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2015-11-09T19:15:41+05:30

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Sum of first n terms of the AP = n² + 8n
Sum of first 1 term = first term, a = 1² + 8×1 = 9
Sum of first 12 terms = 12² + 8×12 = 240

But we also know that sum of first n terms = (n/2)[a + an]
(an = nth term)

So sum of first 12 terms = (12/2)[9 + a₁₂]
⇒ 240 = 6×(9 + a₁₂)
⇒ 9 + a₁₂ = 240/6 = 40
⇒ a₁₂ = 40- 9 
⇒ a₁₂ = 31

12th term is 31.
2 5 2
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2015-11-09T19:17:58+05:30

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Since ,

Sn = n² + 8n

Putting n = 1

S1 = 1 + 8 = 9

so first term is 9 i.e. a

Now putting n = 2

S2 = 4 + 8×2
S2 = 4 + 16 = 20

Now , s2-s1 = a2

a2= 20 - 9 = 11
so , d is a2-a1 = 2
so 12th term will be :-

a12 = a + 11d

a12 = 9 + 11×2

a12 = 9 + 22 = 31
4 3 4
hmm... my teacher told that s2-s1 = d
Anyway, it's upto you... I am not going to argue. I gave a solution, it's upto you what you want to do.
ohkk.. bhaiya m confirm krta hu :) ,
umm.. u r ryt bhaiya s2-s1 = a2
(a1+a2) - a1 = a2