麻烦帮我证明数列，谢谢！

（1）q=a+(p-1)d, p=a+(q-1)d
两式相减：d=-1
a=p+q-1
则 a(p+q)=a+(p+q-1)d=(p+q-1)-(p+q-1)=0
（2）m=(n/2)[2a+(n-1)d], n=(m/2)[2a+(m-1)d]
两式相减：d=-2(m+n)/(mn)
S(m+n)=[(m+n)/2][2a+(m+n-1)d]
=[(m+n)/2][(2n/m)-(n-1)d+(m+n-1)d]
=[(m+n)/2][2a+md]
=-(m+n)
（3）S(n)=S(m
2a=-(m+n-1)d
S(m+n)=0

a(n)=a + (n-1)r. n = 1,2,...
S(n)=na +n(n-1)r/2.

1,
q = a(p) = a + (p-1)r,
p = a(q) = a + (q-1)r.
q-p = r(p-1-q+1)=r(p-q),
(p-q)(r+1)=0.
r=-1.
q = a-p+1,
a = p+q-1.

a(p+q)=a + (p+q-1)r=a-a=0.

2,
n = S(m) = ma + m(m-1)r/2.
m = S(n) = na + n(n-1)r/2.
n-m=a(m-n)+r/2[m^2-n^2 -m+n]=(m-n){a + r/2[m+n-1]},
-1=a+r/2[m+n-1],

S(m+n)=(m+n)a + (m+n)(m+n-1)r/2 = (m+n){a + r/2[m+n-1]}=-(m+n).

3,
ma + m(m-1)r/2 = S(m) = S(n) = na + n(n-1)r/2,
0 = a(m-n) + r/2[m^2 - n^2 - m + n] = (m-n){a + r/2[m+n-1]},
0 = a + r/2[m+n-1],