1.与等差数列的性质结合例1 已知数列{an}为等差数列,若a1+a2+a3+a4=3,an-3+an-2+an-1+an=21,且Sn=48,求n的值. 分析先将第二个条件“倒序”,再将两式相加,目的是用等差数列的性质:am+an=ap+aq<=>m+n=p+q(其中m,n,p,q∈N+),从而减少计算量。解依题意,有 a1+a2+a3+a4=3 an+an-1+an-2+an-3=21将以上两式相加,得 1. Combining with the properties of arithmetic sequences. Example 1 Known series {an} is an arithmetic sequence, if a1+a2+a3+a4=3, an-3+an-2+an-1+an=21, and Sn = 48, find the value of n. Analysis first “reverse” the second condition, and then add the two formulas, the purpose is to use the nature of the arithmetic sequence: am + an = ap + aq <=> m + n = p+q (where m,n,p,q∈N+) reduces the amount of computation. According to the meaning of the solution, there is a1+a2+a3+a4=3 an+an-1+an-2+an-3=21. Add the above two formulas to obtain