Is there an easier way to do this question about series?

[Banker]

Hey guys, the question is 6.b. in the picture : http://imgur.com/FaKUMUZ
Here is what I did to solve it : http://imgur.com/YrIvbTO
I made these two simultaneous equations. 1875 comes from the fact that U1 + U2 = 1500 and U3 + U4 = 375. Then S4 must equal 1500+ 375(1875).
I then found a formula for 'a' and substituted it back into S2, which I further simplified into a polynomial of degree 3 and I got the solutions r = 1, .5 & -.5.
r can't be 1 as then the whole series would stay at whatever 'a' is, so then I concluded that it must be .5 and -.5.
I then substituted .5 and -.5 into one of the equations and solved for 'a' which is 1000 & 3000 respectively.
That is the actual answer, but it seemed a bit long-winded for 4 marks.
Is there an easier way to do it? Any help would be appreciated.

 

[Samy_A]

Hey guys, the question is 6.b. in the picture : http://imgur.com/FaKUMUZ
Here is what I did to solve it : http://imgur.com/YrIvbTO
I made these two simultaneous equations. 1875 comes from the fact that U1 + U2 = 1500 and U3 + U4 = 375. Then S4 must equal 1500+ 375(1875).
I then found a formula for 'a' and substituted it back into S2, which I further simplified into a polynomial of degree 3 and I got the solutions r = 1, .5 & -.5.
r can't be 1 as then the whole series would stay at whatever 'a' is, so then I concluded that it must be .5 and -.5.
I then substituted .5 and -.5 into one of the equations and solved for 'a' which is 1000 & 3000 respectively.
That is the actual answer, but it seemed a bit long-winded for 4 marks.
Is there an easier way to do it? Any help would be appreciated.

Set $a_n=ar^n$.
You know that $a_0+a_1=1500$, $a_2+a_3=375$.
Replace the $a_i$'s by their values expressed in $a,r$ in these two equations.
They then easily let you find $r$, and then $a$.

 

[Banker]

Set $a_n=ar^n$.
You know that $a_0+a_1=1500$, $a_2+a_3=375$.
Replace the $a_i$'s by their values expressed in $a,r$ in these two equations.
They then easily let you find $r$, and then $a$.

Yep, that worked and was much easier than my way, thanks!