The first term of an AP is \( -5 \) and the last term is 45 . If the sum of the terms of the AP is 120 , then find the number of terms and the common difference.
Given:
The first term of an A.P. is $-5$, the last term is 45 and the sum is 120.
To do:
We have to find the number of terms and the common difference of the A.P.
Solution:
Let the number of terms of the given A.P. be $n$, first term be $a$ and the common differnce be $d$.
First term $a=-5$
Last term $l= 45$
Sum of all the terms $S_{n} =120$
We know that,
Sum of the $n$ terms$ S_{n} =\frac{n}{2}( a+l)$
$\Rightarrow 120=\frac{n}{2}( -5+45)$
$\Rightarrow 120=n(20)$
$\Rightarrow n=\frac{120}{20} =6$
Also,
$l=a+( n-1) d$
Therefore,
On subtituting the values of $a$, $l$ and $n$, we get,
$45=-5+( 6-1) d$
$\Rightarrow 5d=45+5=50$
$\Rightarrow d=\frac{50}{5}$
$=10$
Hence, the number of terms is $6$ and the common difference of the given A.P. is $10$.
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