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Method and examples
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Arithmetic Progression |
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Problem 11 of 19 |
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11. The ratio of two arithemetic progression series is : , then find the ratio of their th term.
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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 |
11. The ratio of two arithemetic progression series is 3x+5 : 4x-2 , then find the ratio of their 10 th term.
Let two series be `a, a+d, a+2d, ...` and `A, A+D, A+2D, ...`
Now, `S_n / S_(n') = (3x+5) / (4x-2)`
`=> (n/2 [2a + (n - 1)d]) / (n/2 [2A + (n - 1)D]) = (3x+5) / (4x-2)`
`=> [2a + (n - 1)d] / [2A + (n - 1)D] = (3x+5) / (4x-2) ->(1)`
We know that, `f(n) = a + (n - 1)d`
`f(n)/(F(n)) = [a + (n - 1)d] / [A + (n - 1)D]`
`=> f(10)/(F(10)) = [a + (10 - 1)d] / [A + (10 - 1)D]`
`=> f(10)/(F(10)) = [a + 9d] / [A + 9D]`
`=> f(10)/(F(10)) = [2a + 18d] / [2A + 18D] ->(2)`
Using `(1)` and `(2)`, it is clear that if we put `n = 19` in `(1)` we get the ratio of `(2)`
`=> f(10)/(F(10)) = [2a + 18d] / [2A + 18D]`
`=> f(10)/(F(10)) = (3x+5) / (4x-2),` where `n=19`
`=> f(10)/(F(10)) = 62 / 74`
`=> f(10)/(F(10)) = 31/37`
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