Question

If the $7^{th}$ term of a H.P. is $\frac{1}{10}$ and the ${12}^{th}$ term is $\frac{1}{25}$, then the ${20}^{th}$ term is :

• A. $\frac{1}{37}$
• B. $\frac{1}{41}$
• C. $\frac{1}{45}$
• D. $\frac{1}{49}$

Answer 1

The correct option is D $\frac{1}{49}$

Given that

$T_7=\frac{1}{10}\Rightarrow$

$n^{th}$ term of H.P $=\frac{1}{\left[a+(n-1)d\right]}$

$\frac{1}{10}=\frac{1}{a(7-1)d}$

$\Rightarrow \frac{1}{10}=\frac{1}{a+6d}\Rightarrow a+6d=1\_\_\_\_\_\_\_\_\left(i\right)$

Second

$\frac{1}{25}=\frac{1}{a(12-1)d}$

$\frac{1}{25}=\frac{1}{a+11d}\Rightarrow a+11d=25\_\_\_\_\_\_\_\left(ii\right)$

from eqh 1st and 2nd

$a+6d=10$

$\underset{–}{-a+11d=25}$ subtracting

$-5d=-15$

$d=3$

$d=3$ putiting eqn 1st

$a+6d=10$

$a+6\times 3=10$

$a=10-18$

$a=-8$

and Hence $T_{20}=\frac{1}{-8+(20-1)\times 3}$

$=\frac{1}{-8+19\times 3}$

$=\frac{1}{-8+57}$ $=\frac{1}{49}$

$T_{20}=\frac{1}{49}$

Option $D$ is correct Answer