Plz answer the above mentioned question in photo attached,plz fast,

let G1 be the first term of the GP
G2 be the second term of GP
G3 be the third term of the GP
G1= n(n+1)/2
G2= sqrt10/3 (n(n+1)(2n+1))/6
G3= (n(n+1))^2/ 4
G2/G1= common ratio= r= sqrt10/9 (2n+1).......... (1)
G3/G1= r^2= n(n+1)/2........ (2)
squaring (1) and comparing with (2), we get
10/81 (4n^2+4n+1)= (n^2+n)/2
n^2+n= 20
by solving this quadratic equation we get n=4........... [Ans]