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Method and examples
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Arithmetic Progression |
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Problem 9 of 19 |
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9. For arithmetic progression Sm = n and Sn = m then prove that Sm+n = -(m - n)
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Solution
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Solution provided by AtoZmath.com
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This is demo example. Please click on Find button and solution will be displayed in Solution tab (step by step)
| Arithmetic Progression |
9. For arithmetic progression Sm = n and Sn = m then prove that Sm+n = -(m - n)
For arithmetic progression, `S_n = n/2 [ 2a + (n - 1) d ]`
Now, `S_m = n`
`:. m/2 [ 2a + (m - 1) d ] = n`
`:. [ 2a + (m - 1) d ] = (2n)/m ->(1)`
Now, `S_n = m`
`:. n/2 [ 2a + (n - 1) d ] = m`
`:. [ 2a + (n - 1) d ] = (2m)/n ->(2)`
`(1) - (2) =>`
`(m - 1) d - (n - 1) d = (2n)/m - (2m)/n`
`:. (m - n) d = (2 (n^2- m^2))/(mn) `
`:. (m - n) d = (-2 (m - n)(m + n))/(mn) `
`:. d = (-2 (m + n))/(mn) ->(3)`
Now, `S_(m+n) = (m + n)/2 [ 2a + (m + n - 1) d ]`
`= (m + n)/2 [ 2a + (m - 1) d + nd ]`
`= (m + n)/2 [ (2n)/m + n ( (-2(m+n))/(mn) ) ]` (because from (1) and (3))
`= (m + n)/2 [ (2n)/m + (-2(m+n))/m ] `
`= (m + n)/2 [ (2n - 2m - 2n)/m ]`
`= (m + n)/2 [ (- 2m)/m ]`
`= (m + n)/2 [ -2 ]`
`= -(m + n)` (Proved)
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