Question

The sum of the first n terms of an AP is given by Sn= (3n²-n). Find its
(CBSE 2005C
(i) nth term, (ii) first term and (iii) common difference.
15n² 3n​

Answer 1

$\mathfrak{\huge{\red{SOLUTION}}}$

given sum of n terms = 3n² + 2n

let first term be a₁, second term be a₂ and so on.....

if n = 1

sum = a₁ (since there is no sum when considering only 1st term)

therefore, a₁ = 3(1)² + 2(1) = 3 + 2 = 5

then, if n = 2

S₂ = a₁ + a₂ = 3(2)² + 2(2) = 12 + 4 = 16

since a₁ = 5

∴ a₂ = 16 - 5 = 11

similarly, if n = 3

S₃ = 3(3)² + 2(3) = 27 + 6 = 33

∴ a₃ = 33 - (a₁ + a₂) = 33 - 16 = 17

∴ we gain that common difference = d = 17 - 11 = 11- 5 = 6

∴ let nth term be an

an = a₁ + (n-1)d

( where a₁ = 5 ; d = 6 and n = n)

∴ an = 5 + (n-1)6

= 5 + 6n - 6

= 6n - 1

This is your answer.

Hope it helps you.