Question

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.

Answer 1

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

Answer 2

Have you finished part ii ?

Answer 3

Yes

Answer 4

good, then simply take the ratio of second term and first term :
\[\dfrac{a+4d}{a}\]

plugin \(d = \dfrac{3}{8}a\) and simplify

Answer 5

Ah OK thanks.

Answer 6

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

Answer 7

notice that \(a\) cancels out

Answer 8

so I get 1.5a ?

Answer 9

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

Answer 10

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

Answer 11

Yes, we can factor out \(a\) and cancel it with the \(a\) in the bottom :

\[\text{common ratio}=\dfrac{a+4d}{a} = \dfrac{a+4*\dfrac{3}{8}a}{a}=\dfrac{a\left(1+4*\dfrac{3}{8}\right) }{a}=1+\dfrac{3}{2}=?\]