Question

The sum of first two terms of an infinite G.P. is $5$ and each term is three times the sum of following terms, then the 4th term of the series is $\_\_\_\_.$

• A. $1$
• B. $\frac{1}{4}$
• C. $\frac{1}{16}$
• D. $\frac{1}{8}$

Answer 1

Answer: (C)

$\frac{1}{16}$

$a+ar=5$

each term $=3\left(sumoffollowingterms\right)$

${ar}^{n-1}=3({ar}^n+{ar}^{n+1}+....){ar}^{n-1}=3r^n(a+ar+{ar}^2+....){ar}^{n-1}=\frac{{3ar}^n}{1-r}\frac{3r}{1-r}=13r=1-rr=\frac{1}{4}a+\frac{a}{4}=5=>a=4$

Fourth term of G.P $={ar}^3=\frac{4}{4^3}=\frac{1}{16}$