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CAT 2022 Slot 2 — QA Question 5

Arithmetic ProgressionEasy

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

Answer & solution

  • A

    415

  • B

    404

  • 455

  • D

    442

Solution

Easy

Find a closed form for AnA_n (sum of the first nn terms of the AP), then sum AnA_n from 11 to 2525 using the standard formulas ∑n2=n(n+1)(2n+1)6\sum n^2=\tfrac{n(n+1)(2n+1)}{6} and ∑n=n(n+1)2\sum n=\tfrac{n(n+1)}{2}. Keep the factor 2525 floating so it cancels neatly at the end.

1

Closed form for AnA_n. First term a=3a=3, common difference d=4d=4:

An=n2[2⋅3+(n−1)⋅4]=n2(4n+2)=2n2+n.A_n=\frac{n}{2}\big[2\cdot3+(n-1)\cdot4\big]=\frac{n}{2}(4n+2)=2n^2+n.
2

Sum it for n=1n=1 to 2525:

125∑n=125(2n2+n)=125[2⋅25⋅26⋅516+25⋅262].\frac{1}{25}\sum_{n=1}^{25}(2n^2+n)=\frac{1}{25}\Big[2\cdot\frac{25\cdot26\cdot51}{6}+\frac{25\cdot26}{2}\Big].
3

Factor out 2525 and cancel:

=125[25⋅26⋅17+25⋅13]=26⋅17+13=442+13=455.\begin{aligned} &=\frac{1}{25}\big[25\cdot26\cdot17+25\cdot13\big]\\ &=26\cdot17+13=442+13=455. \end{aligned}

125∑n=125An=455\dfrac{1}{25}\displaystyle\sum_{n=1}^{25}A_n=\mathbf{455} — option (c).

CAT 2022 Slot 2 QA Q5: Consider the arithmetic progressions 3, 7, 11, ... and let An dentoe the sum of the first n terms of this prog — Solution | TheCATExam