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Let the sum of the first n terms of an A.P. be 3n2+5n. Then the sum of squares of the first 10 terms of the A.P. is:
(3) 15220
an=Sn-Sn-1= (3n2+5n )- (3(n-1)2+5(n-1) )=6n+2
Check: a1=S1=8=6(1)+2
n=110an2= n=110(6n+2)2=36 n2+24 n+4 10
n=110n2= 10 11 216=385, n=110n=55
Sum =36 385+24 55+40
=13860+1320+40=15220
Answer: (3)
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