Q8
Let \( a _ { n } \) be the \( n ^ { \text {th } } \) term of a decreasing infinite geometric progression. If \( a _ { 1 } + a _ { 2 } + a _ { 3 } = 52 \) and \( a _ { 1 } a _ { 2 } + a _ { 2 } a _ { 3 } + a _ { 3 } a _ { 1 } = 624 \), then the sum of this geometric progression is
Answer: C
Key idea: Take the three terms as \(\dfrac{a}{r}, a, ar\). Then the sum of pairwise products factorises neatly.
\[\frac{a}{r}\cdot a + a\cdot ar + ar\cdot\frac{a}{r} = a\left(\frac{a}{r} + ar + a\right) = a \times 52.\]So \(52a = 624\), giving \(a = 12\).
Then \(\dfrac{12}{r} + 12 + 12r = 52\), so \(12r^2 - 40r + 12 = 0\), i.e. \(3r^2 - 10r + 3 = 0\), giving \(r = 3\) or \(\tfrac13\).
The GP is decreasing and infinite, so \(r = \tfrac13\): the terms are \(36, 12, 4, \ldots\)
\[S_\infty = \frac{36}{1 - \frac13} = 54.\]Hence, the answer is (C) 54.