Answer to Question #289689 in Algebra for Alvine

Question

In a geometric progression, the sum of the second and third term is 6 and the third and fourth term is -12, find the first, the common ratio and the sum of the first 10 terms of the progression

Answer 1

The general term of a GP is:

$$T_n = ar^{n-1}$$

Therefore,

$$T_2 = ar^{2-1} = ar\\ T_3 = ar^{3-1} = ar^2\\ T_4 = ar^{4-1} = ar^3$$

Now,

$$T_2+T_3 = 6 \implies ar+ar^2 =6 \qquad \qquad(i)\\ T_3+T_4 = -12 \implies ar^2+ar^3 =-12 \quad \quad(ii)$$

From (ii):

$$r(ar+ar^2) = -12 \qquad \qquad (iii)$$

put (i) in (iii):

$$6r =-12 \implies \textbf{r=-2}$$

Put r=-2 in (i):

$$a(-2) + a(-2)^2=6\\ -2a +4a=6\\ 2a=6\\ \textbf{a=3}$$

The sum of first n term of a GP is:

$$S_{n}=a \cdot \frac{1-r^{n}}{1-r}\\ \text{ Given n=10, and putting other values to above formula: }\\ \begin{aligned} S_{10} &=3 \cdot \frac{1-(-2)^{10}}{1-(-2)} \\ S_{10} &=3 \cdot(-341) \\ S_{10} &=-1023 \end{aligned}$$