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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)

.....

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Answer 2

Answer:

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$\huge\green{\underline{\pink{\bigstar{\blue{\mathfrak{Your \: Answer}}}}}}}}$

Let a be the first term and d be the common difference of the given AP. Then,

Sn=2n[2a+(n−1)d]

Given,

Sm=n

2m[2a+(m−1)d]=n

2am+m(m−1)d=2n        .......(1)

Sn=m

2n[2a+(n−1)d]=m

2an+n(n−1)d=2m          ..........(2)

On subtracting 2 from 1, we get,

2a(m−n)+[(m2−n2)−(m−n)]d=2(n−m)

(m−n)[2a+(m+n−1)d]=2(n−m)

2a+(m+n−1)d=−2          ..........(3)

Sum of (m+n) terms of the given AP

Sm+n=2m+n[2a+(m+n−1)d]

           =m+n/2(-2)= --(m+n)

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