Question

$15th$ term of A.P.$\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3},..$ is :

• A. $\frac{1}{3}$
• B. $6$
• C. $5$
• D. $119$

Answer 1

The correct option is A $\frac{1}{3}$

Given series $\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3}$ is an A.P as

$A.P=2b=a+c$

$2\left(\frac{1}{3}\right)=\frac{1}{3}+\frac{1}{3}$

$\frac{2}{3}=\frac{2}{3}$

$d=b-a$......(where d=common difference, and a =first term and b=second term)

$d=\frac{1}{3}-\frac{1}{3}$

$d=0$

Use the formula to find the 15th term;

$a_n=a+(n-1)d$

$a^{15}=\frac{1}{3}+(15-1)\left(0\right)$

$a^{15}=\frac{1}{3}+0$

$a^{15}=\frac{1}{3}$

So, the 15th term of the given A.P series is $\frac{1}{3}$.