# The 8th term of an arithmetic progression is zero.Prove that its 38th term is triple of its 18th term.

$\begin{array}{l}\text{Let, }a\ \text{is first term}\ \&\ d\ \text{is common diffrence in AP }\\ \text{Given},\ T_8=0=a+7d\\ 3T_{18}=3\left(a+17d\right)\\ \ \ \ \ \ \ \ \ \ =3a+51d\\ \ \ \ \ \ \ \ \ \ =2\left(a+7d\right)+a+37d\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[a+7d=0\right]\\ \ \ \ \ \ \ \ \ \ =2\times0+a+37d\\ \ \ \ \ \ \ \ \ \ =T_{38}\\ \text{So, we proved that}\ 3T_{18}=T_{38}.\end{array}$

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