A finite Geometric Proportion (G.P.) series contains 10 terms. The sum of the first three terms is 7 and that of the last three terms is 896. Find the first term and the common ratio.
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Let the first term of the G.P. be $a$ and the common ratio be $r$.
Since, the G.P. contains 10 terms, the terms are $a, ar, ar^2, ......, ar^9$.
Given,
Sum of the first three terms = 7
$$\begin{aligned}\Rightarrow a + ar + ar^2 = 7 \\[1em] \Rightarrow a(1 + r + r^2) = 7 \text{ ......(1)}\end{aligned}$$
Sum of the last three terms = 896
$$\begin{aligned}\Rightarrow ar^7 + ar^8 + ar^9 = 896 \\[1em] \Rightarrow ar^7(1 + r + r^2) = 896 \text{ ......(2)}\end{aligned}$$
Dividing equation (2) by equation (1), we get :
$$\begin{aligned}\Rightarrow \frac{ar^7(1 + r + r^2)}{a(1 + r + r^2)} = \frac{896}{7} \\[1em] \Rightarrow r^7 = 128 \\[1em] \Rightarrow r^7 = 2^7 \\[1em] \Rightarrow r = 2.\end{aligned}$$
Substituting $r = 2$ in equation (1), we get :
$$\begin{aligned}\Rightarrow a(1 + 2 + 2^2) = 7 \\[1em] \Rightarrow a(1 + 2 + 4) = 7 \\[1em] \Rightarrow 7a = 7 \\[1em] \Rightarrow a = 1.\end{aligned}$$
Hence, first term = 1 and common ratio = 2.