The sum of an infinite geometric progression is equal to 4, and the sum of the cubes of its terms is equal to .
If the first term and the common ratio of the geometric progression are and respectively, find .
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Let the first term of the geometric progression be a , and the common ratio be r , where − 1 < r < 1 , r = 0
The geometric progression is:
a , a r , a r 2 , a r 3 , …
The sum of this progression to infinity is 4 . Therefore,
1 − r a = 4
a = 4 ( 1 − r ) ⟹ Eq.(1)
Next, the sum of the cubes of all its terms is 1 9 2 :
a 3 + ( a r ) 3 + ( a r 2 ) 3 + ( a r 3 ) 3 + … = 1 9 2 a 3 + a 3 r 3 + a 3 r 6 + a 3 r 9 + … = 1 9 2
Notice that this sum is also the sum of a geometric progression to infinity, where the first term is a 3 and the common ratio is r 3 . This gives
1 − r 3 a 3 = 1 9 2 a 3 = 1 9 2 − 1 9 2 r 3
Substitute Eq.(1):
( 4 ( 1 − r ) ) 3 = 1 9 2 − 1 9 2 r 3 6 4 ( 1 − 3 r + 3 r 2 − 3 r 3 ) = 6 4 ( 3 − 3 r 3 ) 1 − 3 r + 3 r 2 − r 3 = 3 − 3 r 3 2 r 3 + 3 r 2 − 3 r − 2 = 0 2 r 3 − 2 r 2 + 5 r 2 − 5 r + 2 r − 2 = 0 ( r − 1 ) ( 2 r 2 + 5 r + 2 ) = 0 ( r − 1 ) ( 2 r + 1 ) ( r + 2 ) = 0 r = 1 , − 2 1 , − 2
Since − 1 < r < 1 , r = 0 , we know that r = − 2 1
a = 4 ( 1 − ( − 2 1 ) ) = 6 a + 2 r = 6 + 2 ( − 2 1 ) = 6 − 1 = 5