Question

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

Answer 1

It is given that, $a_{26}=0,a_{11}=3,a_n=-\frac{1}{5}$.
$$\begin{gathered} \Rightarrow a+25d=0.....\left(i\right) \\ \Rightarrow a+10d=3.....\left(ii\right) \\ \mathrm{Subtracting}\left(ii\right)\mathrm{from}\left(i\right). \\ 15d=-3 \\ \Rightarrow d=-\frac{1}{5} \\ \Rightarrow a=-25d=5 \\ Now,a_n=a+\left(n-1\right)d \\ \Rightarrow -\frac{1}{5}=5+\left(n-1\right)\left(-\frac{1}{5}\right) \\ \Rightarrow -\frac{1}{5}-5=\left(n-1\right)\left(-\frac{1}{5}\right) \\ \Rightarrow n=27 \end{gathered}$$