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The sum of the first 10 terms of a geometric sequence is 10, and the sum of the first 30 terms of the same sequence is 70. Find the sum of the first 40 terms.

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S10  = a ( 1 - r^10) / ( 1 -r) =  10

S30 =  a ( 1 - r^30) / (1- r)  =   70

a / (1 -r )  =  10 / (1 -r^10)

a/(1- r)   =   70 / (1-r^30)

10 / ( (1 -r^10)  =  70 / ( 1 -r^30)

10 ( 1 - r^30) = 70 (1 -r^10)

10  - 10r^30  = 70 - 70r^10

70r^10 - 10r^30   =  60

7r^10 - r^30  = 6

r^30 - 7r^10  + 6 =  0

Let   r^30  = x^3 ,  7r^10  = 7x

So we have

x^3 - 7x + 6   =  0

1 is a root

Use synthetic division  to find the other roots

1  [ 1  0   -7     6]

1    1    -6

____________

1   1   -6    0

The remaining polynomial  is

x^2 + x  -  6   =  0

(x + 3) ( x -2) = 0

x = 2  is the only feasible  solution

So

r^30  = x^3

r ^10  =  x

r^10  = 2

r = 2^(1/10)

a ( 1 - r*10) / (1 - r)  = 10

a  ( 1 - 2) / ( 1 - 2^(1/10) )  =  10

a  = - 10 ( 1 - 2^(1/10))

a =  10 [ 2^(1/10) - 1)]

S40  =    10 [ 2^(1/10) - 1 ] [ 1 - 2^4 ]  / [ 1 - 2^(1/10) ] =

-10 [ 1 - 2^4 ]  =

-10 (-15)  =

150

Apr 7, 2024