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
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:
A = 1002, 1005, ....., 9999.
9999 = 1002 + (n $-$ 1)3
$\Rightarrow$ (n $-$ 1)3 = 8997 $\Rightarrow$ n = 3000
B = 4-digit numbers divisible by 7
B = 1001, 1008, ......., 9996
$\Rightarrow$ 9996 = 1001 + (n $-$ 1)7
$\Rightarrow$ n = 1286
A $\cap$ B = 1008, 1029, ....., 9996
9996 = 1008 + (n $-$ 1)21
$\Rightarrow$ n = 429
So, no divisible by either 3 or 7
= 3000 + 1286 $-$ 429 = 3857
total 4-digits numbers = 9000
required numbers = 9000 $-$ 3857 = 5143
Explanation:
${1 \over 5}S = {7 \over 5} + {9 \over {{5^3}}} + {{13} \over {{5^4}}} + ....$
On subtracting
${4 \over 5}S = {7 \over 5} + {2 \over {{5^2}}} + {4 \over {{5^3}}} + {6 \over {{5^4}}} + ....$
$S = {7 \over {14}} + {1 \over {10}}\left( {1 + {2 \over 5} + {3 \over {{5^2}}} + ...} \right)$
$S = {7 \over 4} + {1 \over {10}}{\left( {1 - {1 \over 5}} \right)^{ - 2}}$
$ = {7 \over 4} + {1 \over {10}} \times {{25} \over {16}} = {{61} \over {32}}$
$\Rightarrow$ 160S = 5 $\times$ 61 = 305
Explanation:
Sum = ${{27} \over 2}$(209 + 495) = 9504
Number containing 1 at unit place $\matrix{ {\underline 2 } & {\underline 3 } & {\underline 1 } \cr {\underline 3 } & {\underline 4 } & {\underline 1 } \cr {\underline 4 } & {\underline 5 } & {\underline 1 } \cr } $
Number containing 1 at 10th place $\matrix{ {\underline 3 } & {\underline 1 } & {\underline 9 } \cr {\underline 4 } & {\underline 1 } & {\underline 8 } \cr } $
Required = 9504 $-$ (231 + 341 + 451 + 319 + 418)
= 7744
Explanation:
$b_{1}, b_{2}, b_{3}, \ldots, b_{10}$ are in GP common ratio $=2$
Since, $c_{k}=a_{k}+b_{k}, k=1,2,3 \ldots \ldots, 10$
$\therefore c_{2} =a_{2}+b_{2}=12$
$ c_{3} =a_{3}+b_{3}=13$
Now, $\mathrm{C}_{3}-\mathrm{C}_{2}=1$
$ \begin{array}{ll} \Rightarrow & \left(a_{3}-a_{2}\right)+\left(b_{3}-b_{2}\right) \neq 1 \Rightarrow-3+\left(2 b_{2}-b_{2}\right) \neq 1 \\ \Rightarrow & b_{2}=4 \\ \therefore & a_{2}=8 \end{array} $
So, AP is $11,8,5, \ldots$.
Now, $\sum_{k=1}^{10} C_{k}=\sum_{k=1}^{10} a_{k}+\sum_{k=1}^{10} b_{k}$
$ \begin{aligned} &=\left(\frac{10}{2}\right)[22+9(-3)]+2\left(\frac{2^{10}-1}{2-1}\right) \\ &=5(22-27)+2(1023)=2046-25 \\ &=2021 \end{aligned} $
Explanation:
Let ${2^x} = t$
${\log _3}{(t - 5)^2} = {\log _3}2\left( {t - {7 \over 2}} \right)$
${(t - 5)^2} = 2t - 7$
${t^2} - 12t + 32 = 0$
$(t - 4)(t - 8) = 0$
$\Rightarrow$ 2x = 4 or 2x = 8
x = 2 (Rejected)
Or x = 3
${\left( {1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....upto\,\infty } \right)^{{{\log }_{(0.25)}}\left( {{1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....upto\,\infty } \right)}}$
is $l$, then $l$2 is equal to _______________.
