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

Step 1: Let the first term be $a$ and common ratio be $r$
Step 2: Using $a_1a_2 + a_2a_3 + a_3a_1 = 624$: \[ a_1a_2 + a_2a_3 + a_3a_1 = a\cdot ar + ar\cdot ar^2 + ar^2\cdot a = a^2\bigl(r + r^3 + r^2\bigr) = a^2 r(1 + r + r^2) = 624. \tag{2} \] From (1), $a = \dfrac{52}{1 + r + r^2}$. Substituting in (2), \[ \left(\frac{52}{1 + r + r^2}\right)^2 r(1 + r + r^2) = 624 \;\Rightarrow\; \frac{52^2 r}{1 + r + r^2} = 624. \] Thus, \[ 1 + r + r^2 = \frac{52^2 r}{624} = \frac{2704r}{624} = \frac{13}{3}r. \] So, \[ 1 + r + r^2 = \frac{13}{3}r \;\Rightarrow\; 3 + 3r + 3r^2 = 13r \;\Rightarrow\; 3r^2 - 10r + 3 = 0. \] Solving the quadratic: \[ r = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm 8}{6} \Rightarrow r = 3 \text{ or } r = \frac{1}{3}. \] Since the G.P. is decreasing and infinite, we require $|r|<1$, hence $r = \dfrac{1}{3}$. 
Step 3: Find the first term: \[ 1 + r + r^2 = 1 + \frac{1}{3} + \frac{1}{9} = \frac{13}{9}, \] so from (1), \[ a = \frac{52}{1 + r + r^2} = \frac{52}{13/9} = 36. \] 
Step 4: Sum of the infinite G.P.: \[ S_\infty = \frac{a}{1 - r} = \frac{36}{1 - \frac{1}{3}} = \frac{36}{\frac{2}{3}} = 54. \] Therefore, the sum of the geometric progression is \(54\).