The sum of first ‘n’ terms of an arithemetic progression is given by the formula
Sn
= 3n2 + n, then its 3rd term is
A. 14 B. 16
C. 22 D. 42
Answers
EXPLANATION.
⇒ Sₙ = 3n² + n.
As we know that,
⇒ Tₙ = Sₙ - Sₙ₋₁.
Using this formula in the equation, we get.
Put the value of n = n - 1 in the equation, we get.
⇒ Tₙ = 3n² + n - [3(n - 1)² + (n - 1)].
⇒ Tₙ = 3n² + n - [3(n² + 1 - 2n) + n - 1].
⇒ Tₙ = 3n² + n - [3n² + 3 - 6n + n - 1].
⇒ Tₙ = 3n² + n - [3n² - 5n + 2].
⇒ Tₙ = 3n² + n - 3n² + 5n - 2.
⇒ Tₙ = n + 5n - 2.
⇒ Tₙ = 6n - 2.
⇒ 6n - 2 = [Algebraic expression].
Put the value of n = 1 in the equation, we get.
⇒ 6(1) - 2.
⇒ 6 - 2 = 4.
Put the value of n = 2 in the equation, we get.
⇒ 6(2) - 2.
⇒ 12 - 2 = 10.
Put the value of n = 3 in the equation, we get.
⇒ 6(3) - 2.
⇒ 18 - 2 = 16.
Put the value of n = 4 in the equation, we get.
⇒ 6(4) - 2.
⇒ 24 - 2 = 22.
Series = 4, 10, 16, 22, . . . . .
First term = a = 4.
Common difference = d = b - a = c - b.
Common difference = d = 10 - 4 = 6.
As we know that,
General term of an A.P.
⇒ aₙ = a + (n - 1)d.
3rd term = a₃
⇒ a₃ = a + (3 - 1)d.
⇒ a₃ = a + 2d.
Put the value of a = 4 and d = 6 in the equation, we get.
⇒ a₃ = 4 + 2(6).
⇒ a₃ = 4 + 12.
⇒ a₃ = 16.
Option [B] is correct answer.
MORE INFORMATION.
Supposition of an A.P.
(1) = Three terms as : a - d, a, a + d.
(2) = Four terms as : a - 3d, a - d, a + d, a + 3d.
(3) = Five terms as : a - 2d, a - d, a, a + d, a + 2d.
14 jdjdirjdiosjeosjidjr7hfidhncidníogih9eiccodood