Sequences and Series
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
$ \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, $\sum\limits_{i=1}^{18} a_{1}=\frac{18}{2}\left[2 a_{1}+17 d\right]=9[10+68]=702$
If the sum of the first ten terms of the series
${1 \over 5} + {2 \over {65}} + {3 \over {325}} + {4 \over {1025}} + {5 \over {2501}} + \,\,....$
is ${m \over n}$, where m and n are co-prime numbers, then m + n is equal to ______________.
Explanation:
${T_r} = {r \over {{{(2{r^2})}^2} + 1}}$
$ = {r \over {{{(2{r^2} + 1)}^2} - {{(2r)}^2}}}$
$ = {1 \over 4}{{4r} \over {(2{r^2} + 2r + 1)(2{r^2} - 2r + 1)}}$
${S_{10}} = {1 \over 4}\sum\limits_{r = 1}^{10} {\left( {{1 \over {(2{r^2} - 2r + 1)}} - {1 \over {(2{r^2} + 2r + 1)}}} \right)} $
$ = {1 \over 4}\left[ {1 - {1 \over 5} + {1 \over 5} - {1 \over {13}} + \,\,....\,\, + \,\,{1 \over {181}} - {1 \over {221}}} \right]$
$ \Rightarrow {S_{10}} = {1 \over 4}\,.\,{{220} \over {221}} = {{55} \over {221}} = {m \over n}$
$\therefore$ $m + n = 276$
If a1 (> 0), a2, a3, a4, a5 are in a G.P., a2 + a4 = 2a3 + 1 and 3a2 + a3 = 2a4, then a2 + a4 + 2a5 is equal to ___________.
Explanation:
Let G.P. be a1 = a, a2 = ar, a3 = ar2, .........
$\because$ 3a2 + a3 = 2a4
$\Rightarrow$ 3ar + ar2 = 2ar3
$\Rightarrow$ 2ar2 $-$ r $-$ 3 = 0
$\therefore$ r = $-$1 or ${3 \over 2}$
$\because$ a1 = a > 0 then r $\ne$ $-$1
Now, a2 + a4 = 2a3 + 1
ar + ar3 = 2ar2 + 1
$a\left( {{3 \over 2} + {{27} \over 8} - {9 \over 2}} \right) = 1$
$\therefore$ a = ${8 \over 3}$
$\therefore$ a2 + a4 + 2a5 = a(r + r3 + 2r4)
$ = {8 \over 3}\left( {{3 \over 2} + {{27} \over 8} + {{81} \over 8}} \right) = 40$
For a natural number n, let ${\alpha _n} = {19^n} - {12^n}$. Then, the value of ${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}}$ is ___________.
Explanation:
${\alpha _n} = {19^n} - {12^n}$
Let equation of roots 12 & 19 i.e.
${x^2} - 31x + 228 = 0$
$ \Rightarrow (31 - x) = {{228} \over x}$ (where x can be 19 or 12)
$\therefore$ ${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}} = {{31({{19}^9} - {{12}^9}) - ({{19}^{10}} - {{12}^{10}})} \over {57({{19}^8} - {{12}^8})}}$
$ = {{{{19}^9}(31 - 19) - {{12}^9}(31 - 12)} \over {57({{19}^8} - {{12}^8})}}$
$ = {{228({{19}^8} - {{12}^8})} \over {57({{19}^8} - {{12}^8})}} = 4$.
The greatest integer less than or equal to the sum of first 100 terms of the sequence ${1 \over 3},{5 \over 9},{{19} \over {27}},{{65} \over {81}},$ ...... is equal to ___________.
Explanation:
$S = {1 \over 3} + {5 \over 9} + {{19} \over {27}} + {{65} \over {81}}\, + $ ....
$ = \sum\limits_{r = 1}^{100} {\left( {{{{3^r} - {2^r}} \over {{3^r}}}} \right)} $
$ = 100 - {2 \over 3}{{\left( {1 - {{\left( {{2 \over 3}} \right)}^{100}}} \right)} \over {1/3}}$
$ = 98 + 2{\left( {{2 \over 3}} \right)^{100}}$
$\therefore$ $[S] = 98$
Explanation:
Given,
${l_1},{l_2},\,.......,\,{l_{100}}$ are in A.P with common difference ${d_1}$.
So from property of A.P we can say,
${l_2} = {l_1} + {d_1}$
${l_3} = {l_1} + 2{d_1}$
$ \vdots $
${l_{100}} = {l_1} + 99{d_1}$
Also given,
${w_1},{w_2},\,......\,,\,{w_{100}}$ are in A.P with common difference ${d_2}$.
$\therefore$ From the property of A.P we can say,
${w_2} = {w_1} + {d_2}$
${w_3} = {w_1} + 2{d_2}$
$ \vdots $
${w_{100}} = {w_1} + 99{d_2}$
Now, also given,
${d_1}{d_2} = 10$
and ${R_i}$ is a rectangle whose length is ${l_i}$ and width is ${w_i}$ and area ${A_i}$.
$\therefore$ We know, area of rectangle
${A_i} = {l_i} \times {w_i}$
$\therefore$ ${A_{51}} = {l_{51}} \times {w_{51}}$
and ${A_{50}} = {l_{50}} \times {w_{50}}$
Given, ${A_{51}} - {A_{50}} = 1000$
$ \Rightarrow ({l_1} + 50{d_1})({w_1} + 50{d_2}) - ({l_1} + 49{d_1})({w_1} + 49{d_2}) = 1000$
$ \Rightarrow [{l_1}{w_1} + 2500{d_1}{d_2} + 50({l_1}{d_2} + {d_1}{w_1})] - [{l_1}{w_1} + 49 \times 49{d_1}{d_2} + 49({l_1}{d_2} + {w_1}{d_1})] = 1000$
$ \Rightarrow [{(50)^2}{d_1}{d_2} - {(49)^2}{d_1}{d_2}] + (50 - 49)({l_1}{d_2} + {d_1}{w_1}) = 1000$
$ \Rightarrow (99 \times 1){d_1}{d_2} + {l_1}{d_2} + {d_1}{w_1} = 1000$
$ \Rightarrow 99 \times 10 + {l_1}{d_2} + {w_1}{d_1} = 1000$
$ \Rightarrow {l_1}{d_2} + {w_1}{d_1} = 10$
Now,
${A_{100}} - {A_{90}}$
$ = {l_{100}}\,.\,{w_{100}} - {l_{90}}\,.\,{w_{90}}$
$ = ({l_1} + 99{d_1})({w_1} + 99{d_2}) - ({l_1} + 89{d_1})({w_1} + 89{d_2})$
$ = [{l_1}{w_1} + {(99)^2}{d_1}{d_2} + 99({l_1}{d_2} + {w_1}{d_1})] - [{l_1}{w_1} + {(89)^2}{d_1}{d_2} + 89({l_1}{d_2} + {w_1}{d_1})]$
$ = [{(99)^2} - {89^2}]{d_1}{d_2} + 10({l_1}{d_2} + {w_1}{d_1})$
$ = 188 \times 10 \times {d_1}{d_2} + 10 \times 10$
$ = 188 \times 10 \times 10 + 100$
$ = 18800 + 100$
$ = 18900$
Let $a_{1}, a_{2}, a_{3}, \ldots$ be an arithmetic progression with $a_{1}=7$ and common difference 8. Let $T_{1}, T_{2}, T_{3}, \ldots$ be such that $T_{1}=3$ and $T_{n+1}-T_{n}=a_{n}$ for $n \geq 1$. Then, which of the following is/are TRUE ?