Question

If the sum of firt in terms of an AP be n and
the sum of its first in terms be m then show
that the sum of its first on in terms is (m +n)term is -m+n
​

Answer 1

Correct QuestioN :

In an A.P, the sum of m terms of an AP is n and sum of n terms of AP is m, then prove that sum of (m+n) terms of AP is - ( m + n )

To ProvE :

Sm + n = - ( m + n )

SolutioN :

$\tt \dagger \: \: \: \: \: S_n = \dfrac{n}{2}\Bigg[2a+(n- 1 )d \Bigg]$

Where as,

• Sn → Sum of nth terms.
• a → First term.
• d → Common difference.

☛ Case 1.

$\tt \dagger \: \: \: \: \: S_m= \dfrac{m}{2}\Bigg[2a+(m- 1 )d \Bigg]$

$\tt : \implies n= \dfrac{m}{2}\Bigg[2a+(m- 1 )d \Bigg]$

$\tt : \implies n= \dfrac{m}{2}\Bigg[2a+(m- 1 )d \Bigg]$

$\tt : \implies n= \dfrac{m}{2}\Bigg[2a+md- d \Bigg]$

$\tt : \implies \dfrac{ 2n}{m}= \Bigg[2a+md- d \Bigg] \: \: \: - (1)$

☛ Case 2.

$\tt \dagger \: \: \: \: \: S_n = \dfrac{n}{2}\Bigg[2a+(n- 1 )d \Bigg]$

$\tt : \implies m = \dfrac{n}{2}\Bigg[2a+(n- 1 )d \Bigg]$

$\tt : \implies \dfrac{2m}{n}= \Bigg[2a+nd- d \Bigg] \: \: \: - (2)$

Now,

Subtract Equation ( 2 ) From ( 1 ) We get.

$\tt : \implies \dfrac{ 2n}{m} - \dfrac{2m}{n} = \Bigg[2a+md- d \Bigg] - \Bigg[2a+nd- d \Bigg]$

$\tt : \implies \dfrac{ 2n}{m} - \dfrac{2m}{n} = \Bigg[(2a- d )+ md \Bigg] - \Bigg[(2a - d) + nd\Bigg]$

$\tt : \implies \dfrac{ 2n}{m} - \dfrac{2m}{n} = (2a- d )+ md - (2a - d) - nd$

$\tt : \implies \dfrac{ 2n}{m} - \dfrac{2m}{n} = \cancel{ (2a- d )}+ md \cancel{ - (2a - d) } - nd$

$\tt : \implies \dfrac{ 2n}{m} - \dfrac{2m}{n} = md - nd$

$\tt : \implies \dfrac{ 2 {n}^{2} - 2 {m}^{2} }{mn} =d( m - n)$

$\tt : \implies \dfrac{ 2(n + m)(n - m) }{mn} =d( m - n)$

$\tt : \implies \dfrac{ - 2(n + m) }{mn} =d \: \: \: - (3)$

✎ Sum of ( m + n ) th

$\tt \dagger \: \: \: \: \: S_{m + n} = \dfrac{m + n}{2}\Bigg[2a+(m + n- 1 )d \Bigg]$

$\tt : \implies S_{m + n} = \dfrac{m + n}{2}\Bigg[2a+(m - 1 )d + nd \Bigg]$

$\tt : \implies S_{m + n} = \dfrac{m + n}{2}\Bigg[ \dfrac{2n}{m} + nd \Bigg]$

$\tt : \implies S_{m + n} = \dfrac{m + n}{2}\Bigg[ \dfrac{2n}{m} + \cancel{ n} \bigg( \dfrac{ - 2(n + m)}{m \cancel{n}} \bigg)\Bigg]$

$\tt : \implies S_{m + n} = \dfrac{m + n}{2}\Bigg[ \dfrac{2n}{m} + \bigg( \dfrac{ - 2(n + m)}{m} \bigg)\Bigg]$

$\tt : \implies S_{m + n} = \dfrac{m + n}{2}\Bigg[ \dfrac{2n-2(n+m)}{m} \Bigg]$

$\tt : \implies S_{m + n} = \dfrac{m + n}{2}\Bigg[ \dfrac{2n-2n - 2m}{m} \Bigg]$

$\tt : \implies S_{m + n} = \dfrac{m + n}{2}\Bigg[ \dfrac{ - 2m}{m} \Bigg]$

$\tt : \implies S_{m + n} = \dfrac{m + n}{2} \times - 2$

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

✡ Hence Proved.