Question

45. If Sn, the sum of first n terms of an A.P. is given by Sn=3n²-4n find the nth
term.
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Answer 1

EXPLANATION.

Sum of nth terms of an Ap is = 3n² - 4n.

Tn terms of an Ap = Sn - S(n-1).

→ 3n² - 4n - [ 3(n-1)² - 4(n-1) ].

→ 3n² - 4n - [ 3 ( n² + 1 - 2n ) - 4n + 4 ].

→ 3n² - 4n - [ 3n² + 3 - 6n - 4n + 4 ].

→ 3n² - 4n - [ 3n² - 10n + 7 ].

→ 3n² - 4n - 3n² + 10n - 7.

→ 6n - 7.

Let n = 1 = 6(1) - 7 = -1

Let n = 2 = 6(2) - 7 = 5

Let n = 3 = 6(3) - 7 = 11

Let n = 4 = 6(4) - 7 = 17

Series are = -1,5,11,17,.......

First term = a = -1.

Common Difference = d = b - a = 5 - (-1) = 6.

Nth term of an Ap = a + ( n - 1)d.

An = -1 + ( n - 1) 6

An = -1 + 6n - 6

An = 6n - 7

Nth term = 6n - 7.

Answer 2

Step-by-step explanation:

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