {{7^3}}} + {{30} \over {{7^4}}} + \,.....$. Then 4S is equal to
If a1, a2, a3 ...... and b1, b2, b3 ....... are A.P., and a1 = 2, a10 = 3, a1b1 = 1 = a10b10, then a4 b4 is equal to -
$x = \sum\limits_{n = 0}^\infty {{a^n},y = \sum\limits_{n = 0}^\infty {{b^n},z = \sum\limits_{n = 0}^\infty {{c^n}} } } $, where a, b, c are in A.P. and |a| < 1, |b| < 1, |c| < 1, abc $\ne$ 0, then :
If $A = \sum\limits_{n = 1}^\infty {{1 \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} $ and $B = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} $, then ${A \over B}$ is equal to :
The sum 1 + 2 . 3 + 3 . 32 + ......... + 10 . 39 is equal to :
Let x, y > 0. If x3y2 = 215, then the least value of 3x + 2y is
If $\{ {a_i}\} _{i = 1}^n$, where n is an even integer, is an arithmetic progression with common difference 1, and $\sum\limits_{i = 1}^n {{a_i} = 192} ,\,\sum\limits_{i = 1}^{n/2} {{a_{2i}} = 120} $, then n is equal to :
Let $a_{1}, a_{2}, a_{3}, \ldots$ be an A.P. If $\sum\limits_{r=1}^{\infty} \frac{a_{r}}{2^{r}}=4$, then $4 a_{2}$ is equal to _________.
Explanation:
Given
$S = {{{a_1}} \over 2} + {{{a_2}} \over {{2^2}}} + {{{a_3}} \over {{2^3}}} + {{{a_4}} \over {{2^4}}}\, + \,.....\,\infty $
${{{1 \over 2}S = {{{a_1}} \over {{2^2}}} + {{{a_2}} \over {{2^3}}}\, + \,.........\,\infty } \over {{S \over 2} = {{{a_1}} \over 2} + {{({a_2} + {a_1})} \over {{2^2}}} + {{({a_3} + {a_2})} \over {{2^3}}}\, + \,......\,\infty }}$
$ \Rightarrow {S \over 2} = {{{a_1}} \over 2} + {d \over 2}$
$ \Rightarrow {a_1} + d = {a_2} = 4 \Rightarrow 4{a_2} = 16$
If $\frac{1}{2 \times 3 \times 4}+\frac{1}{3 \times 4 \times 5}+\frac{1}{4 \times 5 \times 6}+\ldots+\frac{1}{100 \times 101 \times 102}=\frac{\mathrm{k}}{101}$, then 34 k is equal to _________.
Explanation:
$S = {1 \over {2 \times 3 \times 4}} + {1 \over {3 \times 4 \times 5}} + {1 \over {4 \times 5 \times 6}}\, + \,....\, + \,{1 \over {100 \times 101 \times 102}}$
$ = {1 \over {(3 - 1)\,.\,1}}\left[ {{1 \over {2 \times 3}} - {1 \over {101 \times 102}}} \right]$
$ = {1 \over 2}\left( {{1 \over 6} - {1 \over {101 \times 102}}} \right)$
$ = {{143} \over {102 \times 101}} = {k \over {101}}$
$\therefore$ $34k = 286$
Explanation:
${1 \over {{3^{12}}}} + 5\left( {{{{2^0}} \over {{3^{12}}}} + {{{2^1}} \over {{3^{11}}}} + {{{2^2}} \over {{3^{10}}}}\, + \,.......\, + \,{{{2^{11}}} \over 3}} \right) = {2^n}\,.\,m$
$ \Rightarrow {1 \over {{3^{12}}}} + 5\left( {{1 \over {{3^{12}}}}{{\left( {{{(6)}^2} - 1} \right)} \over {(6 - 1)}}} \right) = {2^n}\,.\,m$
$ \Rightarrow {1 \over {{3^{12}}}} + {5 \over 5}\left( {{1 \over {{3^{12}}}}\,.\,{2^{12}}\,.\,{3^{12}} - {1 \over {{3^{12}}}}} \right) = {2^n}\,.\,m$
$ \Rightarrow {1 \over {{3^{12}}}} + {2^{12}} - {1 \over {{3^{12}}}} = {2^n}\,.\,m$
$ \Rightarrow {2^n}\,.\,m = {2^{12}}$
$ \Rightarrow m = 1$ and $n = 12$
$m\,.\,n = 12$
$ \frac{2^{3}-1^{3}}{1 \times 7}+\frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 \times 11}+\frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 \times 15}+\cdots+ \frac{30^{3}-29^{3}+28^{3}-27^{3}+\ldots+2^{3}-1^{3}}{15 \times 63}$ is equal to _____________.
