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 3 of 19
3. For arithemetic progression f( ) = , f( ) = , then find n such that f(n) = .

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Arithmetic Progression
3. For arithemetic progression f( 5 ) = 25 , f( 11 ) = 49 , then find n such that f(n) = 105 . We know that, `f(n) = a + (n - 1)d` `f(5) = 25` `=> a + (5 - 1)d = 25` `=> a + 4d = 25 ->(1)` `f(11) = 49` `=> a + (11 - 1)d = 49` `=> a + 10d = 49 ->(2)` Solving `(1)` and `(2)`, we get `a = 9` and `d = 4` Let `n` be the term such that `f(n) = 105` We know that, `f(n) = a + (n - 1)d` `9 + (n - 1)(4) = 105` `(n - 1)(4) = 96` `n - 1 = 24` `n = 25`