Sm=n , Sn = m then S(m+n) = ?

Question

vikas askiitian expert · Answer

let common differnece of this series is d & first term is a then

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

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

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

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

subtracting both equations ,

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

d = -2[n+m/nm] ...................3

now , S_m+n = (m+n)/2 [ 2a + (m+n-1)d ]

S_m+n (2/m+n) = 2a + (m+n-1)d .......................4

eq 4 - eq 2

S_m+n (2/m+n) - 2m/n = d(m)

putting value of d from eq 3 we get

S_m+n (2/m+n) = -2 or

S_m+n = -(m+n)

this is the required result

Dev Kar · Answer

Sm=n m/2(2a+(m-1)d)=n 2a+md-d=2n/m......eqn 1. Sn=m Similarly, 2a+nd-d=2m/n....eqn2. Subtracting 2 from 1, we get- d=-2(m+n)/mn Sub. this value of d in eqn 1. 2a+md-d=2n/m [2a+m*-2(m+n)+2(m+n)]/mn=2n/m Making 2a the subject of the eqn, we get- 2a=[2n(sq)+2m(sq)+2mn-2m-2n]/mn.......(sq) stands for square, ex:2n(sq)=2n*n....pls note! S(m+n)=(m+n)/2*[2a+(m+n-1)d] Now substitute the value of 2a and d we got earlier in the above eqn- You will get- (m+n)/2*[2n*n+2m*m+2mn-2m-2n-2m*m-2n*n-4mn+2m+2n]/mn..... The final eqn you get on simplification is- (m+n)/2*(-2mn)/m The Answer is -m-n.... Be very carefull while simplification as its a lenghty problem...

mycroft holmes · Answer

ankit singh · Answer

Let a is the first term and d is the common difference .
S_n=m=\frac{n}{2}[2a+(n-1)d]\\\\S_m=n=\frac{m}{2}[2a+(m-1)d]
now, S_n-S_m=2a[\frac{n}{2}-\frac{m}{2}]+\frac{n^2}{2}-\frac{m^2}{2}-\frac{d}{2}[n-m] = m - n \\
(m - n) = -2a(m-n)/2 -(m-n)(m+n)/2+(m-n)d/2
1 = -2a/2 - (m+n)/2 + d/2
1 = -1/2 {2a + (m+n-1)d} ---------(1)

S_{m+n} =\frac{m+n}{2}[2a+(m+n-1)d] \\
from equation (1)
S_{m+n} = -(m+n)
hence, proved //

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Sm=n , Sn = m then S(m+n) = ?

Sm=n , Sn = m then S(m+n) = ?

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