Question

Let first term of given A.P. be a and common

difference be d.

\ am = a + (m – 1)d = 1

n ...(i) ½

and an = a + (n – 1)d = 1

m ...(ii) ½

On subtracting (ii) from (i) we get

(m – n)d = 1 1

n m

m n

mn − = − 1

or, d = 1

mn

and a = 1

mn

Now Smn = mn

mn mn 2 mn 2 1 1 1 · ( + − ) 



 





= mn

mn

mn

2 mn mn

2 1

+ − 



 





Smn = mn

2 mn

1

+ 1 



 





= 1

2

[ ] mn + 1

Hence, the sum of first mn terms =

1

What is the question

​

Answer 1

Answer:

Where is the question? In question u have given full solution