Question

If the sum of first m terms of an ap be n and the sum of it's first n term be m then show that the sum of it's first m+n terms is _(m+n)

Answer 1

Hope it helps....!!!!

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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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#Be Brainly✌️

Answer 2

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$let \: a \: be \: the \: first \: term \: and \: d \: be \: the \: common \: difference \: of \: the \: given \: ap \\ s ^{m} \implies \: n \implies \frac{m}{2}[2a + (m - 1)d]= n \\ \implies2am + m(m - 1)d = 2n.......... (1) \\ and \: s ^{n} \implies \: m \implies \frac{n}{2} [2a + (n - 1)d] = m \\ \implies2an + n(n - 1)d = 2m...........(2) \\ on \: subtracting \: (2) \: from \: (1) \: we \: get \\ 2a(m - n) + [(m {}^{2} - n {}^{2} ) - (m - n)]d = 2(n - m ) \\ \implies (m - n)[2a + (m + n - 1)d] = 2(m + n) \\ \implies2a + (m + n - 1)d = - 2 .........(3)\\ sum \: of \: the \: first \: (m + n) \: terms \: of \: the \: given \: ap = \frac{(m + n)}{2} \times [2a + (m + n - 1)d]\\ = \frac{m + n}{2}.( - 2) = - (m + n)$