Sequences and Series
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
Explanation:
$ \begin{aligned} & =7\left(10+10^2+\ldots+10^{99}\right)+50(1+11+\ldots+\overbrace{111 \ldots 1}^{98})+7 \times 99 \\\\ & =70\left(\frac{10^{99}-1}{9}\right)+\frac{50}{9}\left[(10-1)+\left(10^2-1\right)+\ldots+\left(10^{98}-1\right)\right]+7 \times 99 \end{aligned} $
$ \begin{aligned} & =70\left(\frac{10^{99}-1}{9}\right)+\frac{50}{9}\left[10\left(\frac{10^{98}-1}{9}\right)-98\right]+7 \times 99 \\\\ & =\frac{7 \times 10^{100}}{9}-\frac{70}{9}+\frac{50}{9}\left[\frac{10^{99}-1-9}{9}-98\right]+7 \times 99 \end{aligned} $
$ \begin{aligned} & =\frac{7 \times 10^{100}}{9}-\frac{70}{9}+\frac{50}{9}[\overbrace{111 \ldots 1}^{99}-99]+7 \times 99 \\\\ & =\frac{7 \times 10^{100}-70+\overbrace{555 \ldots 50}^{99}}{9}-550+693 \end{aligned} $
$ \begin{aligned} & =\frac{7 \overbrace{555 \ldots 5}^{99}-70+143 \times 9}{9} \\\\ & =\frac{7 \overbrace{55 \ldots 5}^{99}+1210}{9} \end{aligned} $
$ \therefore $ $ m+n=1219 $
If the roots of the equation $k x^3-18 x^2-36 x+8=0$ are in harmonic progression, then $k=$
64
45
81
27
If $f(x)$ is a function such that $f(x+y)=f(x)+f(y)$ and $f(1)=7$, then $\sum_{r=1}^n f(r)=$
$\frac{7 n}{2}$
$\frac{7(n+1)}{2}$
$7 n(n+1)$
$\frac{7 n(n+1)}{2}$
If $i=\sqrt{-1}$, then $\sum_{n=0}^{\infty}\left(\frac{i}{3}\right)^n=$
$\frac{9-3 i}{10}$
$9-3 i$
$9+3 i$
$\frac{9+3 i}{10}$
If $3 x=1+\frac{5}{8}+\frac{5}{8} \cdot \frac{9}{13}+\frac{5}{16}+\ldots$, then $x^4+4 x^3+6 x^2+4 x=$
0
1
4
8
Let $\frac{1}{16}, a$ and $b$ be in GP and $\frac{1}{a}, \frac{1}{b}, 6$ be in AP, where $a, b>0$. Then, $72(a+b)$ is equal to
If $a_1, a_2, \ldots, a_n$ are in HP, then the expression $a_1 a_2+a_2 a_3+\ldots+a_{n-1} a_n$ is equal to
Given, a sequence of 4 numbers, first three of which are in GP and the last three are in AP with common difference 6. If first and last term of this sequence are equal, then the last term is
$ \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