Question

What is the sum of the first seven terms of a geometric progression whose first term is \(729\)? Statements: (I) The seventh term is \(64\). (II) The common ratio of geometric progression is the ratio of the second term to the first term.

• A. Statement (I) alone is sufficient to answer the question, but (II) alone is not sufficient.
• B. Statement (II) alone is sufficient to answer the question, but (I) alone is not sufficient.
• C. Both the statements (I) and (II) are sufficient to answer the question, but neither statement alone is sufficient.
• D. Both the statements (I) and (II) together are not sufficient to answer the question and additional data are required.

Hint:

In G.P. data sufficiency, if any term and the first term are known, use the term formula to find the common ratio first.

Answer 1

The Correct Option is A

Concept: For a geometric progression (G.P.): \[ T_n=ar^{n-1} \] where \(a\) is the first term and \(r\) is the common ratio. Sum of first \(n\) terms: \[ S_n=\frac{a(r^n-1)}{r-1}, \quad r\neq1 \] To find the sum, we must determine the common ratio.

Step 1: Checking Statement (I).
Given: \[ a=729 \] and seventh term: \[ T_7=64 \] Using: \[ T_7=ar^6 \] \[ 64=729r^6 \] \[ r^6=\frac{64}{729} \] \[ r^6=\left(\frac{2}{3}\right)^6 \] \[ r=\frac{2}{3} \] Now sum of first seven terms: \[ S_7=\frac{729\left(1-\left(\frac{2}{3}\right)^7\right)}{1-\frac{2}{3}} \] \[ =\frac{729\left(1-\frac{128}{2187}\right)}{\frac{1}{3}} \] \[ =2187\left(\frac{2059}{2187}\right) \] \[ =2059 \] Thus, Statement (I) alone is sufficient.

Step 2: Checking Statement (II).
Statement (II) gives: \[ r=\frac{T_2}{T_1} \] This is just the definition of common ratio and gives no actual numerical value. So, Statement (II) alone is not sufficient. Hence, Statement (I) alone is sufficient.