Question

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

Answer 1

Let a,d and n be the first term,common difference and number of terms of an AP respectively.
Now,
$a_{26}=a+25d=0$ ...(1)
$a_{11}=a+10d=3$ ...(2)
on doing (1)-(2) , we get
$15d=-3$
$\Rightarrow d=\frac{-1}{5}$
On substituting the value of d in (1), we get
$a+25\times \frac{-1}{5}=0$
$\Rightarrow a=5$
Again, $a_n=a+(n-1)d=\frac{-1}{5}$
On substituting the values of a and d, we get
$5+(n-1)\times \frac{-1}{5}=\frac{-1}{5}$
$\Rightarrow (n-1)\times \frac{-1}{5}=\frac{-1}{5}-5=\frac{-26}{5}$
$\Rightarrow (n-1)=26$
$\Rightarrow n=27$
Therefore,
$d=\frac{-1}{5},n=27$