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CAT 2022 Slot 1 — QA Question 13

Arithmetic ProgressionEasy

For any natural number n, suppose the sum of the first n terms of an arithmetic progression is (n+ 2n2). If the nth term of the progression is divisible by 9, then the smallest possible value of n is

Answer & solution

  • A

    9

  • B

    8

  • C

    4

  • 7

Solution

Easy

The nthn^{\text{th}} term of an AP is the jump in the partial sums: Tn=Sn−Sn−1T_n=S_n-S_{n-1}. Get that formula, then walk nn up until the term is a multiple of 99.

1

Find the general term from Sn=n+2n2S_n=n+2n^2:

Tn=Sn−Sn−1=(n+2n2)−[(n−1)+2(n−1)2]=(n+2n2)−[2n2−3n+1]=4n−1\begin{aligned} T_n&=S_n-S_{n-1}\\ &=(n+2n^2)-\big[(n-1)+2(n-1)^2\big]\\ &=(n+2n^2)-\big[2n^2-3n+1\big]\\ &=4n-1 \end{aligned}
2

Require divisibility by 99.

4n−1≡0(mod9) ⇒ 4n≡1(mod9)4n-1\equiv 0 \pmod 9\ \Rightarrow\ 4n\equiv 1\pmod 9
3

Solve the congruence. The inverse of 44 mod 99 is 77 (since 4⋅7=28≡14\cdot7=28\equiv1), so

n≡7⋅1≡7(mod9)n\equiv 7\cdot 1\equiv 7\pmod 9

The smallest natural value is n=7n=7, and indeed T7=4(7)−1=27=9×3T_7=4(7)-1=27=9\times3.

smallest n=7 — option (d)\text{smallest }n=\mathbf{7}\ \text{— option (d)}

Just list Tn=4n−1: 3,7,11,15,19,23,27,…T_n=4n-1:\ 3,7,11,15,19,23,27,\dots The first multiple of 99 is 2727, at the 7th7^{\text{th}} term.

CAT 2022 Slot 1 QA Q13: For any natural number n, suppose the sum of the first n terms of an arithmetic progression is (n+ 2n 2 ). If — Solution | TheCATExam