12 terms of the arithmetic progression (AP) 9, 17, 25... must be taken to give a sum of 636. This conclusion is drawn from the provided reference, which directly states: "Therefore, 12 terms of the given sequence make sum equal to 636. Q. Q. How many terms of the AP:9,17,25,....."
Here's a breakdown for better understanding:
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The Arithmetic Progression: We have an AP where the first term (a) is 9, and the common difference (d) is 17 - 9 = 8.
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The Goal: We need to find the number of terms (n) such that the sum of those 'n' terms is equal to 636.
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The Formula: The sum (S) of the first 'n' terms of an AP is given by:
S = n/2 * [2a + (n-1)d]
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Applying the Values: We have S = 636, a = 9, and d = 8. So:
636 = n/2 [2(9) + (n-1)8]
1272 = n [18 + 8n - 8]
1272 = n * [10 + 8n]
1272 = 10n + 8n²
8n² + 10n - 1272 = 0
4n² + 5n - 636 = 0Solving the quadratic equation gives us two roots, with only one being a positive integer, which is n = 12.
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Therefore, we need 12 terms to achieve a sum of 636.
| Property | Value |
|---|---|
| First term (a) | 9 |
| Common difference (d) | 8 |
| Sum (S) | 636 |
| Number of Terms (n) | 12 |
