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Method and examples
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Arithmetic Progression |
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Problem 10 of 19 |
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10. For arithmetic progression Sm = n and Sn = m then prove that Sm-n = (m - n)(1 + 2n / m)
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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 |
10. For arithmetic progression Sm = n and Sn = m then prove that Sm-n = (m - n)(1 + 2n / m)
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)`
`:. 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 + 4n))/m ]`
`= (m - n)/2 [ (2(m + 2n))/m ]`
`= (m - n) [ (m + 2n)/m ]`
`= (m - n)(1 + (2n)/m)` (Proved)
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