Progressions 2

Algebra Level 3

The sum of an infinite geometric progression is equal to 4, and the sum of the cubes of its terms is equal to 192 192 .

If the first term and the common ratio of the geometric progression are A A and R R respectively, find A + 2 R A + 2R .


The answer is 5.

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1 solution

Hung Woei Neoh
Jun 1, 2016

Let the first term of the geometric progression be a a , and the common ratio be r r , where 1 < r < 1 , r 0 -1 < r < 1, r \neq 0

The geometric progression is:

a , a r , a r 2 , a r 3 , a,ar,ar^2,ar^3,\ldots

The sum of this progression to infinity is 4 4 . Therefore,

a 1 r = 4 \dfrac{a}{1-r} = 4

a = 4 ( 1 r ) a=4(1-r) \implies Eq.(1)

Next, the sum of the cubes of all its terms is 192 192 :

a 3 + ( a r ) 3 + ( a r 2 ) 3 + ( a r 3 ) 3 + = 192 a 3 + a 3 r 3 + a 3 r 6 + a 3 r 9 + = 192 a^3+(ar)^3+(ar^2)^3+(ar^3)^3 + \ldots = 192\\ a^3 + a^3r^3+ a^3r^6 + a^3r^9 + \ldots = 192

Notice that this sum is also the sum of a geometric progression to infinity, where the first term is a 3 a^3 and the common ratio is r 3 r^3 . This gives

a 3 1 r 3 = 192 a 3 = 192 192 r 3 \dfrac{a^3}{1-r^3} = 192\\ a^3 = 192-192r^3

Substitute Eq.(1):

( 4 ( 1 r ) ) 3 = 192 192 r 3 64 ( 1 3 r + 3 r 2 3 r 3 ) = 64 ( 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 , 1 2 , 2 \left(4(1-r)\right)^3 = 192-192 r^3\\ 64(1-3r+3r^2-3r^3) = 64(3-3r^3)\\ 1-3r+3r^2-r^3 = 3-3r^3\\ 2r^3 +3r^2 - 3r - 2 = 0\\ 2r^3 - 2r^2 + 5r^2 - 5r + 2r - 2 = 0\\ (r-1)(2r^2 + 5r + 2) = 0\\ (r-1)(2r+1)(r+2) = 0\\ r=1,\;-\dfrac{1}{2},\;-2

Since 1 < r < 1 , r 0 -1 < r < 1, r \neq 0 , we know that r = 1 2 r= -\dfrac{1}{2}

a = 4 ( 1 ( 1 2 ) ) = 6 a + 2 r = 6 + 2 ( 1 2 ) = 6 1 = 5 a = 4\left(1-\left(-\dfrac{1}{2}\right)\right) = 6\\ a+2r = 6 + 2\left(-\dfrac{1}{2}\right) = 6-1 = \boxed{5}

Nicely done.! (+1)

Rishabh Tiwari - 5 years ago

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