Explanation:
$S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....$
${S \over 3} = {1 \over 3} + {2 \over {{3^2}}} + {6 \over {{3^3}}} + .....$${{2x} \over 3} = 1 + {1 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + ....$
${{2S} \over 3} = {4 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + .....$
$S = {3 \over 2}\left( {{{4/3} \over {1 - 1/3}}} \right) = 3$
Now, $l = {\left( 3 \right)^{{{\log }_{0.25}}\left( {{{1/3} \over {1 - 1/3}}} \right)}}$
$l = {3^{{{\log }_{\left( {(1/4)} \right)}}\left( {{1 \over 2}} \right)}} = {3^{1/2}} = \sqrt 3 $
$\Rightarrow$ l2 = 3
Explanation:
n should not be multiple of 2, 3, 5 and 17.
Sum of all n = (1 + 3 + 5 + ...... + 99) $-$ (3 + 9 + 15 + 21 + ...... + 99) $-$ (5 + 25 + 35 + 55 + 65 + 85 + 95) $-$ (17)
= 2500 $-$ ${{17} \over 2}$(3 + 99) $-$ 365 $-$ 17
2500 $-$ 867 $-$ 365 $-$ 17
= 1251
Explanation:
${A_{14}} = {1 \over {( - 14)( - 13)......( - 1)(1).......(6)}} = {1 \over {14!.6!}}$
${A_{15}} = {1 \over {( - 15)( - 14)......( - 1)(1).......(5)}} = {1 \over {15!.5!}}$
${A_{13}} = {1 \over {( - 13)......( - 1)(1).......(7)}} = {1 \over {13!.7!}}$
${{{A_{14}}} \over {{A_{13}}}} = {1 \over {14!.6!}} \times - 13! \times 7! = {{ - 7} \over {14}} = - {1 \over 2}$
${{{A_{15}}} \over {{A_{13}}}} = {1 \over {15! \times 5!}} \times - 13! \times 7! = {{42} \over {15 \times 14}} = {1 \over 5}$
$100{\left( {{{{A_{14}}} \over {{A_{13}}}} + {{{A_{15}}} \over {{A_{13}}}}} \right)^2} = 100{\left( { - {1 \over 2} + {1 \over 5}} \right)^2} = 9$
Explanation:
Divide by 8n we get
${{{a_{n + 2}}} \over {{8^n}}} = {{2{a_{n + 1}}} \over {{8^n}}} + {{{a_n}} \over {{8^n}}}$
$ \Rightarrow 64{{{a_{n + 2}}} \over {{8^{n + 2}}}} = {{16{a_{n + 1}}} \over {{8^{n + 1}}}} + {{{a_n}} \over {{8^n}}}$
$64\sum\limits_{n = 1}^\infty {{{{a_{n + 2}}} \over {{8^{n + 2}}}}} = 16\sum\limits_{n = 1}^\infty {{{{a_{n + 1}}} \over {{8^{n + 1}}}}} + \sum\limits_{n = 1}^\infty {{{{a_n}} \over {{8^n}}}} $
$64\left( {P - {{{a_1}} \over 8} - {{{a_2}} \over {{8^2}}}} \right) = 16\left( {P - {{{a_1}} \over 8}} \right) + P$
$ \Rightarrow 64\left( {P - {1 \over 8} - {1 \over {64}}} \right) = 16\left( {P - {1 \over 8}} \right) + P$
$64P - 8 - 1 = 16P - 2 + P$
$47P = 7$
Explanation:
${S_n}(x) = 2{\log _a}x + 3{\log _a}x + 6{\log _a}x + 11{\log _a}x + ......$
${S_n}(x) = {\log _a}x(2 + 3 + 6 + 11 + .....)$
${S_r} = 2 + 3 + 6 + 11$
$ \therefore $ Tn = 2 + (1 + 3 + 5 +......+ (n - 1))
= 2 + ${{n - 1} \over 2}\left[ {2.1 + \left( {n - 2} \right)2} \right]$
= 2 + $\left( {n - 1} \right)\left[ {1 + \left( {n - 2} \right)} \right]$
= n2 - 2n + 3
General term ${T_r} = {r^2} - 2r + 3$
${S_n}(x) = \sum\limits_{r = 1}^n {{{\log }_a}x({r^2} - 2r + 3)} $
${S_{24}}(x) = \sum\limits_{r = 1}^{24} {{{\log }_a}x({r^2} - 2r + 3)} $
${S_{24}}(x) = {\log _a}x\sum\limits_{r = 1}^{24} {({r^2} - 2r + 3)} $
$1093 = 4372{\log _a}x$
${\log _a}x = {1 \over 4}$
