Question

I'll check the answer tomorrow after school :) Help please :)
There's a geometric progression....a7-a3=60 and a7-a5=48 what is S5? q>0

Answer 1

\(a_7=a_1r^6\) and \(a_3=a_1r^2\)
So
\(a_1r^6-a_1r^2=a_1r^2(r^4-1)=60\)
Using similar argument
\(a_1r^6-a_1r^4=a_1r^4(r^2-1)=48\)
It follows that
\(60/48=\frac{a_1r^2(r^4-1)}{a_1r^4(r^2-1)}=\frac{r^2+1}{r^2}\)
Cross multiply we have
\(60r^2=48r^2+48\)
\(12r^2=48\)
\(r^2=4\)
\(r=\pm 2\)
What is the question again? S5? the sum of the first 5 terms?

Answer 2

yea tht's the question

Answer 3

wht is r? is it y2/y1=r? cause we use q..... :)

Answer 4

oh r is the ratio

Answer 5

Ok so plug in \(r=\pm 2\) to \(a_1r^2(r^4-1)=60\) we have
\(a_1\cdot 4\cdot (16-1)=60a_1=60\) so \(a_1=1\)
Now knowing \(a_1=1\) and \(r=\pm 2\) you can find your S5.

Answer 6

thanks ^_^ :)