Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

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

Grade:11

4 Answers

vikas askiitian expert
509 Points
10 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

 

Dev Kar
29 Points
10 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 The Answer is -m-n.... Be very carefull while simplification as its a lenghty problem...
mycroft holmes
272 Points
10 years ago

ankit singh
askIITians Faculty 614 Points
one year 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/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 //

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free