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Arithmetic Progression 



Problem 6 of 19 


6. For arithemetic progression f( ) = , S( ) = , then find f( ) and S( ).










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Arithmetic Progression 
6. For arithemetic progression f( 7 ) = 13 , S( 14 ) = 203 , then find f( 10 ) and S( 8 ).
We have given `f(7) = 13, S_14 = 203` and we have to find `f(10) = ?` and `S(8) = ?`
We know that, `f(n) = a + (n  1)d`
`f(7) = 13`
`=> a + (7  1)d = 13`
`=> a + 6d = 13 >(1)`
We know that, `S_n = n/2 [2a + (n  1)d]`
`S_14 = 203`
`=> 14/2 * [2a + (14  1)d] = 203`
`=> 7 * [2a + 13d] = 203`
`=> 2a + 13d = 29 >(2)`
Solving `(1)` and `(2)`, we get `a = 5` and `d = 3`
We know that, `f(n) = a + (n  1)d`
`f(10) = 5 + (10  1)(3)`
`= 5 + (27)`
`= 22`
We know that, `S_n = n/2 [2a + (n  1)d]`
`:. S_8 = 8/2 * [2(5) + (8  1)(3)]`
`= 4 * [10 + (21)]`
`= 4 * [11]`
`= 44`





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