Find the sum of n terms of an A.P. whose nth terms is given by an = 5 − 6n.
Here, we are given an A.P., whose $n^{\text {th }}$ term is given by the following expression, $a_{n}=5-6 n$
So, here we can find the sum of the $n$ terms of the given A.P., using the formula, $S_{n}=\left(\frac{n}{2}\right)(a+l)$
Where, a = the first term
l = the last term
So, for the given A.P,
The first term (a) will be calculated using in the given equation for nth term of A.P.
$a=5-6(1)$
$=5-6$
$=-1$
Now, the last term (l) or the nth term is given
$a_{e}=5-6 n$
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,
$S_{n}=\left(\frac{n}{2}\right)[(-1)+5-6 n]$
$=\left(\frac{n}{2}\right)[4-6 n]$
$=\left(\frac{n}{2}\right)(2)[2-3 n]$
$=(n)(2-3 n)$
Therefore, the sum of the $n$ terms of the given A.P. is $(n)(2-3 n)$.
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