Explanation:
${T_n} = {{\sum\limits_{k = 1}^n {\left[ {{{(2k)}^3} - {{(2k - 1)}^3}} \right]} } \over {n(4n + 3)}}$
$ = {{\sum\limits_{k = 1}^n {4{k^2} + {{(2k - 1)}^2} + 2k(2k - 1)} } \over {n(4n + 3)}}$
$ = {{\sum\limits_{k = 1}^n {(12{k^2} - 6k + 1)} } \over {n(4n + 3)}}$
$ = {{2n(2{n^2} + 3n + 1) - 3{n^2} - 3n + n} \over {n(4n + 3)}}$
$ = {{{n^2}(4n + 3)} \over {n(4n + 3)}} = n$
$\therefore$ ${T_n} = n$
${S_n} = \sum\limits_{n = 1}^{15} {{T_n} = {{15 \times 16} \over 2} = 120} $
If $\sum\limits_{k=1}^{10} \frac{k}{k^{4}+k^{2}+1}=\frac{m}{n}$, where m and n are co-prime, then $m+n$ is equal to _____________.
Explanation:
$\sum\limits_{k = 1}^{10} {{k \over {{k^4} + {k^2} + 1}}} $
$ = {1 \over 2}\left[ {\sum\limits_{k = 1}^{10} {\left( {{1 \over {{k^2} - k + 1}} - {1 \over {{k^2} + k + 1}}} \right)} } \right.$
$ = {1 \over 2}\left[ {1 - {1 \over 3} + {1 \over 3} - {1 \over 7} + {1 \over 7} - {1 \over {13}}\, + \,...\, + \,{1 \over {91}} - {1 \over {111}}} \right]$
$ = {1 \over 2}\left[ {1 - {1 \over {111}}} \right] = {{110} \over {2\,.\,111}} = {{55} \over {111}} = {m \over n}$
$\therefore$ $m + n = 55 + 111 = 166$
Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ___________.
Explanation:
${d_1} = {{199 - 100} \over 2} \notin I$
${d_2} = {{199 - 100} \over 3} = 33$
${d_3} = {{199 - 100} \over 4} \notin I$
${d_n} = {{199 - 100} \over {i + 1}} \in I$
${d_i} = 33 + 11,\,9$
Sum of CD's $ = 33 + 11 + 9$
$ = 53$
The series of positive multiples of 3 is divided into sets : $\{3\},\{6,9,12\},\{15,18,21,24,27\}, \ldots$ Then the sum of the elements in the $11^{\text {th }}$ set is equal to ____________.
Explanation:
Given series
$\therefore$ 11th set will have $1 + (10)2 = 21$ term
Also upto 10th set total $3 \times k$ type terms will be $1 + 3 + 5\, + \,......\, + \,19 = 100 - $ term
$\therefore$ Set $11 = \{ 3 \times 101,\,3 \times 102,\,......\,3 \times 121\} $
$\therefore$ Sum of elements $ = 3 \times (101 + 102\, + \,...\, + \,121)$
$ = {{3 \times 222 \times 21} \over 2} = 6993$
Let $a, b$ be two non-zero real numbers. If $p$ and $r$ are the roots of the equation $x^{2}-8 \mathrm{a} x+2 \mathrm{a}=0$ and $\mathrm{q}$ and s are the roots of the equation $x^{2}+12 \mathrm{~b} x+6 \mathrm{~b}=0$, such that $\frac{1}{\mathrm{p}}, \frac{1}{\mathrm{q}}, \frac{1}{\mathrm{r}}, \frac{1}{\mathrm{~s}}$ are in A.P., then $\mathrm{a}^{-1}-\mathrm{b}^{-1}$ is equal to _____________.
