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