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Please help with this question (1 Viewer)

scraggy321

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HI everyone,

'If in a geometric sequence the sum of the second and third terms is 20 and the sum of the fourth and fifth terms is 320, find the common ratio and the first term. Assume that the common ratio is positive'.
the answers are: r=4 and a=1
I've tried using simultaneous equations but it just doesnt come out for me, so id really appreciate it if somebody would outline the steps in answering this question.

P.S- just a hi from me to every1 here on the board as i am new.
 

Trev

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I got a different answer to what you stated (a=1, r=4). Though, I see no error in what I have done.
Let series be <sup>a</sup>/<sub>r</sub> + a + ar + ar<sup>2</sup> + ar<sup>3</sup>

2nd and 3rd term = 20
ar+a=20; a(r+1)=20 (1)

3rd and 4th term = 320
ar<sup>2</sup>+ar<sup>3</sup>=320; ar<sup>2</sup>(r+1)=320 (2)

a=20/(r+1)
(1) into (2) gives
[20/(r+1)]*r<sup>2</sup>(r+1)=320
20r<sup>2</sup>=320
r=+/-4; though question states positive only &there4; r=4.
Sub. r=4 into (1) gives a(4+1)=20; a=4.
I don't think the answers you have been given are correct?!
 

Trebla

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Let series be:
a + ar + ar²+ ar³ + ar^4......

.: 2nd term + 3rd term = ar + ar²
= ar(1 + r) = 20
.: 4th term + 5th term = ar³ + ar^4
= ar³(1 + r) = 320

ar³(1 + r) = 320 (1)
ar(1 + r) = 20 (2)

(1) ÷ (2):
r² = 16
.: r = 4 (since it was stated that r was positive)
Subs r = 4 into (2)
4a(1 + 4) = 20
20a = 20
.: a = 1

I don't know what you did Trev, but I ended up with a = 1, r = 4
 

Trebla

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Trev said:
I got a different answer to what you stated (a=1, r=4). Though, I see no error in what I have done.
Let series be <sup>a</sup>/<sub>r</sub> + a + ar + ar<sup>2</sup> + ar<sup>3</sup>

2nd and 3rd term = 20
ar+a=20; a(r+1)=20 (1)

3rd and 4th term = 320
ar<sup>2</sup>+ar<sup>3</sup>=320; ar<sup>2</sup>(r+1)=320 (2)

a=20/(r+1)
(1) into (2) gives
[20/(r+1)]*r<sup>2</sup>(r+1)=320
20r<sup>2</sup>=320
r=+/-4; though question states positive only &there4; r=4.
Sub. r=4 into (1) gives a(4+1)=20; a=4.
I don't think the answers you have been given are correct?!
I know what you did wrong now. You did some weird method where the FIRST TERM WAS a/r NOT a. The answer you have obtained was in fact the second term of the series:
<sup>a</sup>/<sub>r</sub> + a + ar + ar<sup>2</sup> + ar<sup>3</sup>

You had to find a/r, because it was YOUR first term, not a.

I'm only in Year 11 and sorry if it might be a little bit humiliating, but you should always start a series with a as the first term.
 

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