Question

Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4 th by 18.

Answer 1

Let rbe the common ratio and a be the first term of the G.P.

Now, the value of first four term of G.P is,

a 1 =a, a 2 =ar, a 3 =a r 2 and a 4 =a r 3

Third term is greater than first term by 9. So,

a 3 = a 1 +9 a r 2 =a+9 (1)

Second term is greater than fourth term by 18 is,

a 2 = a 4 +18 ar=a r 3 +18 (2)

Now, from equation (1) and (2), we get

a( r 2 −1 )=9(3)

And,

ar( 1− r 2 )=18(4)

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

ar( 1− r 2 ) a( r 2 −1 ) = 18 9 −r=2 r=−2

Substitute the value of r in equation (1), we get

4a=a+9 3a=9 a=3

Now, substitute the value of a and r in a 1 =a, a 2 =ar, a 3 =a r 2 and a 4 =a r 3 .

a 1 =3, a 2 =3×( −2 ), a 3 =3× ( −2 ) 2 and a 4 =3× ( −2 ) 3

Thus, the first four term of G.P is a 1 =3, a 2 =−6, a 3 =12 and a 4 =−24.