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Question

The sum of the first three terms of an A.P. is 9 and the sum of their square is 35. The sum to first n terms of the series can be

A
n(n+1)
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B
n2
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C
n(4n)
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D
n(6n)
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Solution

The correct options are
B n2
D n(6n)
Sum of first three terms of AP = 9
Sum of square of three terms of AP=35
Let AP is a,a+d,a+2d........
with first terms = a
common different = d
a+a+d+a+2d=9
3a+3d=9
a+d=3
a2+(a+d)2+(a+2d)2=35
Put d=3a
a2+32+(6a)2=35
2a2+9+3612a=35
2a212a+10=0
a26a+5=0
(a1)(a5)=0
a=1,5
That given d=2,2
Sum of first n term =n2(2a+(n1)d)
n2(2+2(n1)=n2[a=1,d=2]
Or
n2[102(n1)]
n[5n+1]
n[6n]
B,D are correct

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