- #1

SolCon

- 33

- 1

Alright, I'm having a problem in this question which has 3 parts.

The question is: An arithmetic progression (ap) has first term a and common difference d.

Part 1 of it says to write down expressions for 2nd and 6th terms in terms of a and d.

This is no problem as it is is simply: a (1st term) and a+5d (2nd term).

Part 2 says: The first, second, and sixth terms of this ap are also the first three terms of a geometric progression (gp). Prove that d=3a.

I've got this question done also but not before having seen the initial equation which was:

(a+d)/a = (a+5d)/a+d

This would give us d=3a. However, I don't know why it is written like this and so have actually simply memorized the equation format as "[Small term]/a = [Big term]/[Small term]" . E.g, in another similar question, we are required to express the 5th and 15th term in terms of a and d. This will give us a+4d and a+14d.

So, using the same format:

[Small term]/a = [Big term]/[Small term]

(a+4d)/a = (a+14d)/a+4d.

I would just really appreciate an explanation of this, as I haven't been able to find one.

As for the third part, it says: Given that a=2, find the sum of the first 15 terms of

**each**progression.

Done this but don't have the answers to match (and I could have done it wrong :uhh: ).

The Ap sum is coming as 345. We'll use d=3a (getting d=6) and have the following:

S.n= n/2[2(a)+(n-1)d]

or

S.15= 15/2[2(2)+(15-1)6]

For Gp sum, I'm confused. The gp sum uses infinity formula (s.inf=a/1-r). The common ration (r) we are getting is 4. In part 2 it said that the 1st, 2nd and 6th ap terms are equal to first 3 terms of gp. As a=2 and d=6, 2nd term would be 6+2=8. If we multiply 2 by 4, we get 8, so 'r' is 4.

Then:

s.inf=a/1-r

of

s.inf=2/1-4 which comes as -2/3 which doesn't seem right.

In the other GP formula: Sn=a(1-r^n)/1-r.

We get: S.15=2(1-4^15)/1-4

which is 715,827,882.

Is the ap sum correct? And which of the gp are correct (if any at all)?

Thanks for your help.