Answer:
[tex]\dfrac{6}{7}[/tex].
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
[tex]a_n=ar^{n-1}[/tex]
[tex]a+ar=6[/tex] ...(1)
Each term is 5 times the sum of the succeeding terms.
[tex]a_n=5(a_{n+1}+a_{n+2}+...+\infty)[/tex]
[tex]ar^{n-1}=5(ar^n+ar^{n+1}+...+\infty)[/tex]
[tex]ar^{n-1}=5ar^n(1+r+r^2+...+\infty)[/tex]
Divide both sides by a.
[tex]r^{n-1}=5r^n(\dfrac{1}{1-r})[/tex] [tex][\text{Sum of infinite GP}=\dfrac{a}{1-r}][/tex]
[tex]\dfrac{r^n}{r}=\dfrac{5r^n}{1-r}[/tex]
[tex]\dfrac{1}{r}=\dfrac{5}{1-r}[/tex]
[tex]1-r=5r[/tex]
[tex]1=6r[/tex]
[tex]\dfrac{1}{6}=r[/tex]
The common ratio is 1/6.
Put r=1/6 in (1).
[tex]a+a(\dfrac{1}{6})=6[/tex]
[tex]6a+a=36[/tex]
[tex]7a=36[/tex]
[tex]a=\dfrac{36}{7}[/tex]
Second term [tex]ar=\dfrac{36}{7}\times \dfrac{1}{6}=\dfrac{6}{7}[/tex]
Therefore, the second term is [tex]\dfrac{6}{7}[/tex].
