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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(11⋅2+12⋅3+13⋅4+⋯+199⋅100) equals (P)7 (B)If x is the arithmetic mean between two real numbers a and b,(Q)9y=a2/3⋅b1/3 and z=a1/3⋅b2/3, 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(n−1)2 and(S)100n(n−1)(n−2)6 are in A.P., then the value of n is(T)2

Which of the following is the only CORRECT combination?

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Solution

The correct option is **C** (C)→(R),(D)→(P)

(C)

Let the inserted A.M.s between 14 and 34 be A1,A2,…,A198

14,A1,A2,…,A198,34

A1+A2+⋯+A198=198⎛⎜ ⎜ ⎜⎝14+342⎞⎟ ⎟ ⎟⎠=99

(D)

2n(n−1)2=n+n(n−1)(n−2)6

⇒6(n−1)=6+(n−1)(n−2)

⇒6n−6=6+n2−3n+2

⇒n2−9n+14=0

⇒n=7 or 2

(C)

Let the inserted A.M.s between 14 and 34 be A1,A2,…,A198

14,A1,A2,…,A198,34

A1+A2+⋯+A198=198⎛⎜ ⎜ ⎜⎝14+342⎞⎟ ⎟ ⎟⎠=99

(D)

2n(n−1)2=n+n(n−1)(n−2)6

⇒6(n−1)=6+(n−1)(n−2)

⇒6n−6=6+n2−3n+2

⇒n2−9n+14=0

⇒n=7 or 2

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