if the sum of first m terms of A.P is n and the sum of first n terms is m.then show that the sum of its first (m+n) terms is -(m+n)
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Let a be the first term and d be c.d. of the A P .ThenSm=nn= m/2{2a+ (m-1)d} 2n= 2am+ m( m-1)d. ........(1)andSn= mm= n/2{2a+(n-1)d}2m= 2an+ n(n-1)d. ...........(2)Subtracting eq.(2)- (1), we get2a(m-1)+{m(m-1)- n(n-1)}d = 2n-2m2a(m-n) +{(m^2-n^2)-(m-n)}d = -2(m-n)2a +(m+n-1) d = -2. [On dividing both sides by ( m-n)]………(3) Now,Sm+n=m+n/2{2a +(m+n-1)d}Sm+n=m+n/2(-2) ………[using (3)]Sm+n=-(m+n)
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Let a be the first term and d be c.d. of the A P .Then
Sm=n
n= m/2{2a+ (m-1)d}
2n= 2am+ m( m-1)d. ........(1)
and
Sn= m
m= n/2{2a+(n-1)d}
2m= 2an+ n(n-1)d. ...........(2)
Subtracting eq.(2)- (1), we get
2a(m-1)+{m(m-1)- n(n-1)}d = 2n-2m
2a(m-n) +{(m^2-n^2)-(m-n)}d = -2(m-n)
2a +(m+n-1) d = -2. [On dividing both sides by ( m-n)]………(3)
Now,
Sm+n=m+n/2{2a +(m+n-1)d}
Sm+n=m+n/2(-2) ………[using (3)]
Sm+n=-(m+n)
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Sm=n
n= m/2{2a+ (m-1)d}
2n= 2am+ m( m-1)d. ........(1)
and
Sn= m
m= n/2{2a+(n-1)d}
2m= 2an+ n(n-1)d. ...........(2)
Subtracting eq.(2)- (1), we get
2a(m-1)+{m(m-1)- n(n-1)}d = 2n-2m
2a(m-n) +{(m^2-n^2)-(m-n)}d = -2(m-n)
2a +(m+n-1) d = -2. [On dividing both sides by ( m-n)]………(3)
Now,
Sm+n=m+n/2{2a +(m+n-1)d}
Sm+n=m+n/2(-2) ………[using (3)]
Sm+n=-(m+n)
Read more on Brainly.in - https://brainly.in/question/1048328#readmore
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