Question

arthimetic prgression

Answer 1

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...... :-)