Explanation:
Let $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$ as $\alpha-3 \beta, \alpha-\beta, \alpha+\beta, \alpha+3 \beta$
So sum of roots $2 \alpha-2 \beta=4$ and $2 \alpha+2 \beta=-2$
Clearly $\alpha=\frac{1}{2}$ and $\beta=-\frac{3}{2}$
Now product of roots, $\frac{1}{p} \cdot \frac{1}{r}=\frac{1}{2 a}=-5 \Rightarrow \frac{1}{a}=-10$
and $\frac{1}{q} \cdot \frac{1}{x}=\frac{1}{6 b}=-8 \Rightarrow \frac{1}{b}=-48$
So, $\frac{1}{a}-\frac{1}{b}=38$
Let $a_{1}=b_{1}=1, a_{n}=a_{n-1}+2$ and $b_{n}=a_{n}+b_{n-1}$ for every
natural number $n \geqslant 2$. Then $\sum\limits_{n = 1}^{15} {{a_n}.{b_n}} $ is equal to ___________.
Explanation:
Given,
${a_n} = {a_{n - 1}} + 2$
$ \Rightarrow {a_n} - {a_{n - 1}} = 2$
$\therefore$ In this series between any two consecutives terms difference is 2. So this is an A.P. with common difference 2.
Also given ${a_1} = 1$
$\therefore$ Series is = 1, 3, 5, 7 ......
$\therefore$ ${a_n} = 1 + (n - 1)2 = 2n - 1$
Also ${b_n} = {a_n} + {b_{n - 1}}$
When $n = 2$ then
${b_2} - {b_1} = {a_2} = 3$
$ \Rightarrow {b_2} - 1 = 3$ [Given ${b_1} = 1$]
$ \Rightarrow {b_2} = 4$
When $n = 3$ then
${b_3} - {b_2} = {a_3}$
$ \Rightarrow {b_3} - 4 = 5$
$ \Rightarrow {b_3} = 9$
$\therefore$ Series is = 1, 4, 9 ......
= 12, 22, 32 ....... n2
$\therefore$ ${b_n} = {n^2}$
Now, $\sum\limits_{n = 1}^{15} {\left( {{a_n}\,.\,{b_n}} \right)} $
$ = \sum\limits_{n = 1}^{15} {\left[ {(2n - 1){n^2}} \right]} $
$ = \sum\limits_{n = 1}^{15} {2{n^3} - \sum\limits_{n = 1}^{15} {{n^2}} } $
$ = 2\left( {{1^3} + {2^3} + \,\,...\,\,{{15}^3}} \right) - \left( {{1^2} + {2^2} + \,\,...\,\,{{15}^2}} \right)$
$ = 2 \times {\left( {{{15 \times 16} \over 2}} \right)^2} - \left( {{{15(16) \times 31} \over 6}} \right)$
$ = 27560$
Let for $f(x) = {a_0}{x^2} + {a_1}x + {a_2},\,f'(0) = 1$ and $f'(1) = 0$. If a0, a1, a2 are in an arithmatico-geometric progression, whose corresponding A.P. has common difference 1 and corresponding G.P. has common ratio 2, then f(4) is equal to _____________.
