Consider the arithmetic progressions 3, 7, 11, ... and let An dentoe the sum of the first n terms of this progression. Then the value of 125∑Ann=125
Explanation:
An = 3 + 7 + 11 + ...
⇒ An = n2[2 × 3 + (n - 1) × 4] = n2[4n + 2] = 2n2 + n
Now, 125∑n=125 [2n2 + n]
= 125[2 × 25×26×516 + 25×262]
= 125[25 × 26 × 17 + 25 × 13]
= 26 × 17 + 13 = 455
Hence, option (c).
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