Olympiad Maths Prep

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Problem 85

AMC 10/12, early questions
Algebra Difficulty 3.2 Find the answer

In the arithmetic sequence {an}\{a_n\}, a2+a7=23a_2+a_7=-23, a3+a8=29a_3+a_8=-29.

(1)(1) Find the general formula for the sequence {an}\{a_n\}.

(2)(2) Let the sum of the first nn terms of the sequence {an+1}\{a_n+1\} be SnS_n, find SnS_n.

Official solution

Solution:

(1)(1) Let the common difference of the arithmetic sequence {an}\{a_n\} be dd, then a3+a8(a2+a7)=2d=6a_3+a_8-(a_2+a_7)=2d=-6, d=3\therefore d=-3,

a2+a7=2a1+7d=23\therefore a_2+a_7=2a_1+7d=-23, solving this gives a1=1a_1=-1, \therefore the general formula for the sequence {an}\{a_n\} is an=3n+2a_n=-3n+2;

(2)(2) Sn=n×(1)+n(n1)2×(3)=32n2+32nS_n = n \times (-1) + \frac{n(n-1)}{2} \times (-3) = -\frac{3}{2}n^2 + \frac{3}{2}n.

Therefore, the answers are:

(1)(1) The general formula for the sequence is an=3n+2\boxed{a_n=-3n+2}.

(2)(2) The sum of the first nn terms of the sequence {an+1}\{a_n+1\} is Sn=32n2+32n\boxed{S_n = -\frac{3}{2}n^2 + \frac{3}{2}n}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.