Question

the sum of fiest m terms of an AP is the same as the sum of its first n terms , show that the sum of its first (m+n) terms is zero​

Answer 1

Answer:

LHS = RHS

1.Take out the 'm' and 'n' terms.

2.Take out the (m+n) terms.

3.Get the values by using elimination method on both the equations.

4.Add the differences.

5.Take out common.

6.you will get the answer.

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Answer 2

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

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

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

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

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

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

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

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