Question

If the fourth term of a H.P is $\frac{1}{3}$ and 7th term is $\frac{1}{4}$, then 16th term is

• A. $\frac{1}{5}$
• B. $\frac{1}{6}$
• C. $\frac{1}{7}$
• D. $\frac{1}{8}$

Answer 1

Answer: (C)

$\frac{1}{7}$

4th term of a H.P is $\frac{1}{a+3d}=\frac{1}{3}$
$a+3d=3$
Similarly 7th term is$\frac{1}{a+6d}=\frac{1}{4}$
$a+6d=4$
Subtracting equations and solving we get $a=2,d=\frac{1}{3}$
$a+15d=2+15\times \frac{1}{3}=7$
So 16th term of H.P will be $\frac{1}{7}$