How many terms of the A.P. 9,17,25,………….. must be taken to give a sum of 596.
Answers
Answer:
How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?
Solution:
Sum of the first n terms of an AP is given by Sₙ = n/2 [2a + (n - 1) d] or Sₙ = n/2 [a + l]
Here, a is the first term, d is the common difference and n is the number of terms and l is the last term.
Given,
First term, a = 9
Common difference, d = 17 - 9 = 8
Sum up to nth terms, Sₙ = 596
We know that sum of n terms of an AP
Sₙ = n/2 [2a + (n - 1) d]
636 = n/2 [2 × 9 + (n - 1) 8]
636 = n/2 [18 + 8n - 8]
636 = n/2 [10 + 8n]
636 = n[5 + 4n]
636 = 5n + 4n²
4n² + 5n - 596 = 0
4n² + 53n - 48n - 596 = 0
n (4n + 53) - 12 (4n + 53) = 0
(4n + 53)(n - 12) = 0
Either 4n + 53 = 0 or n - 12 = 0
n = - 53/4 or n =12
n cannot be -53/4 because the number of terms can neither be negative nor fractional, therefore, n = 12