COMEDK · Maths · 5. Sequences and Series
The sum of first three terms of a geometric progression is 16 and the sum of next three terms is 128 . The sum to \(\mathrm{n}\) terms of the geometric progression is
- A \(\dfrac{8}{3}\left(3^n-1\right)\)
- B \(\dfrac{8}{3}\left(2^n-1\right)\)
- C \(\dfrac{16}{7}\left(3^n-1\right)\)
- D \(\dfrac{16}{7}\left(2^n-1\right)\)
Answer & Solution
Correct Answer
(D) \(\dfrac{16}{7}\left(2^n-1\right)\)
Step-by-step Solution
Detailed explanation
Let the first term be \(a\) and common ratio be \(r\). \(a(1+r+r^2) = 16\) ... (1) \(ar^3(1+r+r^2) = 128\) ... (2) Dividing (2) by (1): \(r^3 = 8 \Rightarrow r = 2\) Substituting in (1): \(a(1+2+4) = 16 \Rightarrow a = \dfrac{16}{7}\)…
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