arthimetic prgression
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Hi friend,
Let a be the first term and D be the common difference of the given AP.
Then,
Sm = n
M/2 × [ (2A + (M-1) × D)] = n
=> 2AM + M(M-1) × D = 2n.......(1)
And,
Sn = M
N/2 × [ 2A + (N-1) × D] = m
=> 2AN + N(N-1) × D = 2M.....(2)
On subtracting (2) from (1) we get:
2A(M-N) + [(M²-N²)] - (M-N)] × D = 2(N-M)
=> (M-N) [ 2A + (M+N-1)× D] = 2(N-M)
=> 2A + (M+N-1) × D = -2
=> 2A + (M+N-1) × D = -2......(3)
Sum of the first (M+N) terms of the given AP =
=> (M+N)/2 × [ 2A + (M+N-1) × D]
=> (M+N)/2 × -2
=> -(M+N)
Hence,
The sum of the first (M+N) terms of the given AP is -(M+N).
HOPE IT WILL HELP YOU...... :-)
Let a be the first term and D be the common difference of the given AP.
Then,
Sm = n
M/2 × [ (2A + (M-1) × D)] = n
=> 2AM + M(M-1) × D = 2n.......(1)
And,
Sn = M
N/2 × [ 2A + (N-1) × D] = m
=> 2AN + N(N-1) × D = 2M.....(2)
On subtracting (2) from (1) we get:
2A(M-N) + [(M²-N²)] - (M-N)] × D = 2(N-M)
=> (M-N) [ 2A + (M+N-1)× D] = 2(N-M)
=> 2A + (M+N-1) × D = -2
=> 2A + (M+N-1) × D = -2......(3)
Sum of the first (M+N) terms of the given AP =
=> (M+N)/2 × [ 2A + (M+N-1) × D]
=> (M+N)/2 × -2
=> -(M+N)
Hence,
The sum of the first (M+N) terms of the given AP is -(M+N).
HOPE IT WILL HELP YOU...... :-)
chahul:
can solve 2 more question pls
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