Question

In a H.P. $T_5=\frac{1}{12}$ and $T_{11}$, find $T_{25}$

Answer 1

It is given that $T_5=\frac{1}{12}$ and $T_{11}=\frac{1}{15}$, therefore, the reciprocals are the terms in A.P that are:

$T_5=12$ and $T_{11}=15$

We know that the formula for the $n$th term of an A.P is $T_n=a+(n-1)d$, where $a$ is the first term, $d$ is the common difference.

Thus,

$a+4d=\mathrm{12........}\left(1\right)$

$a+10d=\mathrm{15........}\left(2\right)$

Subtract equation 1 from equation 2 as follows:

$(a-a)+(10d-4d)=15-12\Rightarrow 6d=3\Rightarrow d=\frac{3}{6}=\frac{1}{2}$

Substitute the value of $d$ in equation 1:

$a+4d=12\Rightarrow a+(4\times \frac{1}{2})=12\Rightarrow a+2=12\Rightarrow a=12-2=10$

Now, the first term $a=10$ and the common difference $d=\frac{1}{2}$, therefore,

$T_{25}=10+(25-1)\frac{1}{2}=10+(24\times \frac{1}{2})=10+12=22$

Thus, the $25$th term of H.P is $T_{25}=\frac{1}{22}$.