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Q4:
How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Solution :

Given: An Arithmetic Progression (AP) with the sequence $9, 17, 25, \dots$ and the sum of $n$ terms $S_n = 636$.

To find: The number of terms $n$ required such that the sum of the AP equals $636$.

Step 1: Identify the parameters of the Arithmetic Progression.

The first term ($a$) is the first number in the sequence: $a = 9$.

The common difference ($d$) is the difference between any two consecutive terms: $d = a_2 - a_1 = 17 - 9 = 8$.

The sum of $n$ terms is given as $S_n = 636$.

Step 2: State the formula for the sum of $n$ terms of an AP.

The formula for the sum of the first $n$ terms of an AP is given by:

$S_n = \frac{n}{2} [2a + (n - 1)d]$

[Where $n$ is the number of terms, $a$ is the first term, and $d$ is the common difference.]

Step 3: Substitute the known values into the formula.

$636 = \frac{n}{2} [2(9) + (n - 1)8]$

$636 = \frac{n}{2} [18 + 8n - 8]$

$636 = \frac{n}{2} [8n + 10]$

Step 4: Simplify the equation to form a quadratic equation.

$636 = n(4n + 5)$ [Dividing the terms inside the bracket by 2]

$636 = 4n^2 + 5n$

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

Step 5: Solve the quadratic equation using the quadratic formula.

The quadratic formula is $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=4, b=5, c=-636$.

$n = \frac{-5 \pm \sqrt{(5)^2 - 4(4)(-636)}}{2(4)}$

$n = \frac{-5 \pm \sqrt{25 + 10176}}{8}$

$n = \frac{-5 \pm \sqrt{10201}}{8}$

Since $\sqrt{10201} = 101$:

$n = \frac{-5 \pm 101}{8}$

Step 6: Evaluate the two possible values for $n$.

Case 1: $n = \frac{-5 + 101}{8} = \frac{96}{8} = 12$

Case 2: $n = \frac{-5 - 101}{8} = \frac{-106}{8} = -13.25$

Step 7: Interpret the results.

Since the number of terms ($n$) must be a positive integer, we reject the negative fractional value $n = -13.25$.

Therefore, $n = 12$.

Final Answer: 12 terms must be taken to give a sum of 636.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


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