Question

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

Answer 1

$a_{26}=0a_{11}=3$

$a+25d=0$

$a+10d=3$

$15d=3$

$d=-\frac{1}{5},a=+5$

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

$a+(n-1)d=-\frac{1}{5}$

$5+(n-1)(-\frac{1}{5})=-\frac{1}{5}$

$-n+26=-1$

$n=27$