Question

if the mth term of an H.P. be n and nth term be m, show that the pth term is mn/p
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Answer 1

Step-by-step explanation:

Given that,mthterm of an H.P. is n and nth term is m.

Therefore, mth term of an A.P. is n1 and nth term is m1.

Let, a be the first term and d be the common difference of the A.P

Now, tm=a+(m−1)d=n1          ....(1)

And tn=a+(n−1)d=m1         ....(2)

Subtracting (1) & (2), we get

(m−n)d=n1−m1

⇒d=mn1

Substituting d in (1), we get

a+mn(m−1)=n1       ⇒a=mn1

Therefore, tmn=a+(mn−1)d=mn1+mn

Answer 2

$\underline{\underline{\bf Harmonic \ Progression:}}$

• The progression formed by the reciprocal of Arithmetic Progression.
• It is the sequence of numbers formed by taking the reciprocals of AP
• For example,  

             a,b,c,d... are in AP, then 1/a , 1/b , 1/c , 1/d ... is considered as HP.

• General form :

          1/a , 1/(a + d) , 1/(a + 2d) , 1/(a + 3d) , ...

• Harmonic Mean,

              $\bf = \frac{n}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+...}$

     where n is the number of terms

• nth term of HP is given by the reciprocal of nth term of AP

          $\bf n^{th} \ term \ of \ HP=\frac{1}{a+(n-1)d}$

   where

     a - first term of AP

     d - common difference      

_______________________________              

Step-by-step explanation :

Given,

• mth term of HP = n
• nth term of HP = m

$\implies m^{th} \ term=\frac{1}{a+(m-1)d} \\\\ n=\frac{1}{a+(m-1)d} \\\\ \boxed{\bf a+(m-1)d=\frac{1}{n}} --[1] \\\\\\\\ \implies n^{th} \ term=\frac{1}{a+(n-1)d} \\\\ m=\frac{1}{a+(n-1)d} \\\\ \boxed{\bf a+(n-1)d =\frac{1}{m}}--[2]$

Substract equation [1] from [2]

a + (n - 1)d - [a + (m - 1)d] = 1/m - 1/n

a + nd - d - [a + md - d] = 1/m - 1/n

a + nd - d - a - md + d = (n - m)/mn

       d(n - m) = (n - m)/mn

             d = 1/mn

Substitute  d = 1/mn  in equation [1]

          $a+(m-1)d=\frac{1}{n} \\\\ a+(m-1)\frac{1}{mn}=\frac{1}{n} \\\\ a+\frac{m-1}{mn} =\frac{1}{n} \\\\ \frac{mna+(m-1)}{mn}=\frac{1}{n} \\\\ \frac{mna+m-1}{m} =1 \\\\ mna+m-1=m \\\\ mna-1=0 \\\\ mna=1 \\\\ \bf a=\frac{1}{mn}$

we know,

   $\bf n^{th} \ term \ of \ HP=\frac{1}{a+(n-1)d}$

  $\bf p^{th} \ term \ of \ HP =\frac{1}{\frac{1}{mn}+(p-1)(\frac{1}{mn})}$

                              $\bf =\frac{1}{\frac{1}{mn}+\frac{p-1}{mn} } \\\\ =\frac{1}{\frac{1+p-1}{mn} } \\\\ =\frac{mn}{p}$

  Hence proved!