Q104 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
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$\dfrac{16}{7}\left(3^n-1\right)$
- Correct answerD$\dfrac{16}{7}\left(2^n-1\right)$
Solution
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): $a(1+2+4) = 16 \Rightarrow a = \dfrac{16}{7}$
$S_n = \dfrac{a(r^n-1)}{r-1} = \dfrac{\dfrac{16}{7}(2^n-1)}{1} = \dfrac{16}{7}(2^n-1)$
$a(1+r+r^2) = 16$ ... (1)
$ar^3(1+r+r^2) = 128$ ... (2)
Dividing (2) by (1): $r^3 = 8 \Rightarrow r = 2$
Substituting in (1): $a(1+2+4) = 16 \Rightarrow a = \dfrac{16}{7}$
$S_n = \dfrac{a(r^n-1)}{r-1} = \dfrac{\dfrac{16}{7}(2^n-1)}{1} = \dfrac{16}{7}(2^n-1)$
Hence, the correct answer is option: 'D'.