Method and examples
Arithmetic Progression
1. For given arithemetic progression series 7,3,-1,-5,-9 ,... find 10 th term and addition of first 10 th terms.
2. For arithemetic progression f( 5 ) = 56 , f( 8 ) = 86 then find f( 10 ) and S( 10 ).
3. For arithemetic progression f( 5 ) = 25 , f( 11 ) = 49 , then find n such that f(n) = 105 .
4. For arithemetic progression S( 33 ) = 198 , then find f( 17 ).
5. For arithemetic progression f( 17 ) = 6 , then find S( 33 ).
6. For arithemetic progression f( 7 ) = 13 , S( 14 ) = 203 , then find f( 10 ) and S( 8 ).
7. For arithemetic progression addition of 3 terms is 27 and their multiplication is 648 , then that numbers
8. For arithemetic progression addition of first 17 terms is 24 and addition of first 24 terms is 17 , then find addition of fir ...
9. For arithmetic progression Sm = n and Sn = m then prove that Sm+n = -(m - n)
10. For arithmetic progression Sm = n and Sn = m then prove that Sm-n = (m - n)(1 + 2n / m)
11. The ratio of two arithemetic progression series is 3x+5 : 4x-2 , then find the ratio of their 10 th term.
12. Find the sum of all natural numbers between 100 to 200 and which are divisible by 4 .
13. Find the sum of all natural numbers between 100 to 200 and which are not divisible by 4 .
14. For arithemetic progression addition of three terms is 15 and addition of their squres is 83 , then find that numbers
15. If S1, S2, S3 are sum of n, 2n, 3n terms of arithmetic progression series then prove that S3 = 3(S2 - S1)
16. If Sn is sum of n even terms of arithmetic progression series and Sn' is sum of n odd terms of arithmetic progression series ...
17. For arithemetic progression, addition of three terms is 51 and multiplication of end terms is 273 , then find that numbers
18. For arithemetic progression of four terms, addition of end terms is 14 and multiplication of middle two terms is 45 , then fi ...
19. For arithemetic progression, addition of four terms is 4 and addition of multiplication of end terms and multiplication of mi ...
Problem 10 of 19
10. For arithmetic progression Sm = n and Sn = m then prove that Sm-n = (m - n)(1 + 2n / m)

Solution
Solution provided by AtoZmath.com

Want to know about

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)