Question

if the Mth term of an AP be 1/N and its Nth term be 1/M then show that its (MN)th term is 1

Answer 1

let a be the first term and d be the common difference of the given AP. Then,
TM = a +(M - 1) d
1/N = a+ (M - 1) d
1/N = a+ Md - d ....... equation (1)
Now,
TN = a + (N - 1) d
1/M = a + (N - 1 ) d
1/M = a + Nd - d ........ equation (2)
On subtracting (2) from (1) ,we get:-
1/N = a + Md -d ....eq(1)
1/M = a + Nd -d ....eq(2)

= 1/N - 1/M = Md - Nd
= 1/N - 1/M = d( M- N)
= M-N / MN = d (M-N)
= 1/ MN = d
= d = 1/MN
putting d = 1/MN in eq (1):-
1/N = a + Md - d
1/N = a +( M × 1/MN) - 1/MN
1/N = a + 1/N - 1/MN
1/N - 1/N = a - 1/ MN
a = 1/MN
Now,
tMN= a + (MN -1) d
= 1/MN + (MN - 1) 1/MN
= 1/MN + 1 - 1/ MN
= 1
So, t(MN) = 1 proved...

Answer 2

Answer:

et mth term of AP be ‘Am’ and nth term of AP be ‘An’

Therefore, Am = a+ (m-1)d=1/n ….(i) ,,,,,,,,,,,,,,,,,,,,,,,,,,,,

An = a+(n-1)d=1/m ….(ii)

Subtracting equation (ii) from (i)

d[(m-1)-(n-1)] = 1/n-1/m,

Step-by-step explanation: