UPSC NDA Math Model Paper 2

Every term an HP is the reciprocal of the corresponding terms of an AP.
Let a be the first term and d be the common difference of the AP.
According to the question:
Tm=n
⇒

1

a+(m−1)d

=n
⇒a+(m−1)d=

1

n

..... (1)
Tn=m
⇒

1

a+(n−1)d

=m
⇒a+(n−1)d=

1

m

....... (2)
Subtracting equation (2) from equation (1), we get :
(m−n)d=

1

n

−

1

m

⇒d=

1

mn

.....(3)
Substituting this value of d in any of the equations (1) or (2), we get:
⇒a+(m−1)(

1

mn

)=

1

n

⇒a=

1

n

−

m−1

mn

⇒a=

1

mn

.......(4)
The (m+n)th term of the AP will be:
am+n=a+(m+n−1)d
Using the values in equations (3) and (4), we get:
⇒am+n=

1

mn

+(m+n−1)(

1

mn

)
⇒am+n=

m+n

mn

And the (m+n)th term of the HP will be:
Tm+n=

1

am+n

=

mn

m+n