Question

if the sum of first M terms of an ap is the same as sum of its first n terms, show that the sum of the first (M + n) term is zero. ​

Answer 1

Answer:

Let the first term of an AP be a and common difference be d.

According to question,

Sm= Sn

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

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

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

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

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

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

Now,

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

= m+n/2 ×0 [using (i)]

=0

Hence, Sum of (m+n) terms is equal to 0

Answer 2

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

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

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

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

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

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

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

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