Question

the sum of infinite GP is 5 and the sum of the squares of these terms is 15 find the GP​

Answer 1

Answer

First term= 15/4

common ratio= 1/4

Step-by-step explanation:

a/(1-r)=5             {infinite sum}[equation 1]

sum of squares of the term means it's a GP with common ratio just square of the earlier one, that is, r^2

so

a^2/(1-r^)=15   [equation 2]

solve 1 and 2

Answer 2

Answer:

$\frac{a}{1 - r} = 5 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ..........(1) \\ \\ \frac{a}{1 - {r}^{2} } = 15 \: \: \: \: \: \: \: \: \: \: \: ...............(2) \\ \\ \\ (1) \div (2)... \\ \\ \frac{1 - {r}^{2} }{1 - r} = \frac{1}{3} \\ \\ 3 - 3 {r}^{2} = 1 - r \\ \\ 3 {r}^{2} - r - 2 = 0 \\ \\ 3 {r}^{2} - 3r + 2r - 2 = 0 \\ \\ 3r(r - 1) + 2(r - 1) = 0 \\ \\ r = - \frac{2}{3} \: \: \: \: and \: \: \: r = 1$