Question

If the sum of first m terms of an A:P is same as the sum of its first n terms (m #n), show that the sum of
its first (m + n) terms is zero.
​

Answer 1

#BAL

Sum of m terms=Sum of n terms

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

Cancelling 2 in the denominator on both sides,

we get

m(2a + (m-1)d) - n(2a+ (n-1)d) = 0

 

2am + m^2d - md -2an -n^2d +nd =0

 

2a(m-n) + (m^2 - n^2)d -(m - n)d =0

 

2a(m-n) + ( (m + n) (m- n) ) d - (m - n )d = 0

 

Taking (m-n) common

 

 

2a + ( m + n -1) d = 0 ------------ (1)

 

 

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

 

we know that 2a + (m+n)d is 0 from eqn. 1

 

 

therefore S m+n = 0

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \pink{ \: \: SOLUTION \: \: } \mid}}}}}$

$\underline{ \mathfrak{ \: \: Given : \: }} \\ \\ \: \: \text{Sum of first m terms of an A.P. is same as} \\ \text{the sum \: of first n terms, where m \: is \: not \: } \\ \text{equal \:to n }$

$\underline{ \mathfrak{ \: \: To \: \: show : \mapsto \: }} \\ \\ \text{Sum of first (m+n) terms is equal to zero. }$

$\mathfrak{ \: Suppose,} \\ \\ \: \: \: \: \: \: \: \: \: \text{First term of the A.P. = a} \\ \\ \: \: \: \: \: \: \: \: \: \text{Common difference = d} \\ \\ \: \: \: \: \: \: \: \: \tt{Sum \: \: of \: \: first \: \: m \: \: terms = S_m} \\ \text{And,} \\ \: \: \: \: \: \: \: \: \tt{Sum \: \: of \: \: first \: \: n \: \: terms = S_n}$

$\underline{ \bold{ \: \: A.T.Q., \: \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \huge{ \tt{S_m = S_n } }\\ \\ \tt{ \implies \: \frac{m}{ \cancel{2}} \{2a + (m - 1)d \} = \frac{n}{ \cancel{2}} \{2a + (n - 1)d \} } \\ \\ \tt{ \implies \: m \{2a + (m - 1)d \} = n \{2a + (n - 1)d \}} \\ \\ \tt{ \implies \: m \times 2a + m \times (m - 1)d = n\times 2a + n \times (n - 1)d} \\ \\ \tt{ \implies \: m \times 2a - n \times 2a = \{n \times (n - 1)d \} - \{m \times (m- 1)d \}} \\ \\ \tt{ \implies \:2a \times (m - n) = \{n \times (nd - d) \} - \{m \times (md - d) \}} \\ \\ \tt{ \implies \:2a \times ( m- n) = (n {}^{2}d - nd) - (m{}^{2}d - md) } \\ \\ \tt{ \implies \:2a \times ( m- n) = n {}^{2}d - nd - m{}^{2}d + md } \\ \\ \tt{ \implies \: 2a \times (m - n) = n {}^{2}d - m{}^{2}d - nd +md } \\ \\ \tt{ \implies \: 2a \times (m - n) = d(n {}^{2} - m {}^{2} ) - d(n - m)} \\ \\ \tt{ \implies \: 2a \times (m - n) = d \{( n+ m)(n - m) \} - d(n - m)} \\ \\ \tt{ \implies \: 2a \times (m - n) = d(n - m) \{(n + m) - 1 \} } \\ \\ \tt{ \implies \: 2a \times \cancel{(m - n) }= - \: d \: \cancel{(m - n)} \{(m + n) - 1 \}} \\ \\ \tt{ \implies \: 2a = - d \{(m + n) - 1 \}} \\ \\ \tt{ \implies \: 2a + d \{(m + n) - 1 \} = 0} \\ \\ \tt{ \implies \: 2a + \{(m + n) - 1 \} d= 0 \: \: \: - - - - - > (1)}$

$\mathfrak{ \:Now, \: } \\ \\ \underline{ \mathfrak{ \: Suppose, \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \tt{The \: \: sum \: \: of \: \: first \: \: (m+n) \: \: terms = S_{(m+n)}}$

$\tt{ \therefore{ \: S_{(m+n)} = \frac{( m+ n)}{2}[ 2a + \{(m + n) - 1 \}d}]} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{ = \frac{(m + n}{2} \times 0 \: \: \: \: \: \: \: \: \:[From \: \: (1)] } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{ =0}$

$\: \: \: \: \: \: \: \: \: \tt{ \therefore \: \: S _{(m + n)} = 0 \: \: \: \: \red{ i.e.} \: \: Sum \: \: of \: \: first \: \: } \\ \\ \tt{ (m+n) terms \: \: is \: \: equal \: \: to \: \: zero. } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \pink{\underline{ \text{ \: \: Showed.\: \: }}}$