Question

The sum of first 'm' terms of an AP is 'n' and sum of first 'n' terms is 'm'. Show that the sum of first (m+n) terms is [-(m+n)]

Answer 1

Sm = n

m/2(2a+(m-1)d)=n. (equation no 1)


Sn = m

n/2 ( 2a+(n-1)d)=m (equation no 2),

equation 2 - 1

2an/2 + ( n^2 - n ) d - 2am - ( m^2- m ) d = m - n /2

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

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

2a + ( m + n -1) d = -2 ( equation no 3)

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

Sm+n = m+n / 2 × -2

Sm+n = -(m+n)

Answer 2

n=m/2{2a+(m-1)d}......    [1]
m=n/2{2a+(n-1)d}.....      [2]
[1]-[2]
n-m= m/2{2a+(m-1)d} - n/2{2a+(n-1)d}
      =am + m^2d/2 - md /2 - an - n^2d/2 + nd/2
      =a(m-n) + d/2 [ m^2-m-n^2+n]
      =a(m-n) + d/2 [ (m+n) (m-n) - (m-n) ]
      =a(m-n) + d/2 [ (m-n)(m+n-1) ]
      =(m-n) [ a +d/2(m+n-1) ]
  -(m-n)= (m-n) [a + d/2(m+n-1)]
   -1= 2a+d(m+n-1)/2
   -2= 2a + d(m+n-1)
since, s(m+n) = (m+n)/2 * [2a+ (m+n-1)d ]
s(m+n) = (m+n)/2* -2
s(m+n) = -(m+n)