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


9 years ago

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

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

Sn = 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 , Sm+n = (m+n)/2 [ 2a + (m+n-1)d ]
Sm+n (2/m+n) = 2a + (m+n-1)d               .......................4
eq 4 - eq 2
Sm+n (2/m+n) - 2m/n =  d(m)
putting value of d from eq 3 we get
Sm+n (2/m+n) = -2            or
Sm+n = -(m+n)
this is the required result


9 years ago
							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

Be very carefull while simplification as its a lenghty problem...

9 years ago 9 years ago
							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/21 = -2a/2 - (m+n)/2 + d/21 = -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 //

3 months ago
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