CAT 2025

Slot 2

QA

Question & SolutionQues & Sol

Let $a_{n}$ be the $n^{th}$ term of a decreasing infinite geometric progression. If $a_{1}+a_{2}+a_{3}=52$ and $a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{1}=624$, then the sum of this geometric progression is

1. Concept Used

• Topic: Geometric Progression — Sum of Terms and Sum of Pairwise Products
• Formula: $$S_{\infty} = \frac{a}{1-r}, \quad |r| < 1$$

2. Calculation

Let the first term of the GP be $a$ and the common ratio be $r$. The first three terms are $a,\ ar,\ ar^2$.

Setting up the two equations:

From the sum of the first three terms: $$a + ar + ar^2 = a(1 + r + r^2) = 52 \quad \cdots (1)$$

From the sum of pairwise products: $$a_1 a_2 + a_2 a_3 + a_3 a_1 = a \cdot ar + ar \cdot ar^2 + ar^2 \cdot a$$ $$= a^2r + a^2r^3 + a^2r^2 = a^2r(1 + r + r^2) = 624 \quad \cdots (2)$$

Key insight — divide equation (2) by equation (1):

Notice that equation (2) is $a^2 r(1 + r + r^2)$ and equation (1) is $a(1 + r + r^2)$. Dividing eliminates the entire $(1 + r + r^2)$ factor: $$\frac{a^2r(1+r+r^2)}{a(1+r+r^2)} = \frac{624}{52}$$ $$ar = 12$$

So the second term $a_2 = ar = 12$.

Substituting $a = \frac{12}{r}$ into equation (1): $$\frac{12}{r}(1 + r + r^2) = 52$$ $$12 + 12r + 12r^2 = 52r$$ $$12r^2 - 40r + 12 = 0$$ $$3r^2 - 10r + 3 = 0$$

Solving the quadratic using factorisation: $$(3r - 1)(r - 3) = 0$$ $$r = \frac{1}{3} \quad \text{or} \quad r = 3$$

Applying the convergence constraint:

For an infinite GP to have a finite sum, we need $|r| < 1$. Also, the GP is stated to be decreasing, which further confirms $0 < r < 1$. Therefore: $$r = \frac{1}{3}$$

Finding the first term $a$: $$a = \frac{12}{r} = \frac{12}{\frac{1}{3}} = 36$$

Verification: $a(1 + r + r^2) = 36\left(1 + \frac{1}{3} + \frac{1}{9}\right) = 36 \times \frac{13}{9} = 52$ ✓

Computing the infinite sum: $$S_{\infty} = \frac{a}{1 - r} = \frac{36}{1 - \frac{1}{3}} = \frac{36}{\frac{2}{3}} = 36 \times \frac{3}{2} = 54$$

3. Solution

Answer = Option B ✅

The sum of the infinite geometric progression is 54.