Question

If the sum of first m terms of an ap is n and the first n terms of an ap is m then prove that
sum of its first (m+n) = -(m+n)

Answer 1

Step-by-step explanation:

given

sum of first n terms= m

sum of first m term=n

=> n/2[ 2a+(n-1) d]= m

=> 2a+ (n-1)d= 2m/n......................(1)

=> m/2[2a+(m-1)d]= n

=> 2a+ (m-1)d= 2n/m ......................(2)

(1)- (2)

=>. (n-m)d= 2m/n - 2n/m= 2(m²-n²)/mn

=>. d=.[ 2( m+n) (m-n)]/ mn ( n-m,)

=> d= - 2(m+n)/mn ....................…(3)

substituting in (1) and simplifying

(m+n)= -(m+n)

Proved

Answer 2

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

Hence (m+n)=-(m+n)

proved