Question

If the Mth term of of AP is 1/N and Nth term of AP is 1/M . Show that sum of MN terms 1/2(M+N).......​

Answer 1

Step-by-step explanation:

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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)

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Answer 2

Step-by-step explanation:

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

Therefore,

$Am = a+ (m-1)d= \frac{1}{ n} ….(i). \\ An = a+(n-1)d= \frac{1}{m} ….(ii)$

Subtracting equation (ii) from (i)

$d[(m-1)-(n-1)] = \frac{1}{n} - \frac{1}{m} , \\ </h3><h3></h3><h3>d(m-n) = \frac{m - n}{mn} , \: \\ </h3><h3>d = \frac{1}{mn} ….(iii) \:$

Substituting equation (iii) in (i)

$a+ \frac{m - 1}{mn} = \frac{1}{n} \\ </p><p></p><p>a = \frac{1}{n} [ \frac{1-(m-1)}{m} ], \\ </p><p></p><p>a = \frac{1}{mn} ….(iv)$

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

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

$= > \frac{1}{mn} + \frac{(mn-1)}{mn} \\ = > \frac{1}{mn} [1+(mn-1)] = \frac{mn}{mn} = 1,</p><p></p><p>$

then the mn term = 1

sum of mn term :-

$Amn \: = \frac{mn}{2} ( \frac{2}{mn} + (mn-1) \frac{1}{mn} ) \\ </p><p></p><p>= 1 + \frac{mn}{2} - \frac{1}{2} \\ </p><p></p><p>= \frac{mn}{2} + \frac{1}{2} \\ </p><p></p><p>= \frac{1}{2} (mn + 1) \\ </p><p>hope \: it \: helps \: you$