## anonymous 4 years ago 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 !

1. amistre64

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

2. amistre64

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

3. anonymous

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

4. amistre64

those are option i assume ...

5. amistre64

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 :)

6. amistre64
7. amistre64

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

8. anonymous

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.

9. anonymous

r is common ratio*

10. amistre64

i like that pathway of thinkg alot better :) kudos

11. amistre64

$a+ar=8$ $ar-\frac{4}{3}=ar^2;\ ar=ar^2+\frac 43$ $a+ar^2+\frac 43=\frac {24}3$ $ar^2+a+\frac {-20}3=0$ $a(r^2+1)=-\frac {20}3$ $r^2+1=-\frac {20}{3a}$ $r^2=-\frac {20}{3a}-1$ $r=\sqrt{-\frac {20}{3a}-1}$ that looks fun :)

12. amistre64

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