the question in the book is write the first four terms of the g.s. , if the sum of the first and fourth terms is 195 and the sum of the second and third terms is 60.

asub1+asub1r^3=195 (1)

asub1r+asub1r^2=60 (2)

Dividing (1) by (2)

r=1/2; r=4

Substitute in (2)

asubn=192; asub1=3

Therefore, the four terms are 3, 12, 48, 192.

My question is how did they get r=1/2 and r=4?

# Geometric series

August 2, 2015 - 4:56pm

#1
Geometric series

I've got 4, 1/4 and -1 for r, here is my solution:

$\dfrac{a_1 + a_1 r^3}{a_1 r + a_1 r^2} = \dfrac{195}{60}$

$\dfrac{a_1(1 + r^3)}{a_1(r + r^2)} = \dfrac{13}{4}$

$4(1 + r^3) = 13(r + r^2)$

$4 + 4r^3 = 13r + 13r^2$

$4r^3 - 13r^2 - 13r + 4 = 0$

$r = 4, ~ 1/4, ~ -1$

Thank you sir. Thanks sa help.

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