How many terms of the series 42+39+36+... makes the sum 315?

1.	How many terms of the series 42+39+36+... makes the sum 315?
Answer» Given series is S = 42+39+36+... Let first term of the sequence is a₁ = a = 42. Common difference of the sequence is d = a₂ - a₁ = 39 - 42 = -3. ∴ Sum of n terms of the sequence is S_n = n/2 [2a+(n-1)d] Given that sum = 315 ∴ n/2 [2 x 42 + (n-1) x -3] = 315 ⇒ 42n -3/2 n(n-1) = 315 ⇒ 84n - 3n² + 3n = 630 ⇒ 3n² - 87n + 630 = 0 ⇒ n² - 29n + 210 = 0 ⇒ n² - 14n - 15n + 210 = 0 ⇒ (n - 14) (n - 15) = 0 ⇒ n = 14 or n = 15 Hence, sum of 14^th terms of the series is 315. Also, sum of 15th terms is 315 (∵ a₁₅= a+14d = 0)

Answer»

Given series is S = 42+39+36+...

Let first term of the sequence is a₁ = a = 42.

Common difference of the sequence is d = a₂ - a₁ = 39 - 42 = -3.

∴ Sum of n terms of the sequence is S_n = n/2 [2a+(n-1)d]

Given that sum = 315

∴ n/2 [2 x 42 + (n-1) x -3] = 315

⇒ 42n -3/2 n(n-1) = 315

⇒ 84n - 3n² + 3n = 630

⇒ 3n² - 87n + 630 = 0

⇒ n² - 29n + 210 = 0

⇒ n² - 14n - 15n + 210 = 0

⇒ (n - 14) (n - 15) = 0

⇒ n = 14 or n = 15

Hence, sum of 14^th terms of the series is 315. Also, sum of 15th terms is 315 (∵ a₁₅= a+14d = 0)

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