The 1st term of an arithmetic progression is a and the common difference is d, where d does not equal 0.
(i) Write down expressions, in terms of a and d, for the 5th term and the 15th term.
The 1st term, the 5th term and the 15th term of the arithmetic progression are the first three terms of a geometric progression.
(ii) Show that 3a=8d
I need help with (ii) please! I have the equations for (i) they are a + 4d ; a +14d

- anonymous

- jamiebookeater

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- DanJS

have you done anything yet?

- anonymous

I actually don't even know where to start so no.

- DanJS

first, you have to figure what an arithmetic progression is

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## More answers

- anonymous

well I know what that is already

- DanJS

Arithmetic sequence of numbers ---starts at a, and each consecutive term is 'd' away from the last term

- DanJS

a general form for the nth term is,
\[a _{n} = a _{1} + (n - 1)*d\]

- anonymous

yes, i have the answers for the first question already figured out, its the second one i would like help with please

- DanJS

the nth term = the first term + (n-1) d

- anonymous

ok

- DanJS

do you know what a geometric series is?

- DanJS

instead of an linear change from term to term, geometric series multiplies each by a common ratio r, every term / the one before it = r , and it is constant

- DanJS

The three terms from the last series to use are
a ; a + 4d ; a + 14d

- anonymous

Yes

- DanJS

the common ratio between terms is
r = (a + 4d) / a
and
r = (a + 14d)/(a+4d)

- DanJS

general geometric series to get the nth term
Nth Term = (FIrst Term) * r^n

- DanJS

exponential

- DanJS

really all you have to do is know that the ratio of consecutive terms is always the same value r

- DanJS

set those two ratios equal, and solve for 3a to show it equals 8d

- DanJS

\[\frac{ a+4d }{ a } = \frac{ a+14d }{ a+4d }\] = r

- DanJS

solve for 3a = ... should come to that 8d value

- anonymous

Ok when you say solve for 3a how would I do that exactly (sorry)

- DanJS

just start moving things around and expanding and isolating a to one side and d to the other side

- DanJS

cross multiply first, for example

- DanJS

(a+4d)*(a+4d) = a*(a+14d)

- DanJS

expand all those out from disributing

- DanJS

should be left with what they are looking for

- anonymous

Ok, will try that quickly

- anonymous

Hooray I got the right answer! Thank you very much for your help!

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