Question

if the mth term of an AP is 1/n and nth term of an AP is 1/m then show that the sum of n term is1/2 (mn+l)

Answer 1

$\sf{GIVEN:-}$
$\sf{a_{\tiny{m}}}$ = 1/n

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

$\sf{a_{\tiny{n}}}$ = 1/m

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

From--------( 1 ) &--------( 2 )

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

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

d = 1/nm [ put in -----( 1 ) ]

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

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

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

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

a = 1/mn

we have to find $\sf{S_{mn}}$.

$\sf{S_{mn}}$ = mn/2[2a + (mn - 1)d]

$\sf{S_{mn}}$ = mn/2[2/mn + (mn - 1)*1/mn ]

$\sf{S_{mn}}$ = mn/2[2/mn + (1 - 1/mn)]

$\sf{S_{mn}}$ = mn/2[2/mn - 1/mn + 1 ]

$\sf{S_{mn}}$ = mn/2[(2 - 1)/mn + 1]

$\sf{S_{mn}}$ = mn/2[1/mn + 1]

$\sf{S_{mn}}$ = mn/2[(1 + mn)/mn]

$\sf{S_{mn}}$ = 1/2[1 +mn]