# Thread: Sum of the first 50 terms

1. ## Sum of the first 50 terms

I recently was given a very old Scottish Qualifications Higher paper from 1987. I could do almost every question easily, but when it came to question 10, I was entirely stumped. I have never been taught how to do it (and therefore don't really know what the problem is called so to speak), but here it is anyway. I would be happy if you could shed some light on this.

10. The sum Sn of the first n terms of an arithmetic series is given by
$\displaystyle S_n = 4n - n^2$

a) Find i) $\displaystyle u_1$ and $\displaystyle u_2$, the first 2 terms
ii) the common difference

b) Calculate the sum of the first 50 terms of the series
$\displaystyle u_1 + u_3 + u_5 + \ldots$

2. Originally Posted by scotchegg
I recently was given a very old Scottish Qualifications Higher paper from 1987. I could do almost every question easily, but when it came to question 10, I was entirely stumped. I have never been taught how to do it (and therefore don't really know what the problem is called so to speak), but here it is anyway. I would be happy if you could shed some light on this.

10. The sum Sn of the first n terms of an arithmetic series is given by
$\displaystyle S_n = 4n - n^2$

a) Find i) $\displaystyle u_1$ and $\displaystyle u_2$, the first 2 terms
ii) the common difference

b) Calculate the sum of the first 50 terms of the series
$\displaystyle u_1 + u_3 + u_5 + \ldots$
So we know that

$\displaystyle S_1=4(1)-1^2=3=u_1 \implies u_1=3$

and

$\displaystyle S_2=4(2)-2^2=4 =u_1+u_2 \implies u_2=1$

So now we know that the sequence is gives by

$\displaystyle u_n=5-2n$

but we are asked to sum only the first fifty odd terms to get this new sequence we use $\displaystyle n=2k-1, k=1,2,3,...$

$\displaystyle u_k=5-2(2k-1)=7-4k$

Can you finish from here?