Let $a_n$ be the $n^{th}$ term of a decreasing infinite geometric progression. If $a_1+a_2+a_3 = 52$ and $a_1a_2+a_2a_3+a_3a_1 = 624$ , then the sum of this geometric progression is

✅ Correct Option: 3

Let the terms of the GP be $a, ar, ar^2$ .

From the given conditions:

$a(1 + r + r^2) = 52 \quad \cdots(1)$

$a^2r(1 + r + r^2) = 624 \quad \cdots(2)$

Dividing $(2)$ by $(1)$ :

$\dfrac{a^2r(1+r+r^2)}{a(1+r+r^2)} = \dfrac{624}{52}$

$ar = 12$

$a = \dfrac{12}{r}$

Substituting $a = \dfrac{12}{r}$ in equation $(1)$ :

$\dfrac{12}{r}(1 + r + r^2) = 52$

$12 + 12r + 12r^2 = 52r$

$12r^2 - 40r + 12 = 0$

$3r^2 - 10r + 3 = 0$

$r = \dfrac{10 \pm \sqrt{100 - 36}}{6} = \dfrac{10 \pm 8}{6}$

$r = 3$ or $r = \dfrac{1}{3}$

Since the GP is decreasing, $|r| < 1$ , so $r = \dfrac{1}{3}$ .

$a = \dfrac{12}{1/3} = 36$

Sum of infinite GP $= \dfrac{a}{1 - r}$

$= \dfrac{36}{1 - \frac{1}{3}}$

$= \dfrac{36}{\frac{2}{3}}$

$= 54$

CAT 2018 Slot 2

The value of the sum $7 \times 11 + 11 \times 15 + 15 \times 19 + \dots + 95 \times 99$ is

CAT 2019 Slot 1

If the population of a town is $p$ in the beginning of any year then it becomes $3+2 p$ in the beginning of the next year. If the population in the beginning of $2019$ is $1000$ then the population in the beginning of $2034$ will be

CAT 2022 Slot 2

The average of a non-decreasing sequence of $N$ numbers $a_{1}, a_{2}, \ldots, a_{N}$ is $300$ . If $a_{1}$ is replaced by $6 a_{1}$ , the new average becomes $400$ . Then, the number of possible values of $a_{1}$ is

CAT 2019 Slot 2

The number of common terms in the two sequences: $15, 19, 23, 27,......., 415$ and $14, 19, 24, 29,........,464$ is

CAT 2023 Slot 2

Let both the series $a_1, a_2, a_3, \dots$ and $b_1, b_2, b_3 \dots$ be in arithmetic progression such that the common differences of both the series are prime numbers. If $a_5 = b_9$ , $a_{19} = b_{19}$ and $b_2 = 0$ , then $a_{11}$ equals

CAT 2017 Slot 2

Let $a_1, a_2, a_3, a_4, a_5$ be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with $2a_3$ .

If the sum of the numbers in the new sequence is $450$ , then $a_5$ is

Let ana_nan​ be the nthn^{th}nth term of a decreasing infinite geometric progression. If a1+a2+a3=52a_1+a_2+a_3 = 52a1​+a2​+a3​=52 and a1a2+a2a3+a3a1=624a_1a_2+a_2a_3+a_3a_1 = 624a1​a2​+a2​a3​+a3​a1​=624, then the sum of this geometric progression is

Solution

Related questions:

Keyboard Shortcuts

CAT Past Year Questions (Topic-Wise)

Let $a_n$ be the $n^{th}$ term of a decreasing infinite geometric progression. If $a_1+a_2+a_3 = 52$ and $a_1a_2+a_2a_3+a_3a_1 = 624$ , then the sum of this geometric progression is