COMPASSLearning

Talk to us Register Login

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Sequences and Series

Let

a, ar, ar^{2}, \dots\dots

. be an infinite G.P. If

\sum_{n = 0}^{\infty} ar^{n} = 57

and

\sum_{n = 0}^{\infty} a^{3} r^{3 n} = 9747

, then

a + 18 r

is equal to :

(A)27
(B)46
(C)38
(D)31✓ Correct

Answer: (D)

Solution

n = 0 \sum \infty a r^{n} = \frac{a}{1 - r} = 57

and

n = 0 \sum \infty a^{3} r^{3 n} = \frac{a ^{3}}{1 - r ^{3}} = 9747

.

Dividing the cube of the first by the second:

\frac{a ^{3} / ( 1 - r ) ^{3}}{a ^{3} / ( 1 - r ^{3} )} = \frac{1 - r ^{3}}{( 1 - r ) ^{3}} = \frac{1 + r + r ^{2}}{( 1 - r ) ^{2}} = \frac{5 7 ^{3}}{9747} = \frac{185193}{9747} = 19

1 + r + r^{2} = 19 (1 - r)^{2} = 19 - 38 r + 19 r^{2} ⟹ 18 r^{2} - 39 r + 18 = 0 ⟹ 6 r^{2} - 13 r + 6 = 0

r = \frac{13 \pm 5}{12} ⟹ r = \frac{3}{2} (rejected, ∣ r ∣ < 1) or r = \frac{2}{3}

With

r = \frac{2}{3}

:

a = 57 (1 - r) = 57 \times \frac{1}{3} = 19

.

a + 18 r = 19 + 18 \times \frac{2}{3} = 19 + 12 = 31

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · 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 →