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how many terms of the AP:9,17,25,... must be taken to give a sum of 636
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how many terms of the AP:9,17,25,... must be taken to give a sum of 63...
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how many terms of the AP:9,17,25,... must be taken to give a sum of 63...
Introduction:
To find the number of terms of an arithmetic progression (AP) that must be taken to give a sum of 636, we need to understand the concept of the sum of an AP and solve the problem using the appropriate formula.

Sum of an Arithmetic Progression:
The sum of an arithmetic progression is given by the formula:
S = (n/2)(2a + (n-1)d)
where S is the sum of the terms, n is the number of terms, a is the first term, and d is the common difference.

Given Information:
First term (a) = 9
Common difference (d) = 17 - 9 = 8
Sum of the terms (S) = 636

Using the Sum of AP Formula:
We can rearrange the sum formula to solve for the number of terms (n):
S = (n/2)(2a + (n-1)d)
636 = (n/2)(2(9) + (n-1)(8))
Simplifying the equation:
636 = (n/2)(18 + 8n - 8)
636 = (n/2)(8n + 10)
Dividing both sides by 2:
318 = n(8n + 10)
318 = 8n^2 + 10n
Rearranging the equation to form a quadratic equation:
8n^2 + 10n - 318 = 0

Solving the Quadratic Equation:
To find the number of terms (n), we need to solve the quadratic equation. We can use the quadratic formula:
n = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 8, b = 10, and c = -318.

Using the quadratic formula, we can calculate the values of n.
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how many terms of the AP:9,17,25,... must be taken to give a sum of 636 Related: Exercise 1 - Chapter 5 - Arithmetic Progressions, Class 10, Mathematics?
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