Question

1. The third and the six term of a G.P are 108 and -32 respectively find the sum of first seven term.
2. The first and last term of a G.P are 2048 respectively the sum of the series is 2730. Find the number of terms and common ratio.

Answer 1

1. Write the third and sixth terms of geometric progression:

$$b_3=b_1q^{3-1}=108,\\ b_6=b_1q^{6-1}=-32.\\$$

Solve that system. Divide one by another:

\frac{108}{-32}=\frac{q^2}{q^5},\\ \space\\ q^3=-\frac{8}{27},\\ \space\\ q=-\sqrt[3]\frac{8}{27}=-\frac{2}{3}.

$$b_1=\frac{108}{q^2}=243.$$

Sum of the first seven terms:

$$S_7=\frac{b_1(q^7-1)}{q-1}=\frac{243[(-2/3)^7-1]}{-2/3-1}=\frac{463}{3}.$$

2. This is a "reverse" problem for a sequence with $b_1=b_n=2048$ and $S_n=2730$:

$$S_n=\frac{2048(q^n-1)}{q-1}=2730.$$

Since the first and the last terms are equal, the common ratio (that makes them equal) is 1. Since such a value of a common ratio results in 0 in the denominator of the sum equation, the problem can't be solved.