Question

The 26th, 11th 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

given, $a_{26}=0a_{11}=3a_n=\frac{-1}{5}{a}_{26}=0\Rightarrow a+25d=0\left(1\right)a_{11}=3\Rightarrow a+10d=3\left(2\right)$

From (1) and (2) by elimination
$d=\frac{-1}{5}$
substituting in (1)
$a+25d=0a+25\frac{-1}{5}=0a+(-5)=0a=5a_n=\frac{-1}{5}a+(n-1)d=\frac{-1}{5}5+(n-1)\frac{-1}{5}=\frac{-1}{5}(n-1)\frac{-1}{5}=\frac{-1-25}{5}n=27$