Question

The mth term of an AP is 1/n and nth term is 1/m then show that sum of mn term is 1/2(mn +1)

Answer 1

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)

Step-by-step explanation:

Answer 2

Step-by-step explanation:

HEY MATE................

$am = \frac{1}{n} = a + (m - 1)d........(1) \\ \\ an = \frac{1}{m} = a + (n - 1)d........(2) \\ \\ subtracting \: eqn.2from \: eqn1 \: we \: get \: \\ \\ (m - n)d = \frac{1}{n} - \frac{1}{m} \\ \\ (m - n) d = \frac{m - n}{mn} \\ \\ d = \frac{1}{mn} \\ \\ putting \: this \: value \: of \: d \: in \: eqn \: 1 \: we \: get \: \\ \\ a + (m - 1) \frac{1}{mn} = \frac{1}{n} \\ \\ a + \frac{1}{n} - \frac{1}{mn} = \frac{1}{n} \\ \\ a = \frac{1}{mn} \\ \\ s(mn) = \frac{mn}{2}(2a + (mn - 1)d) \\ \\ s(mn) = \frac{mn}{2} ( \frac{2}{mn} + (mn - 1) \times \frac{1}{mn)} \\ \\ s(mn) = \frac{1}{2} (mn + 1)$

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