Question

the third term of a GP be one fourth of fifth term and their sum is 5/4. Find the 10th term of G.P.

Answer 1

A5=4A3
A4=square root of (4A3*A3)
     =2A3
4A3+A3=5/4
A3=1/4
A4=1/2
r=A4/A3
 =2
A1=A3/(r^2)
A1=1/16
A10=A1*(r^9)
A10=2^5
      =32

Answer 2

third term = $a.r^{2}$

fifth term = $a.r^{4}$

As per question third term  = fifth term

$a.r^{2}$ = $\frac{1}{4}$ $a.r^{4}$

$r^{2}$ = 4

r = ±2

As per another condition

$a.r^{2}$ +  $a.r^{4}$ = $\frac{5}{4}$

$a.r^{2}$ (1 +  $r^{2}$) = $\frac{5}{4}$

Putting value of $r^{2}$ as 4 from above we get

4.a.(1+4) =  $\frac{5}{4}$

a = 1/16

tenth term = $a.r^{9}$

If r = +2

then tenth term is $\frac{1}{16} . (2)^9$ = 32

if r= -2

then tenth term will be  $\frac{1}{16} . (-2)^9$ = -32