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Question

Let Vr denote the sum of the first r terms of an A.P whose first term is r and common difference is (2r1).Let
Tr=Vr+1Vr2 is always

A
an odd number
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B
an even number
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C
a prime number
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D
a composite number
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Solution

The correct option is D a composite number
We have sum of n terms =n2(2a+(n1)d) where a is the first term, n is the number of terms and d is the common difference in an A.P.
From the passage n=r, a=2r and d=(2r1)
Vr=r2[2r+(r1)(2r1)]
r2[2r+2r23r+1]=r2[2r2r+1]
Thus Vr=12[2r3r2+r]=r3r22+r2
Now Tr=Vr+1Vr2
From above Vr=r3r22+r2
Vr+1=(r+1)3(r+1)22+(r+1)2
We have Tr=Vr+1Vr2
Tr=r3r22+r2((r+1)3(r+1)22+(r+1)2)2
On simplifying, we get
Tr=(r+1)3r312((r+1)2r2)+12(r+1r)2
Tr=(r+1r)((r+1)2+r(r+1)+r2)+122
On simplifying, we get
Tr=r2+1+2r+r2+r+r2+12(2r1+1)2
Tr=3r2+2r1=(3r1)(r+1)
We have T1=(31)(1+1)=2.2
T2=(3×21)(2+1)=5.3
T3=(3×31)(3+1)=8.4 which are in A.P
Thus,Tn=(3n1)(n+1)
From the above sequence, we note that
Product of even number and an odd number is Even
Product of odd number and an odd number is odd
Product of even number and an even number is Even
and we see that every term is a composite number.
Hence, their sum is a composite number.
T is a composite number.

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