Question

The sum of the first two terms of an infinite GP is 6 and each term is 5 times the sum of the succeeding termsthen the second term GP is

Answer 1

Answer:

$\dfrac{6}{7}$.

Step-by-step explanation:

It is given that the sum of first two terms of an infinite GP is 6.

nth term of a GP is

$a_n=ar^{n-1}$

$a+ar=6$     ...(1)

Each term is 5 times the sum of the succeeding terms.

$a_n=5(a_{n+1}+a_{n+2}+...+\infty)$

$ar^{n-1}=5(ar^n+ar^{n+1}+...+\infty)$

$ar^{n-1}=5ar^n(1+r+r^2+...+\infty)$

Divide both sides by a.

$r^{n-1}=5r^n(\dfrac{1}{1-r})$    $[\text{Sum of infinite GP}=\dfrac{a}{1-r}]$

$\dfrac{r^n}{r}=\dfrac{5r^n}{1-r}$

$\dfrac{1}{r}=\dfrac{5}{1-r}$

$1-r=5r$

$1=6r$

$\dfrac{1}{6}=r$

The common ratio is 1/6.

Put r=1/6 in (1).

$a+a(\dfrac{1}{6})=6$

$6a+a=36$

$7a=36$

$a=\dfrac{36}{7}$

Second term $ar=\dfrac{36}{7}\times \dfrac{1}{6}=\dfrac{6}{7}$

Therefore, the second term is $\dfrac{6}{7}$.