If the sum of four numbers in GP is 60 and the arithmetic mean of the first and last numbers is 18 then the numbers are

Step-by-step solution

Let four terms in a GP be a r^3, a r, a r and a r^3 . According to the given condition, a r^3+a r+ a r + a r^3 =60 ....(i) and a r^3+ a r^3 2 =18 a r^3+ a r^3 =36 .....(ii) Now, from Eq. (i), we have (a r+ a r )+a r^3+ a r^3 =60 a (r+ 1 r )+36=60 [from Eq. (ii)] a (r+ 1 r )=24 .....(iii) On dividing Eq. (iii) by Eq. (ii), we get a (r^3+ 1 r^3 ) a (r+ 1 r ) = 36 24 (r+ 1 r ) (r^2+ 1 r^2 -1 ) r+ 1 r = 3 2 2 (r^2+ 1 r^2 -1 )=3 2 (r^4+1-r^2 ) r^2 =3 array lc & 2 r^4+2-2 r^2=3 r^2 & 2 r^4-5 r^2+2=0 & 2 r^4-4 r^2-r^2+2=0