If 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)
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Answers
EXPLANATION.
→ Mth term of an Ap = 1/n .......(1)
→ Nth term of an Ap = 1/m .......(2)
→ To show that sum of mn terms is
1/2 ( mn + 1 ).
→ Nth term of an Ap = An = a + ( n - 1 )d
→ Tm = a + ( m - 1 )d = 1/n ......(1)
→ Tn = a + ( n - 1 )d = 1/m ......(2)
→ From equation (1) and (2) we get,
subtract both equation.
→ ( m - n)d = 1/n - 1/m
→ ( m - n)d = m - n / nm
→ d = 1/nm.
→ put the value of d = 1/nm in equation (1)
we get,l
→ a + ( m - 1 )1/nm = 1/n
→ a + m/nm - 1/nm = 1/n
→ a + 1/n - 1/nm = 1/n
→ a - 1/nm = 0
→ a = 1/nm
→ sum of mnth terms.
→ sum of Nth terms of an Ap.
→ Sn = n/2 [ 2a + ( n - 1 )d ]
→ Smn = mn/2 [ 2*1/mn + ( mn - 1 ) 1/nm ]
→ Smn = mn/2 [ 2/mn + mn/mn - 1/mn ]
→ Smn = mn/2 [ 1/mn + 1 ]
→ Smn = mn/2 [ 1 + mn/mn ]
→ Smn = 1/2 [ 1 + mn ]
Answer:
EXPLANATION.
→ Mth term of an Ap = 1/n .......(1)
→ Nth term of an Ap = 1/m .......(2)
→ To show that sum of mn terms is
1/2 ( mn + 1 ).
→ Nth term of an Ap = An = a + ( n - 1 )d
→ Tm = a + ( m - 1 )d = 1/n ......(1)
→ Tn = a + ( n - 1 )d = 1/m ......(2)
→ From equation (1) and (2) we get,
subtract both equation.
→ ( m - n)d = 1/n - 1/m
→ ( m - n)d = m - n / nm
→ d = 1/nm.
→ put the value of d = 1/nm in equation (1)
we get,l
→ a + ( m - 1 )1/nm = 1/n
→ a + m/nm - 1/nm = 1/n
→ a + 1/n - 1/nm = 1/n
→ a - 1/nm = 0
→ a = 1/nm
→ sum of mnth terms.
→ sum of Nth terms of an Ap.
→ Sn = n/2 [ 2a + ( n - 1 )d ]
→ Smn = mn/2 [ 2*1/mn + ( mn - 1 ) 1/nm ]
→ Smn = mn/2 [ 2/mn + mn/mn - 1/mn ]
→ Smn = mn/2 [ 1/mn + 1 ]
→ Smn = mn/2 [ 1 + mn/mn ]
→ Smn = 1/2 [ 1 + mn ]
Step-by-step explanation: