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Method and examples
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Arithmetic Progression |
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Problem 6 of 19 |
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6. For arithemetic progression f( ) = , S( ) = , then find f( ) and S( ).
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Solution
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Solution provided by AtoZmath.com
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This is demo example. Please click on Find button and solution will be displayed in Solution tab (step by step)
| 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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