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 9 of 19
9. For arithmetic progression Sm = n and Sn = m then prove that Sm+n = -(m - n)

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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)