## angela210793 5 years ago 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

1. watchmath

$$a_7=a_1r^6$$ 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?

2. angela210793

yea tht's the question

3. angela210793

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

4. watchmath

oh r is the ratio

5. watchmath

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.

6. angela210793

thanks ^_^ :)