Question

If the mth term of an A.P. is 1/n and the nth term is 1/ m , show the sum of its first(mn) terms is 1/ 2 (mn + 1).

Answer 1

Let ‘a’ and ‘d’ be the first term and common difference of the given A.P.
GIVEN : mth term=1/n & nth term=1/m.

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

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

On subtracting eq. (2) from (1),
1/n-1/m = (m - n )d
(m-n) /mn = (m-n)d
d = (m-n) /mn / (m - n)
d=1/mn………………………..(3)

Putting the value of d in equation (1)
1/n = a+(m-1)d
1/n = a + (m-1) / mn
1/n = a+ m/ mn - 1/mn
1/n =  a+ 1/ n - 1/mn
1/n - 1/n + 1/mn = a
1/mn = a
a=1/mn………………….(4)

Sum of mn terms,
S(mn) = mn/2 [2a +(mn -1)d]

[ Sn = n/2[2a +(n -1)d]
S(mn) =  mn/2 [2/mn +(mn -1)×1/mn]

[ From eq 3 & 4]
S(mn) =  mn/2 [2/mn + (mn/mn -1/mn)
S(mn) =  mn/2 [2/mn +(1 -1/mn)]
S(mn) =  mn/2[ 2/mn - 1/mn +1]
S(mn) =  mn/2 × [1/mn +1]
S(mn) = mn/2mn + mn/2
S(mn) = ½ + mn/2
S(mn) = ½(1+mn)

S(mn) = ½ (mn +1)

HOPE THIS WILL HELP YOU....

Answer 2

Answer:

Let 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,

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

d = 1/mn ….(iii)

Substituting equation (iii) in (i)

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

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

a = 1/mn ….(iv)

Now Amn i.e the mnth term of AP = a+(mn-1)d,

Substitute equation (iii) and (iv) in Amn,

1/mn+(mn-1)/mn = 1/mn[1+(mn-1)] = mn/mn = 1,

then the mn term = 1

sum of mn term :-

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

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

= mn /2 + 1/2

= 1/2 (mn + 1)