COMPASSLearning

Talk to us Register Login

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Sequences and Series

If the sum of series

\frac{1}{1 \cdot ( 1 + d )} + \frac{1}{( 1 + d ) ( 1 + 2 d )} + \dots\dots + \frac{1}{( 1 + 9 d ) ( 1 + 10 d )}

is equal to 5, then 50d is equal to :

(A)20
(B)5✓ Correct
(C)15
(D)10

Answer: (B)

Solution

Each term telescopes:

\frac{1}{( 1 + ( k - 1 ) d ) ( 1 + k d )} = \frac{1}{d} (\frac{1}{1 + ( k - 1 ) d} - \frac{1}{1 + k d})

. Summing

k = 1

to

10

:

\frac{1}{d} (1 - \frac{1}{1 + 10 d}) = \frac{10}{1 + 10 d} = 5 \Rightarrow 1 + 10 d = 2

. Thus

10 d = 1

and

50 d = 5

— option (2).

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Single correct option

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 →