Question

Find the fourth term from the end in an A.P.-11, -8,-5,.....,49.

Answer 1

$a=-11$,

$d=3$

$a_n=49$

$a_n=a+(n-1)d$

$49=-11+(n-1)\times 3$

$\frac{60}{3}=n-1$

$n-1=20$

$n={21}^{st}$ term

Fourth term from the end

$={18}^{th}$ term

$a_{18}=a+(18-1)d$

$=-11+17\times 3$

$=-11+51$

$a_{18}=40$.