Question

if the mth term of an ap be 1/n and its nth term be 1/m then show that its (mn)th terms is 1

Answer 1

am=a+(m-1)d
1/n=a+(m-1)d. eq1
an=a+(n-1)d
1/m=a+(n-1)d. eq2
eq1-eq2
1/n-1/m=a+(m-1)d-a-(n-1)d
m-n/mn=dm-d-dn+d
(m-n)/mn=d(m-n)
d=1/mn
am=a+(n-1)d
1/n=a+dn-d
1/n=a+n/mn-1/mn
1/n-1/n=a-1/mn
a-1/mn=0
a=1/mn
amn=a+(mn-1)d
=1/mn+(mn-1)1/mn
=1/mn+mn/mn-1/mn
=0+1
=1
MNth term of an AP=1

Answer 2

Answer:

Step-by-step explanation:

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

a + (m – 1)d = 1/n .........(i)

a + (n – 1)d= 1/m ........(ii)

On solving (i) and (ii) we get

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

(m-n)d = m-n/mn

d = 1/mn

Therefore,

1m 1mm11

a

nmnmnmn





mn

mn11

S2(mn1)

2mnmn









=

1

2

[2 + (mn – 1)] =

1

2

(mn + 1)