1. ## Arithmetic progression

Hey guys..

I have a question here regarding arithmetic progressions.. English isn't my mother tongue and neither do I study Math in English, thus I might not use the correct mathematical expressions in the translation, but I hope it will be understandable enough.

A finite arithmetic progression has an even number of terms. The sum of the odd terms is 150, and the sum of the even terms is 174. Moreover, it's also given that the last term is the initial term plus 44.
Find the common difference (d) of the progression, and the number of terms it has.

2. Originally Posted by loui1410
A finite arithmetic progression has an even number of terms. The sum of the odd terms is 150, and the sum of the even terms is 174. Moreover, it's also given that the last term is the initial term plus 44.
Find the common difference (d) of the progression, and the number of terms it has.
Okay,
$a,a+d,a+2d,...,a+(n-1)d$ where the number of terms $n$ is even. The sum of the even terms is: $(a+d)+(a+3d)+...+(a+(n-1)d) = \underbrace{(a+a+...+a)}_{n/2} + d(1+3+...+(n-1))$. Now $1+3+...+(n-1)= (n/2)^2$. So the arithmetic sum is $\frac{na}{2}+\frac{n^2d}{4} = 150$.

The odd sums are $a+(a+2d)+...+(a+(n-2)d) = (a+a+...+a)+(2+4+...+(n-2))$. But $2+4+...+(n-2) = (1/4)n(n-2)$. So the arithemtic sum is $\frac{na}{2}+\frac{n(n-2)d}{4}=174$.

And finally $a+(n-1)d = a + 44 \implies (n-1)d = 44$.

Can you do it from here?