# The sum of the first three terms of G.P. is 7 and sum of their squares is 21. Calculate first five terms of the G.P.

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### 3 Answers

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

For the beginning, we'll note the terms of the G.P. as:

a, a*r, a*r^2, ...

From enunciation, we know that the sum of the first 3 terms is:

a + a*r + a*r^2 = 7

We'll factorize and we'll get:

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

Also, the sum of the squares of the first 3 terms is:

a^2 + (a*r)^2 + (a*r^2)^2 = 21

We'll factorize and we'll get:

a^2(1 + r^2 + r^4) = 21 (2)

We'll form the ratio (2)/(1)^2:

a^2(1 + r^2 + r^4)/[a(1 + r + r^2 )]^2 = 21/7^2

We'll eliminate like terms:

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

We'll form the square at numerator:

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

Now, the result is a difference of squares:

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

With this result, we'll go back into the ratio (2)/(1)^2:

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

We'll eliminate like brackets:

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

We'll cross multiply:

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

We'll move all terms to the left side:

7 + 7r^2 - 7r - 3 - 3r - 3r^2 = 0

We'll combine like terms:

4r^2 - 10r + 4 = 0

We'll divide by 2 :

2r^2 - 5r + 2 = 0

We'll apply the quadratic formula:

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

r1 = (5+3)/4

r1 = 2

r2 = (5-3)/4

r2 = 1/2

For r = 2, we'll calculate the first term of the g.p. from the relation (1):

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

a(1 + 2 + 4 ) = 7

7a = 7

a = 1

So, the g.p. is:

1 , 1*2 , 1*2*2 , ..........

For r = 1/2, the first term is:

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

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

a*(7/4) = 7

a = 4

The g.p. is:

4 , 4*(1/2), 4*(1/2)^2,..................

Let the 1st 3 terms of GP be a,ar and ar^2, wgere r is the common ratio of the GP.

Then by data a+ar+ar^2 = 7 and a^2+(ar)^2+(ar^2) = 21

a(1+r+r^2) =7 Or

a(1-r^3)/(1-r) = 7......................(1)

a^2(1+r^2+r^4) = 21

a^2{1-(r^2)^3}/(1-r^2) = 21..............(2)

We eliminate a by squaring (1) and dividing by (2):

a^2 {(1-r^3)/(r-1)}^2 / {(a^2(1-r^6)/(1-r^2)} = 7^2/21 = 7/3

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

= {1-r^3)(1+r)/[(1-r)(1+r^3)] = 7/3

= (1+r+r^2)/(1-r+r^2) = 7/3, as 1-r^3 = (1-r)(1+r+r^2) and 1-r^3 = (1+r)(1-r+r^2).

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

0 = (7-3)r^2 -(7+3)r 7-3

4r^2-10-4 = 0. Divide by 2.

2r^2-5r+2 = 0

(2r-1)(r-2) = 0

r = 1/2 or r= 2.

Taking r =2 and substituting in (1): a+ar+ar^2 = 7, we get:

a+2a+4a = 7

7a = 7or

a = 1.

Therefore a = 1 and a = 2 and 1,2 qnd 4 are the terms whose sum is 1+2+4 = 7 and square sum = 1^1+2^2+4^2 = 1+4+21.

When r = 1/2, we can find that a = 4 and ar =2 and ar^2 = 1 which is the same 3 terms of GP in reverse order.

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