Question

If mth term of an H.P is n , while nth term is m , find its (m+n)th term

Answer 1

@ikram002p @ganeshie8

Answer 2

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

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

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

after that i am unable to proceed

Answer 3

We need to find a / { (m + n - 1) * d } ---- (1)
We are given that:
1/{a+(m-1)d} = n or
a+(m-1)d = 1/n ---- (2)
1/{a+(n-1)d} = m or
a+(n-1)d = 1/m ---- (3)

From (2) and (3) solve for a and d in terms of m and n.
Substitute in (1).

Answer 4

my textbook says
(m-n)d = m-n/mn
How?

Answer 5

Subtract (3) from (2). That will get rid of 'a' and and you can solve for 'd'.

Answer 6

The general term for a Harmonic progression is: \[a_n = \frac{1}{a+(n-1)d}\]
Now from the equations you are given,
\[\frac{1}{a+(n-1)d} = m\]
\[\frac{1}{a+(m-1)d} = n\]
From those, you can get,
\[m[a+(n-1)d] = n[a+(m-1)d]\]

You'll get this equation, after solving that: \[an - nd = am - md\]
I think it must be given that \(m \neq n\)
Thus, \(a=d\)

And then we have to find \[\frac{1}{a+(m+n-1)d}\]
Which is:
\[\frac{1}{a+dm+nd -d}\]
But \(a=d\) so we have,
\[a_{m+n} = \frac{1}{d(m+n)}\]
From the equations you have given, we can substitute the value of \(a\) in the other equation, that is, we can find the value of \(d\) from there.
That is, you get the equation: \[\frac{1}{n} - (m-1)d = \frac{1}{m} - (n-1)d\]
That gives:
\[\frac{1}{n} - md + d = \frac{1}{m} - nd + d\]
Cancel the \(d\)

Rearrange the equation with fractions on one side and you'll get:\[d = \frac{1}{mn}\]
Substitute this value in the original answer.

Understood? :)

Answer 7

:) Reading it thanks for the effort bro , god bless you

Answer 8

Yup thankss a ton!

Answer 9

:)

Answer 10

Probably the best explanation i have ever seen on OS

Answer 11

Thanks!

Answer 12

After this
an−nd=am−md
how you came to the conclusion that a=d

Answer 13

Try canceling \((m-n)\) from both sides.

Answer 14

means?

Answer 15

\(an-nd=am-md\)
\(an - nd - am + md = 0\)
\(d(m-n) - a(m-n) = 0\)
\((d-a)(m-n) = 0\)

Thus, either, \(m=n\) or \(a=d\)
But, \(m\neq n\) [Should be specified usually, or the question talks about the same terms]
So, \(a=d\) :)

Answer 16

Oh I seee!