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Sum of First N Terms of AP

CAT 2022 Slot 2 · Quantitative Ability · Medium · Algebra

This is a medium Quantitative Ability question from the CAT 2022 Slot 2 paper. It tests Algebra. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Consider the arithmetic progression 3,7,11,…3, 7, 11, \dots and let AnA_n denote the sum of the first nn terms of this progression. Then the value of 125∑n=125An\frac{1}{25} \sum_{n=1}^{25} A_n is

  1. A.

    404

  2. B.

    442

  3. C.

    455

  4. D.

    415

Answer

C

Explanation

AP has first term a=3a = 3, common difference d=4d = 4. An=n2[2(3)+(n−1)4]=n2[4n+2]=n(2n+1)=2n2+nA_n = \frac{n}{2} [2(3) + (n-1)4] = \frac{n}{2} [4n + 2] = n(2n + 1) = 2n^2 + n. We need 125∑n=125(2n2+n)=125(2∑n=125n2+∑n=125n)\frac{1}{25} \sum_{n=1}^{25} (2n^2 + n) = \frac{1}{25} \left( 2 \sum_{n=1}^{25} n^2 + \sum_{n=1}^{25} n \right). ∑n=125n=25×262=325\sum_{n=1}^{25} n = \frac{25 \times 26}{2} = 325. ∑n=125n2=25×26×516=5525\sum_{n=1}^{25} n^2 = \frac{25 \times 26 \times 51}{6} = 5525. Sum = 2(5525)+325=11050+325=113752(5525) + 325 = 11050 + 325 = 11375. Value = 1137525=455\frac{11375}{25} = 455.

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