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Arithmetic Progressions
EXERCISE 5.3
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

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

Step-by-Step Solution

Key Concept: Use the formula for the sum of the first n terms of an arithmetic progression: \(S_n = \frac{n}{2}[2a + (n-1)d]\), where \(a\) is the first term and \(d\) is the common difference. Solve the resulting quadratic equation for \(n\).
1. Identify the first term and common difference of the AP.
\[a = 9, \quad d = 17-9 = 8\]

2. Write the sum of the first \(n\) terms using the AP sum formula.
\[S_n = \frac{n}{2}[2a + (n-1)d]\]
Substituting \(a\) and \(d\):
\[S_n = \frac{n}{2}[2\times9 + (n-1)\times8]\]
\[S_n = \frac{n}{2}[18 + 8n - 8]\]
\[S_n = \frac{n}{2}[8n + 10]\]
Simplify:
\[S_n = n\left(4n + 5\right)\]

3. Set the sum equal to the given value (636) and form the quadratic equation.
\[n(4n + 5) = 636\]
\[4n^2 + 5n - 636 = 0\]

4. Solve the quadratic equation.
Compute the discriminant:
\[\Delta = 5^2 - 4\times4\times(-636) = 25 + 10176 = 10201\]
\[\sqrt{\Delta} = 101\]
Using the quadratic formula:
\[n = \frac{-5 \pm 101}{2\times4}\]
Positive root:
\[n = \frac{-5 + 101}{8} = \frac{96}{8} = 12\]
The negative root is discarded because the number of terms cannot be negative.

5. Hence, the required number of terms is \(n = 12\).

Correct Answer: 12
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