$x = {a^{1/4}}$ .....(i)
${S_{12}}(2x) = {\log _a}(2x)\sum\limits_{r = 1}^{12} {({r^2} - 2r + 3)} $
$265 = 530{\log _a}(2x)$
${\log _a}(2x) = {1 \over 2}$
$2x = {a^{1/2}}$ ....(ii)
From (i) and (ii), we get
$2{a^{{1 \over 4}}} = {a^{{1 \over 2}}}$
$ \Rightarrow $ ${\left( {2{a^{{1 \over 4}}}} \right)^4} = {\left( {{a^{{1 \over 2}}}} \right)^4}$
$ \Rightarrow $ 16$a$ = $a$2
$ \Rightarrow $ $a = 16$
Explanation:
Solving, we get $a = {1 \over {12}}$ or $a = - {1 \over 4}$ [rejected]
if $a = {1 \over {12}} \Rightarrow b = {1 \over 9}$
$ \therefore $ $72(a + b) = 72\left( {{1 \over {12}} + {1 \over 9}} \right) = 14$
Explanation:
G.P. from the set will be 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192 .....
So common terms are 16, 256, 4096.
Explanation:
gcd(N, 18) = 3
Hence N is an odd integer which is divisible by 3 but not by 9.
4 digit odd multiples of 3
1005, 1011, ..........., 9999 $ \to $ 1500
4 digit odd multiples of 9
1017, 1035, ..........., 9999 $ \to $ 500
Hence number of such N = 1000
sequence $-$16, 8, $-$4, 2, ...... satisfy the equation
4x2 $-$ 9x + 5 = 0, then p + q is equal to __________.
Explanation:
$ \Rightarrow (x - 1)(4x - 5) = 0$
$ \Rightarrow $ A. M. $ = {5 \over 4}$, G. M. = 1 (As A. M. $ \ge $ G. M)
Again, for the series
$-$16, 8, $-$4, 2 ..........
${p^{th}}$ term ${t_p} = - 16{\left( {{{ - 1} \over 2}} \right)^{p - 1}}$
${q^{th}}$ term ${t_p} = 16{\left( {{{ - 1} \over 2}} \right)^{q - 1}}$
Now, A. M. = ${{{t_p} + {t_q}} \over 2} = {5 \over 4}$ & G. M. = $\sqrt {{t_p}{t_q}} = 1$
$ \Rightarrow {16^2}{\left( { - {1 \over 2}} \right)^{p + q - 2}} = 1$
$ \Rightarrow {( - 2)^8} = {( - 2)^{(p + q - 2)}}$
$ \Rightarrow p + q = 10$
Explanation:
$ \therefore $ Side lengths are in G.P.
${T_n} = {{12} \over {{{\left( {\sqrt 2 } \right)}^{n - 1}}}}$
$ \therefore $ Area $ = {{144} \over {{2^{n - 1}} }}$ < 1
$ \Rightarrow {2^{n - 1}} > 144$
Smallest n = 9
Explanation:
$a + ar + a{r^2} + a{r^3} = {{65} \over {12}}$ ..........(1)
${1 \over a} + {1 \over {ar}} + {1 \over {a{r^2}}} + {1 \over {a{r^3}}} = {{65} \over {18}}$
${1 \over a}\left( {{{{r^3} + {r^2} + r + 1} \over {{r^3}}}} \right) = {{65} \over {18}}$ ...............(2)
Doing ${{(1)} \over {(2)}},$
${a^2}{r^3} = {{18} \over {12}} = {3 \over 2}$
Also given, ${a^3}{r^3} = 1 \Rightarrow a\left( {{3 \over 2}} \right) = 1 \Rightarrow a = {2 \over 3}$
${4 \over 9}{r^3} = {3 \over 2} \Rightarrow {r^3} = {{{3^3}} \over {{2^3}}} \Rightarrow r = {3 \over 2}$
$\alpha = a{r^2} = {2 \over 3}.{\left( {{3 \over 2}} \right)^2} = {3 \over 2}$
$2\alpha = 3$
Explanation:
$ \therefore $ Common difference, $d = {{b - a} \over {m + 1}} = {{240} \over {m + 1}}$
$ \therefore $ 4th A.M. = a + 4d
= 3 + 4 $ \times $ ${{240} \over {m + 1}}$
Also there are 3 G.M between 3 and 243
$ \therefore $ Common ratio (r) = ${\left( {{b \over a}} \right)^{{1 \over {n + 1}}}}$
where n = number of G.M inserted.
