Question

Pls answer this question and also give the steps
wrong answers will be reported....

Answer 1

=================== HEYA THERE ==================




----------------  HERE IS YOUR ANSWER ----------------



mth term=1/n and nth term=1/m.



that is why , 

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

and

                     a+(n-1)d=1/m...........(2).


subtracting equation (1) by (2) we get,


md-d-nd+d=1/n-1/m


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


=>d=1/mn. 




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 let k be the mn th term of the kP


k+(mn-1)d=1/mn+1+(-1/mn)=1 





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# brainly2222



@ computer science 

Answer 2

Let a be the first term and d be the common difference.

Given that mth term of an AP am = 1/n.

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

Given that nth term of an AP an = 1/m.

an = a + (n - 1) * d = 1/m  ---------------------- (2)

On solving (1) & (2), we get

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

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

--------------------------

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

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

d = 1/mn.


Substitute d = 1/mn in (1), we get

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

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

a = 1/n - (m - 1)/mn

a = 1/mn.


Hence, a = 1/mn and d = 1/mn.


Therefore sum of mn terms = mn/2(2a + (mn - 1) * d)

                                               = mn/2(2/mn + (mn - 1)/mn)

                                               = mn/2(2/mn + mn/mn - 1/mn)

                                               = mn/2(2/mn + 1 - 1/mn)

                                               = mn/2(1/mn + 1)

                                               = mn/2(mn + 1)/mn

                                               =  1/2(mn + 1).


Hope this helps!