Question

5. If the sum of first m terms of an A – P be n and the sum of its first n terms be m, then show

that the sum of its first (m + n)th term is – (m + n)​

Answer 1

Answer:

Given ,sum of first m terms of AP is n

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

=[2am+m(m-1)d]=2n.......(1);

Also,sum of first n terms is m

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

=> [2an+n(n-1)=2m......(2);

=2a(m-n)+[(m²-n²)-(m-n)d]=2(n-m)

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

=2a+(m+n-1)d=-2

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

Hence the sum of its first (m+n)terms is -(m+n)

Answer 2

Given :

• the sum of first m terms of Ap be n

• The sum of its first n terms be m

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To prove :

The sum of its first (m+n)th term = -(m+n)

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Formula :

Sum of nth terms =

$\\ \red \bigstar \sf \: sn = \frac{n}{2} (2a + (n - 1)d)$

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Solution :

• Let a be the first term and d is the common difference of the given A.P .Then,

$\\ \implies \sf \: sn = \frac{n}{2} (2a + (n - 1)d)$

Given,

$\\ \implies\sf {S_{m}=n}$

$\\ \implies\sf{\dfrac{m}{2}\bigg( 2a+(n-1)d\bigg)=n}$

$\\ \implies\sf{2am+m(m-1)=2n \: \: \: ....(1)}$

$\\ \implies\sf{S_{n} =m}$

$\\ \implies\sf{\dfrac{n}{2} \bigg(2a+(n-1)d\bigg)=m}$

$\\ \implies \sf{2an+n(n-1)d=2m \: \: \: ....(2)}$

On subtracting equations (2) from (1) we get,

$\\ \implies\sf{2a(m-n) +\bigg((m^2-n^2)-(m-n)\bigg)=2(n-m) }$

$\\ \implies \sf \: (m - n) \bigg(2a + (m + n - 1)d \bigg) = 2(n - m)$

$\\ \implies \sf \: 2a + (m + n - 1) = - 2$

$\\ \implies \sf \: 2a + (m + n - 1) = - 2 \: \: .....(3)$

Sum of (m+n)th terms of the given AP

$\\ \implies\sf{S_{m+n} =\dfrac{m+n}{2} \bigg(2a+(m+n-1)d\bigg)}$

$\\ \implies\sf{S_{m+n} =\dfrac{m+n}{2} \times (-2)}$

$\\ \implies\sf{S_{m+n} =-(m+n)}$

$\\ \implies\boxed{\red\bigstar\sf{S_{m+n} =-(m+n)}}$

Hence the sum of (m+n)th term is -(m+n)