Question

The sum of the first n terms of an AP is given by Sn = 3n2 – 4n. Determine the AP and the 12th
term.

Answer 1

an = sn - s(n-1)
sn = 3n2 - 4n.
s(n-1) = 3(n-1)^(2) - 4(n-1)
s(n-1) = 3(n2- 2n+1) -4n-4
= 3n2 - 6n + 3 - 4n - 4. (-6n + -4n = -10n)(-4+3=-1)

therefore sn - s(n-1) =

3n2 - 4n - { 3n2 - 10n - 1}
3n2 - 4n - 3n2 +10n + 1.

- 4n + 10n +1
= 6n + 1.
now applying value n = 0;1;2;3;....

A.P: 1 ; 7 ; 13 .............

Now in A.P ;
a= 1
d=6
a12=?
a12 = a + 11d
= 1 + 11*6
= 1+66
12th term = 67.
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