## anonymous one year ago I need some help with the last question of an exercise thats too long to type out so I will attach a picture. The answers for all the questions before are: (i) = a+4d; a+14d (iii) This is the one I need help with and I know that the answer is 2.4 but I can't get there.

1. anonymous

I am sorry I made a mistake the answer is 2.5. still need help though

2. ganeshie8

Have you finished part ii ?

3. anonymous

Yes

4. ganeshie8

good, then simply take the ratio of second term and first term : $\dfrac{a+4d}{a}$ plugin $$d = \dfrac{3}{8}a$$ and simplify

5. anonymous

Ah OK thanks.

6. ganeshie8

good, then simply take the ratio of second term and first term : $\text{common ratio}=\dfrac{a+4d}{a}$ plugin $$d = \dfrac{3}{8}a$$ and simplify

7. ganeshie8

notice that $$a$$ cancels out

8. anonymous

so I get 1.5a ?

9. ganeshie8

plugging $$d = \dfrac{3}{8}a$$ gives $\text{common ratio}=\dfrac{a+4d}{a} = \dfrac{a+4*\dfrac{3}{8}a}{a}=1+\dfrac{3}{2}=?$

10. anonymous

Oh ok! So the a's cross out to 1 which is added?

11. ganeshie8

Yes, we can factor out $$a$$ and cancel it with the $$a$$ in the bottom : $\text{common ratio}=\dfrac{a+4d}{a} = \dfrac{a+4*\dfrac{3}{8}a}{a}=\dfrac{a\left(1+4*\dfrac{3}{8}\right) }{a}=1+\dfrac{3}{2}=?$