The $ 26^{\text {th }}, 11^{\text {th }} $ and the last term of an AP are 0,3 and $ -\frac{1}{5} $, respectively. Find the common difference and the number of terms.
Given:
The 26th, 11th and last term of an A.P. are $0, 3$ and $-\frac{1}{5}$, respectively.
To do:
We have to find the common difference and the number of terms.
Solution:
Let the first term, common difference and the number of terms of the given A.P. be $a, d$ and $n$ respectively.
We know that,
nth term of an A.P. $a_n=a+(n-1)d$
Therefore,
$a_{26}=a+(26-1)d$
$0=a+25d$
$a=-25d$.....(i)
$a_{11}=a+(11-1)d$
$3=a+10d$
$3=-25d+10d$ (From (i))
$3=-15d$
$d=\frac{3}{-15}$
$d=\frac{-1}{5}$....(ii)
Last term $l=a+(n-1)d$
$-\frac{1}{5}=-25(\frac{-1}{5})+(n-1)\frac{-1}{5})$ (From (i) and (ii))
$1=-25+(n-1)$
$1+25+1=n$
$n=27$
The common difference and the number of terms are $\frac{-1}{5}$ and $27$ respectively.
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