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Question 488218: The sum of all terms of an infinite geometric progression is 12, and each term is three times the sum of all terms that follow it. What is the first term of the sequence?
Please help. I'm not sure if I'm following it correctly but here's what I've got so far:
Formula: S(infinity)=a1/1-r
where,
S(infinity)=12
a1=3(a2+a3+a4+..an)
And that's about it. I don't know what to do afterwards. :(
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! this one's a bear, but i think i have it.
you are given that:
sum of an infinite series is 12.
each term in the series is equal to 3 times the sum of the infinite series starting with the term that comes after it.
we'll let your series be represented by:
a1, a2, a3, a4, .................................a[infinity].
the sum of the series starting with a1 is equal to:
a1/(1-r) = 12
the sum of the series starting with a2 is equal to:
a2/(1-r)
since the sum of the infinite series starting with a1 is equal to (the sum of the infinite series starting with a2 plus a1), then we can derive the formula of:
a2/(1-r) + a1 = 12
we now have:
a1/(1-r) = 12 (equation 1)
and we have:
a2/(1-r) + a1 = 12 (equation 2)
we also know that each term in the series is equal to 3 times the sum of the infinite series starting with the next term in the series.
this means that:
a1 = 3 * (a2/(1-r)
we can substitute for a1 in equation 2 to get:
a2/(1-r) + 3a2/(1-r) = 12 (equation 4)
we now have:
a1/(1-r) = 12 (equation 1)
a2/(1-r) + 3a2/(1-r) = 12 (equation 4)
since they are both equal to 12, we can make the expressions on the left side of equation 1 and equation 4 equal to each other to get:
a1/(1-r) = a2/(1-r) + 3a2/(1-4)
if we multiply both sides of this equation by (1-r), we get:
a1 = a2 + 3a2 which becomes:
a1 = 4a2
if we solve this equation for a2, then we get:
a2 = (1/4) * a1
this means that our common ratio is equal to (1/4).
if our common ratio is equal to (1/4), then the equation of:
a1/(1-r) = 12 becomes:
a1/(1 - (1/4)) = 12 which becomes:
a1/(3/4) = 12 which becomes:
a1 = (3/4) * 12 which becomes:
a1 = 9
we now have:
a1 = 9
r = (1/4)
because a2 = a1*r, we can derive a2 to get:
a2 = (9/4)
we can derive the infinite sum starting with (9/4) by using the formula:
(9/4)/ ( 1-r) = S
since r = (1/4), this formula becomes:
(9/4) / (3/4) = S
Solving this formulas for S gets us:
S = 3
the infinite sum of the geometric series starting with a2 is equal to 3.
a1 should be equal to 3 times this and it is.
we have:
a1 = 9
r = (1/4)
the question is:
what is the first term of the sequence?
the answer is:
the first term of the sequence is 9.
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