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Arithmetic Progression 



Problem 10 of 19 


10. For arithmetic progression Sm = n and Sn = m then prove that Smn = (m  n)(1 + 2n / m)










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Arithmetic Progression 
10. For arithmetic progression Sm = n and Sn = m then prove that Smn = (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_(mn) = (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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