    Question

# Let n≥1 be a positive integer. Suppose, between two distinct positive real numbers k and l,n arithmetic means a1,a2,...,an and n harmonic means h1,h2,....,hn are inserted. For i=1,2,....,n and j=1,2,....,n, let p(i,j)=aihj then

A
p(i,j)=kl for some i and j
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B
If n is even, there are an even number of ordered pairs (i,j), such that p(i,j)=kl
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C
If n is odd, there are an even number of ordered pairs (i,j), such that p(i,j)=kl
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D
Whenever n is odd there is an i such that p(i,i)=kl
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Solution

## The correct options are A If n is even, there are an even number of ordered pairs (i,j), such that p(i,j)=kl B p(i,j)=kl for some i and j D Whenever n is odd there is an i such that p(i,i)=klIf between two distinct positive real numbers a and b, n arithmetic means a1,a2........an are inserted. Then we can write as: ⇒k,a1,a2,a3,.........an,l, .....(1)Where all the numbers collectively form an A.P. For an A.P., the last term l=a+(n−1)d, where a is the first term, n is total number of terms and d is the common difference between terms.For the given Arithmetic progression, the first term is k, the total number of terms are n+2 and the last term is l.So we can write, l=k+(n+2−1)dA→dA=l−kn+1Hence from the given A.P., we can write any general term as ⇒ai=k+(i−1)dA⇒ai=k+(i−1)(l−k)n+1⇒ai=k(n+1)+(i−1)(l−k)n+1Now if n harmonic means are inserted between k and l, we can write as: ⇒1k,1h1,1h2,....1hn,1l ...(2)As We know that by inverting the term of a harmonic mean, the new set of terms becomes an Arithmetic progression. Hence the term in eq.(2) are now in arithmetic progression. For an A.P., the last term l=a+(n−1)d, where a is the first term, n is total number of terms and d is the common difference between terms.Hence 1l=1k+(n+2−1)dH⇒dH=k−lkl(n+1)So we can write that 1hj=1k+(j−1)dH⇒1hj=1k+(j−1)(k−l)kl(n+1)⇒1hj=l(n+1)+(j−1)(k−l)kl(n+1)⇒hj=kl(n+1)l(n+1)+(j−1)(k−l)Now P(i,j)=ai,hj⇒P(i,j)=k(n+1)+(i−1)(l−k)l(n+1)+(j−1)(k−l) .klSo if P(i,j)=kl, Then k(n+1)+(i−1)(l−k)=l(n+1)+(j−1)(k−l)⇒(l−k)(i+j−2)=(l−k)(n+1)⇒i+j=n+3 ....(3)Hence we can say that P(i,j)=kl for some (i,j), related with the equation i+j=n+3→ Now if n is even in eq.(3), then i+j=even+odd=oddAs i,j can be maximum at (n+2), because there are only (n+2) terms. n is even, so is n+2, So there will be even number of ordered pairs of (i,j), such that P(i,j)→ Now if n is odd then i+j=odd+odd=even=n+3As i,j can be maximum at (n+2), because there are only (n+2) terms. n is odd, so is (n+2), So there will be odd number of ordered pairs of (i,j), such that P(i,j)`So when n is odd, then n+3 is even, and in that case there will lie i=n+32,j=n+32So id n is odd, then at some point where i=j. we can say that P(i,i)=kl.Hence correct options are A,B and C.  Suggest Corrections  0      Similar questions  Related Videos   Geometric Progression
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