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How Many Terms of the Ap. 9, 17, 25 … Must Be Taken to Give a Sum of 636? - Mathematics

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How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

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Solution

Let there be n terms of this A.P.

For this A.P., a = 9

d = a2 − a1 = 17 − 9 = 8

`S_n = n/2[2a+(n-1)d]`

`636 = n/2[2xxa+(n-1)8]`

`636 = n/2[18+(n-1)8]`

636 = [9 + 4n − 4]

636 = (4n + 5)

4n2 + 5n − 636 = 0

4n2 + 53n − 48n − 636 = 0

(4n + 53) − 12 (4n + 53) = 0

(4n + 53) (n − 12) = 0

Either 4+ 53 = 0 or n − 12 = 0

`n = (-53)/4` or n = 2

n cannot be `(-53)/4`

As the number of terms can neither be negative nor fractional, therefore, n = 12 only.

Concept: Sum of First n Terms of an AP
  Is there an error in this question or solution?

APPEARS IN

NCERT Class 10 Maths
Chapter 5 Arithmetic Progressions
Exercise 5.3 | Q 4 | Page 113
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