Sequences and Series
Let $a, b, c$ and $d$ be positive real numbers such that $a+b+c+d=11$. If the maximum value of $a^{5} b^{3} c^{2} d$ is $3750 \beta$, then the value of $\beta$ is
Let $x_{1}, x_{2}, \ldots, x_{100}$ be in an arithmetic progression, with $x_{1}=2$ and their mean equal to 200 . If $y_{i}=i\left(x_{i}-i\right), 1 \leq i \leq 100$, then the mean of $y_{1}, y_{2}, \ldots, y_{100}$ is :
If $\mathrm{S}_{n}=4+11+21+34+50+\ldots$ to $n$ terms, then $\frac{1}{60}\left(\mathrm{~S}_{29}-\mathrm{S}_{9}\right)$ is equal to :
Let the first term $\alpha$ and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to
Let $\mathrm{a}_{\mathrm{n}}$ be the $\mathrm{n}^{\text {th }}$ term of the series $5+8+14+23+35+50+\ldots$ and $\mathrm{S}_{\mathrm{n}}=\sum_\limits{k=1}^{n} a_{k}$. Then $\mathrm{S}_{30}-a_{40}$ is equal to :
Let $S_{K}=\frac{1+2+\ldots+K}{K}$ and $\sum_\limits{j=1}^{n} S_{j}^{2}=\frac{n}{A}\left(B n^{2}+C n+D\right)$, where $A, B, C, D \in \mathbb{N}$ and $A$ has least value. Then
If $\operatorname{gcd}~(\mathrm{m}, \mathrm{n})=1$ and $1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+(2021)^{2}-(2022)^{2}+(2023)^{2}=1012 ~m^{2} n$ then $m^{2}-n^{2}$ is equal to :
The sum of the first $20$ terms of the series $5+11+19+29+41+\ldots$ is :
The sum $\sum\limits_{n = 1}^\infty {{{2{n^2} + 3n + 4} \over {(2n)!}}} $ is equal to :
The sum of 10 terms of the series
${1 \over {1 + {1^2} + {1^4}}} + {2 \over {1 + {2^2} + {2^4}}} + {3 \over {1 + {3^2} + {3^4}}}\, + \,....$ is
If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , respectively, then the sum of common ratios of all such GPs is
If ${a_n} = {{ - 2} \over {4{n^2} - 16n + 15}}$, then ${a_1} + {a_2}\, + \,....\, + \,{a_{25}}$ is equal to :
For three positive integers p, q, r, ${x^{p{q^2}}} = {y^{qr}} = {z^{{p^2}r}}$ and r = pq + 1 such that 3, 3 log$_yx$, 3 log$_zy$, 7 log$_xz$ are in A.P. with common difference $\frac{1}{2}$. Then r-p-q is equal to
$\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{2^{2}}-\frac{1}{2 \cdot 3}+\frac{1}{3^{2}}\right)+\left(\frac{1}{2^{3}}-\frac{1}{2^{2} \cdot 3}+\frac{1}{2 \cdot 3^{2}}-\frac{1}{3^{3}}\right)+$
$\left(\frac{1}{2^{4}}-\frac{1}{2^{3} \cdot 3}+\frac{1}{2^{2} \cdot 3^{2}}-\frac{1}{2 \cdot 3^{3}}+\frac{1}{3^{4}}\right)+\ldots$
is $\frac{\alpha}{\beta}$, where $\alpha$ and $\beta$ are co-prime, then $\alpha+3 \beta$ is equal to __________.
Explanation:
$S = \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{4}-\frac{1}{6}+\frac{1}{9}\right) + \left(\frac{1}{8}-\frac{1}{12}+\frac{1}{18}-\frac{1}{27}\right) + \left(\frac{1}{16}-\frac{1}{24}+\frac{1}{36}-\frac{1}{54}+\frac{1}{81}\right) + \ldots$
The first few terms of the series are :
$S = \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots\right) - \left(\frac{1}{3} + \frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \ldots\right) + \left(\frac{1}{9} + \frac{1}{18} + \frac{1}{36} + \ldots\right) - \left(\frac{1}{27} + \frac{1}{54} + \ldots\right) + \ldots$
We can now see that each group of terms forms a geometric series with a common ratio of $\frac{1}{2}$:
$S = \frac{\frac{1}{2}}{1-\frac{1}{2}} - \frac{\frac{1}{3}}{1-\frac{1}{2}} + \frac{\frac{1}{9}}{1-\frac{1}{2}} - \frac{\frac{1}{27}}{1-\frac{1}{2}} + \ldots$
The series can be rewritten as :
$S = 2 \left(\frac{1}{{2}} - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \ldots\right)$
Now, we can simplify and rewrite the series inside the parentheses as:
$S = 2 \left[\frac{1}{{2}} + (- \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \ldots\right)]$
The series inside the parentheses is an infinite geometric series with the first term $a = - \frac{1}{3}$ and the common ratio $r = -\frac{1}{3}$:
$S = 2 \left(\frac{1}{{2}} +\frac{a}{1-r}\right) = 2 \left(\frac{1}{{2}} + \frac{- \frac{1}{3}}{1+\frac{1}{3}}\right) = 2 \left(\frac{1}{{2}} + \frac{- \frac{1}{3}}{\frac{4}{3}}\right) $
= $2\left( {{1 \over 2} - {1 \over 3} \times {3 \over 4}} \right)$
$ = 2\left( {{1 \over 2} - {1 \over 4}} \right) = 2\left( {{1 \over 4}} \right) = {1 \over 2}$
Thus, the sum of the series is $\frac{1}{2}$, and $\alpha = 1$ and $\beta = 2$ are co-prime.
Therefore, $\alpha + 3\beta = 1 + 6 = 7$
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 :