Question

The number of terms in the geometric progression \[ 3,\; \frac32,\; \frac34,\dots \] that are needed to give a sum \[ \frac{3069}{512} \] is

• A. \(8\)
• B. \(9\)
• C. \(10\)
• D. \(12\)

Hint:

For decreasing G.P., use the standard sum formula and simplify carefully.

Answer 1

The Correct Option is C

Concept: Sum of \(n\) terms of G.P.: \[ S_n=\frac{a(1-r^n)}{1-r} \] Here: \[ a=3,\quad r=\frac12 \] Substitute: \[ S_n=\frac{3\left(1-\left(\frac12\right)^n\right)}{1-\frac12} \] \[ =6\left(1-\frac1{2^n}\right) \] Given: \[ 6\left(1-\frac1{2^n}\right)=\frac{3069}{512} \] \[ 1-\frac1{2^n}=\frac{3069}{3072} \] \[ \frac1{2^n}=\frac3{3072}=\frac1{1024} \] \[ 2^n=1024=2^{10} \] \[ n=10 \] Thus, \[ \boxed{10} \]