Question

The ${26}^{th}$, ${11}^{th}$ and last term of an A.P. are $0,3and\frac{-1}{5}$ respectively. Find the common difference and the number of terms of the A.P

Answer 1

$a_{26}$

$a_n=\frac{-1}{5}$

$a_{11}=3$

$a+25d=0$ ...(1)

$a+10d=3$ ...(2)

$d=\frac{-1}{5}\Rightarrow$ from 1 and 2

$a+25d=0$

$a+25\left(\frac{-1}{5}\right)=0$

$a\Rightarrow a_n=\frac{-1}{5}$

$5+(n-1){\left(\frac{-1}{5}\right)}^2\frac{-1}{5}$

$n-1=126$

$\Rightarrow n=127$