Question

Let $a_n$ be the $n^{\text{th}}$ 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:

• A. \(57\)
• B. \(63\)
• C. \(54\)
• D. \(60\)

Hint:

For an infinite geometric progression with first term $a$ and common ratio $r$ (where $\lvert r \rvert<1$): \[ S_\infty = \frac{a}{1 - r}. \] Also, using relationships between sums and products of initial terms can help form equations in $a$ and $r$.

Answer 1

The Correct Option is C

Approach: Skip the symmetric algebra entirely. Three consecutive G.P. terms are $\dfrac{a}{r},\,a,\,ar$ about a middle term $a$. This middle-term framing makes the second equation collapse instantly.

Step 1: Call the three terms $\dfrac{t}{r},\ t,\ tr$, where $t$ is the middle term. Their sum is \[ \frac{t}{r}+t+tr = 52. \]

Step 2: Now the sum of pairwise products. Pair them up: \[ \frac{t}{r}\cdot t + t\cdot tr + tr\cdot\frac{t}{r} = t^2\!\left(\frac{1}{r}+r+1\right) = t\cdot\underbrace{t\!\left(\frac1r+1+r\right)}_{=\,52} = 52\,t = 624. \] So $t = \dfrac{624}{52} = 12$. The middle term is $12$ directly.

Step 3: Put $t=12$ back into the sum: \[ \frac{12}{r}+12+12r = 52 \;\Rightarrow\; \frac{12}{r}+12r = 40 \;\Rightarrow\; 12 + 12r^2 = 40r \;\Rightarrow\; 3r^2 - 10r + 3 = 0. \] This gives $r=3$ or $r=\tfrac13$; for a decreasing infinite G.P. we keep $r=\tfrac13$.

Step 4: The first term is $a=\dfrac{t}{r}=\dfrac{12}{1/3}=36$, so \[ S_\infty = \frac{36}{1-\tfrac13} = 54. \] Final answer: $54$.