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How Many Terms of the A.P. 9, 17, 25, . . . Must Be Taken So that Their Sum is 636? - Mathematics

How many terms of the A.P. 9, 17, 25, . . . must be taken so that their sum is 636?

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Solution

In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the number of terms.

A.P. 9, 17, 25

So here, let us find the number of terms whose sum is 636. For that, we will use the formula,

`S_n = n/2 [2a + (n -1)d]`

Where; a = first term for the given A.P.

d = common difference of the given A.P.

= number of terms

The first term (a) = 9

The sum of n terms (Sn) = 636

Common difference of the A.P. (d) = `a_2 - a_1`

= 17 - 9

= 8

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

`636 = n/2 [2(9) + (n - 1)(8)]`

`636 = (n/2)[18 + (8n - 8)]`

`636 = (n/2)(10 + 8n)`

`636(2) = 10n + 8n^2`

So, we get the following quadratic equation,

`8n^2 + 10n - 1272 = 0`

4n^2 + 5n - 636 = = 0

On solving by splitting the middle term, we get,

`4n^2 - 48n + 53n - 636 = 0`

4n(n - 12) - 53(n - 12) = 0

(4n - 53)(n - 12) = 0

Further,

4n - 53 = 0

`n = 53/4`

or

n - 12 = 0

n = 12

Since, the number of terms cannot be a fraction, the number of terms (n) is 12

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 10 Maths
Chapter 5 Arithmetic Progression
Exercise 5.6 | Q 10.3 | Page 51
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