Math, asked by Khushikanyal13, 9 months ago

If sum of first m terms of Ap is
the same as the sum of its
n terms show that the sum of its
first (m+n) terms is zero
Plz answer ☺ ☺​

Answers

Answered by baskykishore2006
1

Answer:

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

ma+m(m-1)d=na+n(n-1)d

ma-na=n(n-1)d-m(m-1)d

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

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

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

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

divide by m-n on both the sides

a=  -1(n+m-1)d

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

a m+n =0

Step-by-step explanation:

Answered by TRISHNADEVI
3

 \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.\:  \: }}}

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