The $11^{ ext {th }}$ term from the end of the A．P．： $10,7,4, \ldots$. ， $-62$ is ：

Question

The $11^{	ext {th }}$ term from the end of the A．P．： $10,7,4, \ldots$. ， $-62$ is ：

Professor PC Thomas Classes · Answer

(c) : Given A.P. in the reverse order, is $-62, \ldots 4,7,10$. So, $a=-62, d=3, a_{n}=10$\ As $a_{n}=a+(n-1) d$\ For $n=11, a_{11}=-62+(11-1) 	imes 3$\ $=-62+30=-32$\ $	herefore \quad$ The required term is -32 .\