Let Vr denote the sum of the first r terms of an A.P whose first term is r and common difference is (2r−1).Let Tr=Vr+1−Vr−2 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+(n−1)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=(2r−1) ∴Vr=r2[2r+(r−1)(2r−1)] ⇒r2[2r+2r2−3r+1]=r2[2r2−r+1] Thus Vr=12[2r3−r2+r]=r3−r22+r2 Now Tr=Vr+1−Vr−2 From above Vr=r3−r22+r2 Vr+1=(r+1)3−(r+1)22+(r+1)2 We have Tr=Vr+1−Vr−2 Tr=r3−r22+r2−((r+1)3−(r+1)22+(r+1)2)−2 On simplifying, we get
Tr=(r+1)3−r3−12((r+1)2−r2)+12(r+1−r)−2 ⇒Tr=(r+1−r)((r+1)2+r(r+1)+r2)+12−2 On simplifying, we get Tr=r2+1+2r+r2+r+r2+12(−2r−1+1)−2 Tr=3r2+2r−1=(3r−1)(r+1) We have T1=(3−1)(1+1)=2.2 T2=(3×2−1)(2+1)=5.3 T3=(3×3−1)(3+1)=8.4 which are in A.P Thus,Tn=(3n−1)(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.