Question

Sum of first n terms of an arithmetic sequence is 3n^2-6n.calculate first term and common difference

Answer 1

Solution:-

$\to \bf \: s_n = 3 {n}^{2} + 6 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (given)$

Sum of nth term is :-

$\to \bf \: s_n = \frac{n}{2} (2a + (n - 1)d)$

Now , we get

$\to \bf \: \frac{n}{2} (2a + (n - 1)d) = 3 {n}^{2} + 6n$

$\to \bf \: \frac{n}{2} (2a + (n - 1)d )= n(3n + 6)$

$\to \bf \: \frac{ \not \: n}{2} (2a + (n - 1)d) = \not n(3n + 6)$

$\to \bf \: 2a + (n - 1)d = 2(3n + 6)$

$\to \bf \: 2a + (n - 1)d = 6n + 12$

$\to \bf \: 2a + (n - 1)d = 6n + 18 - 6$

$\to \bf \: 2a + (n - 1)d = 6(n - 1) + 18$

$\to \bf \: 2a +d =18 + 6$

Comparing both side , we get

=> d = 6

=> 2a = 18

=> a = 9

Also find nth ferm , we get

$\bf \: a_n = a + (n - 1)d$

=> 9 + ( n - 1 ) × 6

=> 9 + 6n - 6

=> 3 + 6n

nth term = 3 + 6n