# Thread: series and sequences question

1. ## series and sequences question

The sum of the first two terms of an arithmetic series is 47.
The thirtieth term of this series is -62. Find the first term of the series and the common difference.
I'm stuck on the first bit. I'm not quite sure how to work out which two terms equal the sum of 47....thanks.

2. Hello,
Originally Posted by girlpower1991
The sum of the first two terms of an arithmetic series is 47.
The thirtieth term of this series is -62. Find the first term of the series and the common difference.
I'm stuck on the first bit. I'm not quite sure how to work out which two terms equal the sum of 47....thanks.
Let $\displaystyle \delta$ be the common difference and $\displaystyle a_1, ~a_2,~\text{etc}$ the terms of the series.

$\displaystyle a_2=a_1+ \delta$

But we know that $\displaystyle a_1+a_2=47$

Thus $\displaystyle 2a_1+ \delta=47$

The thirtieth term is $\displaystyle -62=a_{30}=a_1+29 \delta$

So we have the system :

$\displaystyle \left\{\begin{array}{ll} 2a_1+\delta=47 \quad (1) \\ a_1+29 \delta=-62 \quad (2) \end{array} \right.$

$\displaystyle 2*(2)-(1) ~:~ 2(a_1+29 \delta)-(2a_1+\delta)=2*(-62)-47$

The $\displaystyle a_1$ simplify and we get :
$\displaystyle 58 \delta- \delta=-124-47$

$\displaystyle 57 \delta=-171$

$\displaystyle \boxed{\delta=-3}$