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

Question

Shashank Singh · Answer

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]

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

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

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