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In the infinite sequence a1, a2, a3,...., an, each term afte

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by metalhead » Fri Jan 29, 2010 4:18 pm
This is a geometric series, where each element is represented by a.r^n-1 . a = first element ; n = 1,2,3,etc for 1st, 2nd , 3rd elements ; r = base of the progression (in this case equal to 2, because each subsequent element is TWICE the previous one).
So, this gives - a5 = a.2^4 = 16a ; a2 = a.2^1 = 2a.
So, 16a - 2a = 14a is equal to 12.
a1 = a = first element = 6/7
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by jeffedwards » Fri Jan 29, 2010 4:52 pm
If you don't want to worry about the confusing r^n-1 formula, you can also try what I did.

First I wrote each term out to a5. Then underneath I wrote their value, in terms of a1

a1, a2, a3, a4, a5
a, 2a, 4a, 8a, 16a

As you can see each term is twice the previous term.

Then, I plugged in the terms and solved for a

(a5 - a2) → (16a - 2a) → 14a

So we get
14a =12
a= 12/14
or
a= 6/7
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by ajith » Fri Jan 29, 2010 10:38 pm
imane81 wrote:Anyone can help on this one, thank you in advance..

In the infinite sequence a1, a2, a3,...., an, each term after the first is equal to twice the previous term. If a5-a2=12, what is the value of a1?

A. 4
B.24/7
C.2
D.12/7
E.6/7
In a series where there is a first term a and each term after the first is derived by the multiplying previous term by a factor r

n-th term = ar^(n-1)

in this case a*2^4 -a*2 =12

14 a = 12

a= 6/7

[For the same series sum of n terms = a((r^n)-1)/(r-1) ]
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