Question

If the mth term of an AP is 1/n and the nth term is 1/m Thame show that the sum of MNthterms is mn + 1/2

Answer 1

am = 1/n , an = 1/m

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

now, 1/n - 1/m = a+ (m-1)d -{a+ (n-1)d}

(m-n)/mn = md-d - nd +d

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

d = 1 / (mn)

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

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

1 /n = {amn + m-1} / mn

m = amn+m-1

a = 1 / mn

amn = a+ (mn-1)d

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

= {1 + mn -1} / mn

= mn / mn

amn= 1

Answer 2

Your question needs a correction.

Correct Question : If the mth term of an AP is 1/n and the nth term is 1/m. Then show that the sum of mn terms is 1 / 2 mn + 1 / 2 i.e. 1 / 2( mn + 1 ).


Let a be the first term and d be the common difference between of the A.P., then

$a_m =\dfrac{1}{n}\quad and \quad a_n =\dfrac{1}{m}$

$\implies a + (m - 1)d = \dfrac{1}{n \: } \: \: \: \: \: \: \: \: \: \: \: - - :(1) \\ \\ \quad and \quad a + ( n- 1)d = \dfrac{1}{m} \: \: \: \: \: \: - - :(2)$



On subtracting ( ii ) from ( i ) :

= > a + ( m - 1 )d - [ a + ( n - 1 )d ] = 1 / n - 1 / m

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

= > ( m - 1 )d - ( n - 1 )d = ( m - n ) / mn

= > d[ m - 1 - n + 1 ] = ( m - n ) / mn

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

= > d = 1 / mn



Substituting the value of d in ( 1 ) :

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

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

= > a + 1 / n - 1 / mn = 1 / n

= > a = 1 / mn



Hence,
$S_{mn}=\dfrac{mn}{2} [2a+(mn-1)d]$

$S_{mn}=\dfrac{mn}{2} \bigg[ \bigg\{2\times \dfrac{1}{mn} \bigg\} + ( mn - 1 )\bigg\{\dfrac{1}{mn}\bigg\} \bigg] [tex]S_{mn}=\dfrac{1}{2}(mn+1)$


Hence proved.
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