Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2025 June Paper 13 Q6

A Level / CIE / P1

CIE 9709 2025 June Paper 13 Paper · Question 6

题目

Problem

An arithmetic progression has first term aa and common difference 2. The NNth term is 55 and the sum of the first 3N3N terms is 5760. Find the values of NN and aa.

[6]
题目中文翻译

一个等差数列的首项为 aa,公差为 2。第 NN 项为 55,前 3N3N 项的和为 5760。求 NN 和 aa 的值。

[6]

解答

解法一

思路

展开

先用第 NN 项求出 aa 与 NN 的关系,再代入前 3N3N 项的和。项数只能是正整数。

答题过程

展开

The term condition gives a+2(N−1)=55a+2(N-1)=55, so a=57−2Na=57-2N. The sum condition gives

5760=3N2(2a+2(3N−1))=3N(N+56).5760=\frac{3N}{2}\bigl(2a+2(3N-1)\bigr) =3N(N+56).

Thus N2+56N−1920=(N−24)(N+80)=0N^2+56N-1920=(N-24)(N+80)=0. Since N>0N>0, N=24N=24 and a=57−48=9a=57-48=9.