Problem:
Find the value of a2+a4+a6+⋯+a98 if a1,a2,a3,… is an arithmetic progression with common difference 1, and a1+a2+a3+⋯+a98=137.
Solution:
The sum of an arithmetic progression is the product of the number of terms and the arithmetic average of the first and last terms. Therefore,
a1+a2+⋯+a98=982(a1+a98)=49(a1+a98)=137.
Similarly,
a2+a4+⋯+a98=492(a2+a98)=492(a1+1+a98)=(492(a1+a98))+(249)=(2137)+(249)=93.
OR
Separating the terms of odd and even subscripts, let S0=a1+a3+⋯+a97 and Se=a2+a4+⋯+a98. Then each of these series has 49 terms, and since an+1=an+1 for all n≥1, we have So=Se−49. Furthermore, Se+So=137. Substituting for So in the last equation, and solving, for Se, one finds that Se=93.
The problems on this page are the property of the MAA's American Mathematics Competitions