$ \therefore $ r = ${\left( {{{243} \over 3}} \right)^{{1 \over {3 + 1}}}} = 3$
Given,
4th A.M = 2nd G.M
$ \Rightarrow 3 + 4 \times {{240} \over {m + 1}} = 3{(3)^2}$
$ \Rightarrow {{960} \over {m + 1}} = 24$
$ \Rightarrow m = 39$
Explanation:
As sum of GP upto infinity = ${a \over {1 - r}}$
$ \therefore $ ${1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....\infty $ = ${{{1 \over 3}} \over {1 - {1 \over 3}}}$ = ${1 \over 2}$
$ \therefore $ ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}$
= ${\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left( {{{16} \over {100}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left( {{4 \over {10}}} \right)^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left[ {{{\left( {{{10} \over 4}} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${\left[ {{{\left( {2.5} \right)}^{ - 2}}} \right]^{{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}$
= ${{{\left( {2.5} \right)}^{ - 2{{\log }_{2.5}}\left( {{1 \over 2}} \right)}}}$
= ${{{\left( {{1 \over 2}} \right)}^{ - 2}}}$ = 4
Explanation:
d1 = 4
Second A.P. is 2, 9, 16, 23, ..... 709
d2 = 7
First common term = 23
Common difference of new A.P using the common terms of the two given A.P's is d = L.C.M. (4, 7) = 28
Last term $ \le $ 407
$ \Rightarrow $ 23 + (n – 1) (28) $ \le $ 407
$ \Rightarrow $ n $ \le $ 14.7
$ \therefore $ n = 14
Explanation:
= ${1 \over 4}\sum\limits_{n = 1}^7 {\left( {2{n^3} + 3{n^2} + n} \right)} $
= ${1 \over 2}\sum\limits_{n = 1}^7 {{n^3}} $ + ${3 \over 4}\sum\limits_{n = 1}^7 {{n^2}} $ + ${1 \over 4}\sum\limits_{n = 1}^7 n $
= ${1 \over 2}{\left( {{{7\left( {7 + 1} \right)} \over 2}} \right)^2}$ + ${3 \over 4}\left( {{{7\left( {7 + 1} \right)\left( {14 + 1} \right)} \over 6}} \right)$ + ${1 \over 4}{{7\left( 8 \right)} \over 2}$
= (49)(8) + (15$ \times $7) + (7)
= 392 + 105 + 7 = 504
Explanation:
= $\sum\limits_{k = 1}^{20} {{{k\left( {k + 1} \right)} \over 2}} $
= $\sum\limits_{k = 1}^{20} {{{{k^2}} \over 2}} + \sum\limits_{k = 1}^{20} {{k \over 2}} $
= ${1 \over 2} \times {{20 \times 21 \times 41} \over 6} + {1 \over 2} \times {{20 \times 21} \over 2}$
= 1540
${T_r}$ is always
The sum ${V_1}$+${V_2}$ +...+${V_n}$ is
Which one of the following statements is correct ?
Which one of the following statements is correct ?
Which one of the following statements is correct ?
Which one of the following is a correct statement?
Let bi > 1 for I = 1, 2, ......, 101. Suppose logeb1, logeb2, ......., logeb101 are in Arithmetic Progression (A.P.) with the common difference loge2. Suppose a1, a2, ......, a101 are in A.P. such that a1 = b1 and a51 = b51. If t = b1 + b2 + .... + b51 and s = a1 + a2 + ..... + a51, then