1.IF Pth term of an A.P is q and qth tem of an AP is p then prove that (p+q) term is 0

let a & d are the first term & common differnece respectively then

pth term = a+(p-1)d = q                       ...............1

qth term = a+(q-1)d = p                     ...............2

1-2

 q-p = (p-q)d

d=-1

now 

    (p+q)th term = a + (p+q-1)d

putting value of a from eq 1

         (p+q)th term = q - (p-1)d + (p+q-1)d                  (d =-1)

                            =0                            

Approved

The pth term of an AP is = a+(p-1)d

where a is the first term of the AP and  d is the common difference

as pth term is equal to q therefore q=a+(p-1)d .....1

similarly qth term is equal to p

p=a+(q-1)d .......2

solve the two equations to get the value of a and d. It will come

a=p+q-1

d=-1

now (p+q)th term is =a+(p+q-1)d

put the values of a and d

=(p+q-1)+(p+q-1)(-1)

=(p+q-1)-(p+q-1)

=0