sum of first fifteen terms of AP is 570 and its 12th term is 62 .find sum of first 30 terms and sum of Nth term

Suppose a be the first term and d be the common difference of an A.P.
So, 12th term = t12 = a+12-1d = 62
a+11d = 62 ...(i)

Sn=n2[2a+(n-1)d], nth term of an AP, a= a + (n - 1) d
Sn=n2[2a+(n1)d]Sn=n2[2a+(n1)d]
Sum of first 15 terms = S15 = 1522a+15-1d = 570
1522a+14d = 570152×2a+7d = 570a+7d = 57015a+7d = 38 ...(ii)
Solving (i) and (ii) we get;
a+11d-a-7d = 62-384d = 24d = 244 = 6
So from (i) we have;
a+11×6 = 62a+66 = 62a = 62-66 = -4
So sum of 30 terms = 3022×-4+30-1×6 = 15×166 = 2490
And sum of nth term = n22a+n-1d

  • 0

sum of nth term u could give two ..equations according to the question first equation is:

1)n/2*(2a+(n-1)*d)

2)n/2*(a+a1) a1=nth term..

  • -1
What are you looking for?