Question

16. If the sum of first m terms of an A.P.is
same as the sum of its first n termss
show that the sum of its first (m+n)terms
is zero​

Answer 1

Answer:

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

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

Sm=Sn(given)

-›m/2[2a+(m-1)d]=n/2[2a+(n-1)d]

Multiply 2 on both the sides

-›2×m/2[2a+(m-1)d]=2×n/2[2a+(n-1)d]

-›m[2a+(m-1)d]=n[2a+(n-1)d]

-›2am+(m-1)dm=2an+(n-1)dn

-›2am+dm²-dm=2an+dn²-dn

-›2am-2an=dn²-dm²-dn+dm

-›2a(m-n)=d(n²-m²-n+m)

-›2a(m-n)=d[(n²-m²)-(n-m)]

-›2a(m-n)=(-d)[(m²-n²)-(m-n)]

-›2a(m-n)=(-d)[(m-n)(m+n)-(m-n)]

-›2a(m-n)=(-d)(m-n)(m+n-1)

-›2a=(-d)(m+n-1)

-›2a=(-dm)-dn+d

-›2a+dm+dn-d=0

-›2a+(m+n-1)d=0

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

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

S(m+n)=(m+n)/2×0

S(m+n)=0

Hence proved