COMPASSLearning

Talk to us Register Login

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Sequences and Series

Let

a_{1}, a_{2}, \dots\dots, a_{n}

be in A.P. If

a_{5} = 2 a_{7}

and

a_{11} = 18

, then

12 (\frac{1}{a _{10} + a _{11}} + \frac{1}{a _{11} + a _{12}} + \dots .. \frac{1}{a _{17} + a _{18}})

is equal to

Answer: 8

Solution

From

a_{5} = 2 a_{7}

,

a_{1} + 4 d = 2 (a_{1} + 6 d)

gives

a_{1} + 8 d = 0

; combined with

a_{11} = a_{1} + 10 d = 18

this yields

d = 9

and

a_{1} = - 72

, so

a_{10} = 9

and

a_{18} = 81

. Each term rationalises as

\frac{1}{a _{k} + a _{k + 1}} = \frac{a _{k + 1} - a _{k}}{d}

, so the bracketed sum telescopes to

\frac{a _{18} - a _{10}}{d} = \frac{9 - 3}{9} = \frac{2}{3}

. Multiplying by

12

gives

8

.

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Integer answer

Practise more: all JEE Main PYQs on Sequences and Series →

More JEE Main questions on Sequences and Series

Let $a_{1}, a_{2}, a_{3}, \dots$ . be a G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64$ and $a_{1} + a_{3} + a_{5} = \frac{813}{7}$ .

Then $a_{3} + a_{5} + a_{7}$ is equal to :
JEE Main 2026View solution →
Let $a_{1} = 1$ and for $n \geq 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}}$ . Then $\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})$ is equal to
JEE Main 2026View solution →
The positive integer n, for which the solutions of the equation
$x (x + 2) + (x + 2) (x + 4) + \dots + (x + 2 n - 2) (x + 2 n) = \frac{8 n}{3}$

are two consecutive even integers, is :-
JEE Main 2026View solution →
Let $a_{1}, \frac{a _{2}}{2}, \frac{a _{3}}{2 ^{2}}, \dots, \frac{a _{10}}{2 ^{9}}$ be a G.P. of common ratio $\frac{1}{2}$ . If $a_{1} + a_{2} + \dots + a_{10} = 62$ , then $a_{1}$ is equal to :
JEE Main 2026View solution →
If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is
JEE Main 2026View solution →
Suppose $a, b, c$ are in A.P. and $a^{2}, 2 b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = 1$ , then $9 (a^{2} + b^{2} + c^{2})$ is equa …
JEE Main 2026View solution →