Explanation:
Given,
$f(x) = {a_0}{x^2} + {a_1}x + {a_2}$
$f'(0) = 1$
$f'(1) = 0$
a0, a1, a2 are in A. G. P
Common difference of $AP = 1$
Common ratio of $GP = 2$
A.P terms = a, a + 1, a + 2
G.P terms = y, ry, r2y
$\therefore$ AGP terms = ay, (a+1)ry, (a+2)r2y
$\therefore$ ${a_0} = ay$
${a_1} = (a + 1)ry = (a + 1)2y$
${a_2} = (a + 2){r^2}y = (a + 2)4y$
Now, $f'(x) = 2x{a_0} + {a_1}$
$\therefore$ $f'(0) = {a_1} = 1$
and $f'(1) = 2{a_0} + {a_1} = 0$
$ \Rightarrow 2{a_0} + 1 = 0$
$ \Rightarrow {a_0} = - {1 \over 2}$
$\therefore$ $ay = - {1 \over 2}$
and $(a + 1)2y = 1$
$ \Rightarrow 2ay + 2y = 1$
$ \Rightarrow 2 \times \left( { - {1 \over 2}} \right) + 2y = 1$
$ \Rightarrow 2y = + \,2$
$ \Rightarrow y = + \,1$
$\therefore$ $a = - {1 \over 2}$
$\therefore$ ${a_2} = (a + 2)4y$
$ = \left( { - {1 \over 2} + 2} \right) \times 4\,.\,1$
$ = 6$
$\therefore$ $f(x) = - {1 \over 2}{x^2} + x + 6$
$\therefore$ $f(4) = - {1 \over 2}{(4)^2} + 4 + 6$
$ = - 8 + 10$
$ = 2$
Let 3, 6, 9, 12, ....... upto 78 terms and 5, 9, 13, 17, ...... upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to ________.
Explanation:
1st AP :
3, 6, 9, 12, ....... upto 78 terms
t78 = 3 + (78 $-$ 1)3
= 3 + 77 $\times$ 3
= 234
2nd AP :
5, 9, 13, 17, ...... upto 59 terms
t59 = 5 + (59 $-$ 1)4
= 5 + 58 $\times$ 4
= 237
Common term's AP :
First term = 9
Common difference of first AP = 3
And common difference of second AP = 4
$\therefore$ Common difference of common terms
AP = LCM (3, 4) = 12
$\therefore$ New AP = 9, 21, 33, .......
tn = 9 + (n $-$ 1)12 $\le$ 234
$ \Rightarrow n \le {{237} \over {12}}$
$ \Rightarrow n = 19$
$\therefore$ ${S_{19}} = {{19} \over 2}\left[ {2.9 + (19 - 1)12} \right]$
$ = 19(9 + 108)$
$ = 2223$
Let for n = 1, 2, ......, 50, Sn be the sum of the infinite geometric progression whose first term is n2 and whose common ratio is ${1 \over {{{(n + 1)}^2}}}$. Then the value of
${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $ is equal to ___________.
Explanation:
${S_n} = {{{n^2}} \over {1 - {1 \over {{{(n + 1)}^2}}}}} = {{n{{(n + 1)}^2}} \over {n + 2}} = ({n^2} + 1) - {2 \over {n + 2}}$
Now ${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $
$ = {1 \over {26}} + \sum\limits_{n = 1}^{50} {\left\{ {({n^2} - n) + 2\left( {{1 \over {n + 1}} - {1 \over {n + 2}}} \right)} \right\}} $
$ = {1 \over {26}} + {{50 \times 51 \times 101} \over 6} - {{50 \times 51} \over 2} + 2\left( {{1 \over 2} - {1 \over {52}}} \right)$
$ = 1 + 25 \times 17(101 - 3)$
$ = 41651$
Let A = {1, a1, a2 ....... a18, 77} be a set of integers with 1 < a1 < a2 < ....... < a18 < 77.
Let the set A + A = {x + y : x, y $\in$ A} contain exactly 39 elements. Then, the value of a1 + a2 + ...... + a18 is equal to _____________.
Explanation:
It means all other sums are already present in these 39 values, which is only possible in case when all numbers are in A.P.
Let the common difference be '$d$'.
$77=1+19 \mathrm{~d} \Rightarrow d=4$
So, $\sum\limits_{i=1}^{18} a_{1}=\frac{18}{2}\left[2 a_{1}+17 d\right]=9[10+68]=702$