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

Open in App

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)=kl

If 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+1

Hence 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+1

Now 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) .kl

So 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=odd

As 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+3

As 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+32

So 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.

0

View More