In an AP, the sum of first n terms is ( 3n²/2) + ( 13n/2), Find it's 15th term.
Answers
EXPLANATION.
Sum of nth terms : 3n²/2 + 13n/2.
As we know that,
We can write equation as,
⇒ Sₙ = (3n² + 13n)/2.
As we know that,
Formula of : Tₙ = Sₙ - Sₙ₋₁.
Put the value of n = n - 1 in equation, we get.
⇒ Sₙ₋₁ = [3(n - 1)² + 13(n - 1)]/2.
⇒ Sₙ₋₁ = [3(n² + 1 - 2n) + 13n - 13]/2.
⇒ Sₙ₋₁ = [3n² + 3 - 6n + 13n - 13]/2.
⇒ Sₙ₋₁ = [3n² + 7n - 10]/2.
We can write as,
⇒ Tₙ = Sₙ - Sₙ₋₁.
⇒ Tₙ = [(3n² + 13n)/2] - [(3n² + 7n - 10)/2].
⇒ Tₙ = [(3n² + 13n - 3n² - 7n + 10)/2].
⇒ Tₙ = [(6n + 10)/2].
⇒ Tₙ = [2(3n + 5)/2].
⇒ Tₙ = 3n + 5.
Algebraic Expression = 3n + 5.
Put the value of n = 1 in equation, we get.
⇒ T₁ = 3(1) + 5 = 8.
Put the value of n = 2 in equation, we get.
⇒ T₂ = 3(2) + 5 = 11.
Put the value of n = 3 in equation, we get.
⇒ T₃ = 3(3) + 5 = 14.
Series = 8, 11, 14, . . . . .
First term = a = 8.
Common difference = d = b - a = 11 - 8 = 3.
As we know that,
General term of an ap.
⇒ Tₙ = a + (n - 1)d.
To find : 15th term.
⇒ T₁₅ = a + (15 - 1)d.
⇒ T₁₅ = a + 14d.
Put the values in the equation, we get.
⇒ T₁₅ = 8 + 14(3).
⇒ T₁₅ = 8 + 42.
⇒ T₁₅ = 50.
15th term of an ap = 50.