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Arithmetic Progressions

Question 4:

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

Answer:

Let there be n terms of this A.P.

For this A.P., a = 9

d = a2a1 = 17 − 9 = 8

636 = n [9 + 4n − 4]

636 = n (4n + 5)

4n2

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