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Plz answer the above mentioned question in photo attached,plz fast,
let G1 be the first term of the GPG2 be the second term of GPG3 be the third term of the GPG1= n(n+1)/2G2= sqrt10/3 (n(n+1)(2n+1))/6G3= (n(n+1))^2/ 4G2/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 get10/81 (4n^2+4n+1)= (n^2+n)/2n^2+n= 20by solving this quadratic equation we get n=4........... [Ans]
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