NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.3
Question 4

Question. 4

How many terms of the AP \(9, 17, 25, \ldots\) must be taken to give a sum of 636?

Answer:

12.

By putting \(a = 9\), \(d = 8\), \(S = 636\) in the formula \(S = \dfrac{n}{2}[2a + (n - 1)d]\), we get the quadratic equation \(4n^2 + 5n - 636 = 0\).

On solving, we get \(n = -\dfrac{53}{4}, 12\). Out of these two roots only one root \(12\) is admissible.

Detailed Answer with Explanation:

Step 1: Identify \(a\), \(d\) and \(S_n\).

The AP is \(9, 17, 25, \ldots\).

First term: \(a = 9\)

Common difference: \(d = 17 - 9 = 8\)

Required sum: \(S_n = 636\)

Step 2: Use the sum formula for an AP.

For an AP, the sum of first \(n\) terms is

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

Substitute \(a = 9\), \(d = 8\), \(S_n = 636\):

\[636 = \dfrac{n}{2}[2\cdot 9 + (n - 1)8].\]

Step 3: Simplify inside the bracket.

\[2a = 2 \cdot 9 = 18\]

\[(n - 1)d = 8(n - 1) = 8n - 8\]

So,

\[2a + (n - 1)d = 18 + 8n - 8 = 8n + 10.\]

Thus,

\[636 = \dfrac{n}{2}(8n + 10).\]

Step 4: Form the quadratic equation in \(n\).

Multiply both sides by 2:

\[1272 = n(8n + 10).\]

Expand the right side:

\[1272 = 8n^2 + 10n.\]

Bring all terms to one side:

\[8n^2 + 10n - 1272 = 0.\]

Divide the whole equation by 2 to simplify:

\[4n^2 + 5n - 636 = 0.\]

Step 5: Solve the quadratic equation.

Use the quadratic formula for \(4n^2 + 5n - 636 = 0\):

\[n = \dfrac{-5 \pm \sqrt{5^2 - 4\cdot 4\cdot (-636)}}{2\cdot 4}.\]

Compute the discriminant:

\[D = 25 + 4\cdot 4\cdot 636 = 25 + 10176 = 10201 = 101^2.\]

So,

\[n = \dfrac{-5 \pm 101}{8}.\]

This gives two values:

\[n = \dfrac{-5 + 101}{8} = \dfrac{96}{8} = 12,\]

\[n = \dfrac{-5 - 101}{8} = \dfrac{-106}{8} = -\dfrac{53}{4}.\]

Step 6: Choose the valid value of \(n\).

Since \(n\) is the number of terms, it must be a positive integer. The value \(-\dfrac{53}{4}\) is negative and not acceptable.

Therefore, we take \(n = 12\).

Conclusion: We must take 12 terms of the AP to get a sum of 636.

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