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Arithmetic and geometric progressions: MI 2024 Promo Q1

Millennia Institute2024 PromoPromo Q1●○○ Foundational4 marks

What this question tests

Sequences & Series (AP).

Question

. The sum, SnS_n, of the first nn terms of the sequence u1u_1, u2u_2, u3u_3, … is given by Sn=n2+kn,S_n = n^2 + kn, where kk is a non-zero constant.

  1. Find unu_n in terms of nn and kk. [2]

  2. Hence determine if the sequence is an arithmetic progression. [2]

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QnSolution

Q1(i) [2]

Sn=n2+knS_n=n^2+kn Sn−1=(n−1)2+k(n−1)=n2−2n+1+kn−kun=Sn−Sn−1=n2+kn−(n2−2n+1+kn−k)=n2+kn−n2+2n−1−kn+k=2n−1+k\begin{aligned} S_{n-1}&=(n-1)^2+k(n-1)\\ &=n^2-2n+1+kn-k\\ u_n&=S_n-S_{n-1}\\ &=n^2+kn-(n^2-2n+1+kn-k)\\ &=n^2+kn-n^2+2n-1-kn+k\\ &=2n-1+k \end{aligned}

1(ii) [2] un−1=2(n−1)−1+k=2n−2−1+k=2n−3+kun−un−1=2n−1+k−(2n−3+k)=2n−1+k−2n+3−k=2\begin{aligned} u_{n-1}&=2(n-1)-1+k\\ &=2n-2-1+k\\ &=2n-3+k\\ u_n-u_{n-1}&=2n-1+k-(2n-3+k)\\ &=2n-1+k-2n+3-k\\ &=2 \end{aligned} Since un−un−1=2u_n-u_{n-1}=2 which is a constant independent of nn, the sequence is an arithmetic progression.

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