Question

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)

Answer 1

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)

Answer 2

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)