Sum of n terms: $5 + 9 + 13 + 17 + 21 + 25 + . . . . .$

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Solution

Given terms are in $A P$ with First term $a = 5$ and $d = 9 - 5 = 4$
So sum of n terms is $\frac{n}{2} [2 a + (n - 1) d] \frac{n}{2} [2 (5) + (n - 1) 4] \frac{n}{2} [10 + 4 n - 4] \frac{n}{2} [4 n - 6]$

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