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Question

Each entry of List I is to be matched with one entry of List II.

List IList II (A)100(112+123+134++199100) equals (P)7 (B)If x is the arithmetic mean between two real numbers a and b,(Q)9y=a2/3b1/3 and z=a1/3b2/3, then y3+z3xyz equals(C)If 198 arithmetic means are inserted between 14 and 34, then(R)99the sum of these arithmetic means is(D)If n is a positive integer such that n,n(n1)2 and(S)100n(n1)(n2)6 are in A.P., then the value of n is(T)2

Which of the following is the only CORRECT combination?

A
(A)(R),(B)(T)
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B
(A)(Q),(B)(P)
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C
(A)(S),(B)(P)
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D
(A)(T),(B)(R)
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Solution

The correct option is A (A)(R),(B)(T)
(A)
100(112+123+134++199100)
=100[(112)+(1213)++(1991100)]
=100[11100]
=99

(B)
y3+z3xyz=a2b+ab2(a+b2)(a×b)=2

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