Asked by Sinachi
Two geometric progressions have equal sums to infinity. Their first terms are 27 and 36 respectively. If the common ratio of the first is 3/4, find the common ratio of the other
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Let the common ratio of the second geometric progression be 'k'.
The sum to infinity of the first progression is:
S1 = a1/(1-r1) = 27/(1-3/4) = 108
The sum to infinity of the second progression is:
S2 = a2/(1-k) = 36/(1-k)
Given that S1 = S2, we have:
27/(1-3/4) = 36/(1-k)
Simplifying this expression, we get:
k = 1 - (27/36)(1-3/4) = 9/16
Therefore, the common ratio of the second geometric progression is 9/16.
The sum to infinity of the first progression is:
S1 = a1/(1-r1) = 27/(1-3/4) = 108
The sum to infinity of the second progression is:
S2 = a2/(1-k) = 36/(1-k)
Given that S1 = S2, we have:
27/(1-3/4) = 36/(1-k)
Simplifying this expression, we get:
k = 1 - (27/36)(1-3/4) = 9/16
Therefore, the common ratio of the second geometric progression is 9/16.
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