Answer to Question #148695 in Algebra for Dayana

Question

In a geometric sequence, the second term exceeds the fourth term by 1/2 .
The third term exceeds the fifth term by 1/8. Find the sum to infinity.

Answer 1

Let $a$ and $r$ be the first term and common ratio of the geometric sequence.

We know that n-th term of the geometric sequence is

$Tn=a.r^(n-1)$

So,$T2=a.r,T3=a.r^2,T4=a.r^3,T5=a.r^4$

Now given that,$T2-T4=1/2.........(1)$ &

$T3-T5=1/8........(2)$

From (1) , $ar-ar^3=1/2$

=> $ar(1-r^2)=1/2.......(3)$

From (2), $ar^2-ar^4=1/8$

=> $ar^2(1-r^2)=1/8........(4)$

Now divide (4) by (3),we get

$ar^2(1-r^2)/ar(1-r^2)=(1/8)/(1/2)$

=> $r=1/4$

From (3),putting the value of $r$ we get,

$a(1/4-1/64)=1/2$

=> $a.(15/64)=1/2$

=> $a=32/15$

Now we know that sum of the infinite geometric sequence is $=a/(1-r)$ .

Therefore sum of the required geometric sequence is $=(32/15)/(1-(1/4))$

$=(32/15)/(3/4)$

$=(32/15)×(4/3)$

$=128/45$