Question

In a H.P. $T_4=\frac{1}{11}$ and $T_{14}=\frac{3}{23}$ then find $T_{19}$

Answer 1

It is given that $T_4=\frac{1}{11}$ and $T_{14}=\frac{3}{23}$, therefore, the reciprocals are the terms in A.P that are:

$T_4=11$ and $T_{14}=\frac{23}{3}$

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,

$a+3d=\mathrm{11........}\left(1\right)$

$a+13d=\frac{23}{3}.........\left(2\right)$

Subtract equation 1 from equation 2 as follows:

$(a-a)+(13d-3d)=\frac{23}{3}-11\Rightarrow 10d=\frac{23-33}{3}\Rightarrow 10d=-\frac{10}{3}\Rightarrow 30d=-10\Rightarrow d=-\frac{10}{30}=-\frac{1}{3}$

Substitute the value of $d$ in equation 1:

$a+3d=11\Rightarrow a+(3\times -\frac{1}{3})=11\Rightarrow a-1=11\Rightarrow a=11+1=12$

Now, the first term $a=12$ and the common difference $d=-\frac{1}{3}$, therefore,

$T_{19}=12+(19-1)(-\frac{1}{3})=12+(18\times -\frac{1}{3})=12-6=6$

Hence, the $19$th term of H.P is $T_7=\frac{1}{6}$.