NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.2
Question 8

Question. 8

An AP consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.

Answer:

64

Detailed Answer with Explanation:

Step 1: Recall the nth term formula of an AP.

For an arithmetic progression (AP), the nth term is given by:

\[a_n = a + (n - 1)d\]

where \(a\) is the first term and \(d\) is the common difference.

Step 2: Use the information about the 3rd term.

The 3rd term is 12, so:

\[a_3 = a + 2d = 12\]

This gives our first equation:

\[a + 2d = 12 \quad ...(1)\]

Step 3: Use the information about the 50th (last) term.

The AP has 50 terms and the last term is 106. So the 50th term is:

\[a_{50} = a + 49d = 106\]

This gives our second equation:

\[a + 49d = 106 \quad ...(2)\]

Step 4: Solve the two equations to find \(d\).

Subtract equation (1) from equation (2):

\[(a + 49d) - (a + 2d) = 106 - 12\]

\[a + 49d - a - 2d = 94\]

\[47d = 94\]

So,

\[d = \dfrac{94}{47} = 2\]

Step 5: Find the first term \(a\).

Substitute \(d = 2\) into equation (1):

\[a + 2d = 12\]

\[a + 2(2) = 12\]

\[a + 4 = 12\]

\[a = 12 - 4 = 8\]

Step 6: Find the 29th term.

Use the nth term formula with \(n = 29\):

\[a_{29} = a + (29 - 1)d = a + 28d\]

Substitute \(a = 8\) and \(d = 2\):

\[a_{29} = 8 + 28 \times 2\]

\[a_{29} = 8 + 56 = 64\]

Conclusion: The 29th term of the AP is 64.

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