Question

if the sum of first M terms of an AP is the same as the sum of first N terms show that the sum of first (M +N)and terms is zero .​

Answer 1

Step-by-step explanation:

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Answer 2

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

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

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

$\mathfrak{ \: Suppose,} \\ \\ \: \: \: \: \: \: \: \: \: \text{First term of the A.P. = a} \\ \\ \: \: \: \: \: \: \: \: \: \text{Common difference = d} \\ \\ \: \: \: \: \: \: \: \: \tt{Sum \: \: of \: \: first \: \: M \: \: terms = S_M} \\ \text{And,} \\ \: \: \: \: \: \: \: \: \tt{Sum \: \: of \: \: first \: \: N \: \: terms = S_N}$

$\underline{ \bold{ \: \: A.T.Q., \: \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \huge{ \tt{S_M = S_N } }\\ \\ \tt{ \implies \: \frac{m}{ \cancel{2}} \{2a + (M - 1)d \} = \frac{n}{ \cancel{2}} \{2a + (N - 1)d \} } \\ \\ \tt{ \implies \: M \{2a + (M - 1)d \} = N \{2a + (N - 1)d \}} \\ \\ \tt{ \implies \: M \times 2a + M \times (M - 1)d = N \times 2a + N \times (N - 1)d} \\ \\ \tt{ \implies \: M \times 2a - N \times 2a = \{N \times (M - 1)d \} - \{M \times (M- 1)d \}} \\ \\ \tt{ \implies \:2a \times (M - M) = \{N \times (Nd - d) \} - \{M \times (Md - d) \}} \\ \\ \tt{ \implies \:2a \times ( M- N) = (N {}^{2}d - Nd) - (M{}^{2}d - Md) } \\ \\ \tt{ \implies \:2a \times ( M- N) = N {}^{2}d - Nd - M{}^{2}d + Md } \\ \\ \tt{ \implies \: 2a \times (M - N) = N {}^{2}d - M{}^{2}d - Nd +Md } \\ \\ \tt{ \implies \: 2a \times (M - N) = d(N {}^{2} - M {}^{2} ) - d(N - M)} \\ \\ \tt{ \implies \: 2a \times (M - N) = d \{(N+ M)(N - M) \} - d(N - M)} \\ \\ \tt{ \implies \: 2a \times (M - N) = d(N - M) \{(N + M) - 1 \} } \\ \\ \tt{ \implies \: 2a \times \cancel{(M - N) }= - \: d \: \cancel{(M - N)} \{(M + N) - 1 \}} \\ \\ \tt{ \implies \: 2a = - d \{(M + N) - 1 \}} \\ \\ \tt{ \implies \: 2a + d \{(M + N) - 1 \} = 0} \\ \\ \tt{ \implies \: 2a + \{(M + N) - 1 \} d= 0 \: \: \: - - - - - > (1)}$

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

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

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