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How many terms of the AP 3, 7, 11, 15, … will make the sum 406? (a) 10 (b) 12 (c) 14 (d) 20

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How many terms of the AP 3, 7, 11, 15, … will make the sum 406? 

(a) 10 (b) 12 (c) 14 (d) 20 

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Here, a = 3 and d = (7-3) = 4

Let the sum of n terms be 406. 

Then, we have:

Hence, 14 terms will make the sum 406.

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