The sum of the first $n$ terms of the sequence $a_{n}$ is defined by the formula $S_{n} = 2 n^{2} + 3 n$ . Prove that it is an arithmetic progression.

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Solution

$S_{n} = 2 n^{2} + 3 n$ is the sum of $n$ terms sequence $a_{n}$
for $n = 1$ $a_{1} = 2 {(1)}^{2} + 3 (1)$
$a_{1} = 5$
For $n = 2 S_{2} = a_{1} + a_{2} = 2 {(2)}^{2} + 3. (2) 5 + a_{1} + d = 8 + 6 d = 14 - 10 = 4$
So, if it is $A P$ then $a, a + d, a + 2 d$
$= 2 {(3)}^{2} + 3. (3) 3 a + 3 d = 2 \times (9) + 9 = 27 a + d = 9 a s a + d = 9$
Hence proved.

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