If the sum of first term n of AP is ( $3 n^{2} - 4)$ then find of its first term and common different of term.

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Solution

$S_{n} = 3 n^{2} - 4$
$S_{1} = 3 (1)^{2} - 4 = - 1 = a$
$S_{2} = a_{1} + a_{2}$
$= 3 (2)^{2} - 4 = 8$
So $a + a + d = 8$
$\Rightarrow 2 a + d = 8$
$\Rightarrow d = 8 - 2 (- 1)$
$\Rightarrow d = 8 + 2 = 10$
So $a = - 1$ & $d = 10$ .

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If the sum of first term n of AP is (3n2−4) then find of its first term and common different of term.

Sn=3n2−4S1=3(1)2−4=−1=aS2=a1+a2=3(2)2−4=8So a+a+d=8⇒2a+d=8⇒d=8−2(−1)⇒d=8+2=10So a=−1 & d=10.

If the sum of first term n of AP is ( $3 n^{2} - 4)$ then find of its first term and common different of term.

$S_{n} = 3 n^{2} - 4$
$S_{1} = 3 (1)^{2} - 4 = - 1 = a$
$S_{2} = a_{1} + a_{2}$
$= 3 (2)^{2} - 4 = 8$
So $a + a + d = 8$
$\Rightarrow 2 a + d = 8$
$\Rightarrow d = 8 - 2 (- 1)$
$\Rightarrow d = 8 + 2 = 10$
So $a = - 1$ & $d = 10$ .