Question

the sum of first three terms of a GP is 7 and sum of their squares is 21 find the first five terms of the GP​

Answer 1

Answer:

4, 2, 1

Step-by-step explanation:

Let a1, a2, a3 be terms in a G.P

a1+a2+a3 = 7

a1^2 + a2^2 + a3^2 = 21

But we know that:

a1= a1

a2= a1*r

a3= a1*r^2

==> a1+ a2 + a3

    = a1+ a1r+ a1r^2 = 7

     = a1(1+r + r^2) = 7

Now let us square:

==> a1^2 (r^2 + r + 1)^2= 49........(1)

Also:

==> a1^2 + a2^2 + a3^2 =

     = a1^2 + (a1r)^2 + a1^2*r^4 = 21

      = a1^2 (1 + r^2 + r^4) = 21  .........(2)

Let us divide (2) by (1):

==> (1+r^2+ r^4)/(r^2 +r + 1 )^2 = 21/49

==> (1+r^2 + r^4)/ (r^2 + r+1)^2 = 3/7

But  r^4 + r^2 + 1= (r^2 +1)^2 - r^2= (r^2+1+r)(r^2+1-r)

Now substitute:

==? (r^2 + r+ 1)(r^2 -r + 1)/(r^2 + r + 1)^2 = 3/7

==> (r^2 -r + 1)/(r^2 + r + 1) = 3/7

==> Cross multiply:

==> 7(r^2 -r +1) = 3(r^2 + r + 1)

==> 7r^2 - 7r + 7 = 3r^2 + 3r + 3

Group similar:

==> 4r^2 -10r + 4 = 0

Divide by 2:

==> 2r^2 - 5r +2 = 0

==> r1= [5 + sqrt(25 -16)/4=

          = [5 + 3]]/4 = 8/4 = 2

==> r2= 5-3/4= 2/4 = 1/2

For r= 2

a1(1+r+r^2 ) = 7

a1(1+2+4) = 7

a1= 7/7= 1

a2= 2

a3= 4

==> 1,2,4  is the G.P\

For r= 1/2:

a1(1+1/2+ 1/4) = 7

a1= 7 * 4/7= 4

a2= 2

a3= 1

4, 2, 1  is the G.P