Answer to Question #237129 in Algebra for Mr. Lee

Question

How many terms of progression 3, 5, 7, 9 . . . must be taken

in order that their sum will be 2600?

Answer 1

Let us use the following formula for the sum of $n$ terms of arithmetic progression: $S_n=\frac{2a_1+d(n-1)}2n.$ In our case, $a_1=3,\ d=2,\ S_n=2600.$ It follows that $2600=\frac{6+2(n-1)}2n=(3+n-1)n=(n+2)n.$

Taking into account that $2600=52\cdot 50,$ we conclude that $n=50.$ Therefore, 50 terms must be taken.