Question

arithmetic progression
 if the sum of first p terms of an a.p is equal to the sum of first q terms then show that the sum of its first (p+q) terms is zero where pis not equal to q            

Answer 1

Let the first term = a
common difference = d 
Sum of first n terms is S(n) = (n/2) [ 2a + (n-1) d ]

S(p) = S(q)
⇒ (p/2) [ 2a + (p-1)d ] = (q/2) [ 2a + (q-1)d ] 
⇒ 2ap + (p²-p)d = 2aq + (q²-q)d 
⇒ 2ap + (p²-p)d - 2aq - (q²-q)d =0
⇒ 2ap + p²d - pd - 2aq - q²d - qd =0
⇒ 2a(p-q) + (p²-q²)d - (p-q)d = 0 
⇒ 2a(p-q) + [ (p²-q²) - (p-q) ]d = 0 
⇒ 2a(p-q) + [ (p-q)(p+q) - (p-q) ]d = 0 
⇒ 2a(p-q) + (p-q)[ (p+q) - 1]d = 0 
⇒ (p-q) [ 2a + (p+q-1)d ] = 0
⇒ 2a + ( p+q-1)d = 0            (as p and q are not equal, p-q is not equal to 0)

Sum of (p+q) terms is

S(p+q) = [ (p+q)/2 ] [ 2a + (p+q-1)d ] 
           = [ (p+q)/2 ] × 0 
           = 0 (proved)  

Answer 2

                  S(p) = S(q)
p/2 [2a + (p-1) d] = q/2 [2a + (q - 1) d]
   2ap + (p - 1)dp =  2aq + (q - 1)dq
2ap - 2aq + p²d - pd - q²d + qd = 0
  2a(p - q) + d(p² - q²) - d(p - q) = 0
         (p - q) [2a + d(p + q) - d] = 0
                   [2a + d(p + q) - d] = 0
                   [2a + d(p + q - 1)] = 0 ...............(1)
  
S(p+q) = (p+q)/2 [2a + d(p + q - 1)]
           = (p+q)/2 [0]   .........[from (1) ]
           = 0