Question

If the second term of gp is 2 and sum of its infinite term is 8 then gp is

Answer 1

Let first term be a and common ratio be r.

$given \: \: T2 = 2 \\ ar = 2 - - - (1)\\also \: \: given \\ S \infty = 8 \\ \frac{a}{1 - r} = 8 \\ a = 8 - 8r \\ substituting \: \: from \: \: (1) \\ \frac{2}{r} = 8 - 8r \\ 4 {r}^{2} - 4r + 1 = 0 \\ {(2r - 1)}^{2} = 0 \\ r = \frac{1}{2} \\ a = \frac{2}{r} = 4 \\ \\ therefore \: \: gp \: \: is \\ 4,2,1, \frac{1}{2} .......$

Answer 2

Step-by-step explanation:

Question:-

If the second term of gp is 2 and sum of its infinite term is 8 then gp is

The second term in G.P is 2 i.e.,

$\bf \red{a \times r ^ (n-1) = 2 }$

where 'n' is the second term(2)..

so we got ar = 2 ------ Equation (1)

Sum of infinite terms in G.P is 8 i.e.,

$\bf \red{ \frac{a}{1-r \: } = 8 -------} Equation(2)$

$\bf \: From \: equ.(1) \: we \: got \: \red{r = \frac{2}{r} } \: \:$

so substitute 'r' value in eq (2).

$\bf \: we \: got \: 'a' \: value \red{ = 4.}$