Question

The second term of geometric sequence exceeds the third term by 4/3 and the sum of the first two terms is 8. Find the possible values of the first three terms.

Thank for the solution !

Answer 1

a,b,c are what we have to work with; and its geometric ...

b = c+ 4/3

b c
--- = ----- ; common ratio
a c+4/3

a+b = 8

b = 8-a = c+4/3

hmmm

Answer 2

do we have an answer guide to chk with in the end?

Answer 3

6 , 2 , 2/3 or 16/3 , 8/3 , 4/3

Answer 4

those are option i assume ...

Answer 5

3c b
----- = ---
3c+4 a


3ca = b(3c+4)

hard to tell which route to take ....

\[\frac{a-\cfrac{9ac^2}{9c^2+12c+16}}{1-\cfrac{3c}{3c+4}}=8\]
a = 8-b = 8-(c+4/3)

\[\frac{(8-c-\frac{4}{3})-\cfrac{9(8-c-\frac{4}{3})c^2}{9c^2+12c+16}}{1-\cfrac{3c}{3c+4}}=8\]
would then solve for C :)

Answer 6

wolframalpha simplifies that to:

http://www.wolframalpha.com/input/?i=%28%2820%2F3-c%29-%289c%5E2%2820%2F3-c%29%29%2F%289c%5E2%2B12c%2B16%29%29%2F%281-%283c%29%2F%283c%2B4%29%29-8%3D0

Answer 7

im sure theres a simpler way to di this, i just cant think of one at the moment ....

Answer 8

I have another method . In general, the equation for 'find the number of term' is \[ar ^{n-1}\] , where a is the first term, r is common difference, n is number of term. So a, ar, ar^2 , what we need to work with. From question, we can two equation, 1) a+ar =8 ; 2) ar = ar^2 + 4/3 .Use simultaneous equation, you are able to solve.

Answer 9

r is common ratio*

Answer 10

i like that pathway of thinkg alot better :) kudos

Answer 11

\[a+ar=8\]
\[ar-\frac{4}{3}=ar^2;\ ar=ar^2+\frac 43\]
\[a+ar^2+\frac 43=\frac {24}3\]
\[ar^2+a+\frac {-20}3=0\]
\[a(r^2+1)=-\frac {20}3\]
\[r^2+1=-\frac {20}{3a}\]
\[r^2=-\frac {20}{3a}-1\]
\[r=\sqrt{-\frac {20}{3a}-1}\]

that looks fun :)

Answer 12

\[r=\sqrt{\frac {20}{3a}-1}\]perpetuated a - lol