If the sum of the first m terms of an AP is n and first n terms is m then show that sum of its first (m+n) terms is -(m+n)
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Answered by
940
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)
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)
Answered by
17
Answer:
-(m+n)
Step-by-step explanation:
n/2{2a+(n-1)d}=m
⇒2a+(n-1)d=2m/n .....(i)
similarly
2a+(m-1)d=2n/m .....(ii)
subtracting (ii) from (i)
(n-1-m+1)d=2m/n-2n/m
⇒d= -2(m+n)/mn .....(iii)
from (i) and (iii) we have
2a=2m/n - (n-1) × -2(m+n)/mn
⇒2a = 2m/n + 2(n-1)(m+n)/mn .....(iv)
S(m+n)= m+n/2{ 2m/n + 2(n-1)(m+n)/mn + (m+n-1) × -2(m+n)/mn}
=m+n/2{2m/n + 2(m+n)/mn × (n-1-m-n+1)}
=m+n/2{2m/n -(2m²+2mn)/mn}
=m+n/2{(2m²-2m²-2mn)/}
=m+n/2 × -2mn/mn
= -(m+n)
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