NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.2
Question 7

Question. 7

Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Answer:

178

Detailed Answer with Explanation:

Step 1: Use the general term of an AP.

For an arithmetic progression (AP), the \(n\)th term is given by

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

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

Step 2: Translate the given information.

11th term is 38:

\[a_{11} = a + 10d = 38\]

16th term is 73:

\[a_{16} = a + 15d = 73\]

Step 3: Form equations and subtract.

From 11th term: \(a + 10d = 38\)  ...(1)

From 16th term: \(a + 15d = 73\)  ...(2)

Subtract (1) from (2):

\[(a + 15d) - (a + 10d) = 73 - 38\]

\[5d = 35\]

So,

\[d = \dfrac{35}{5} = 7\]

Step 4: Find the first term \(a\).

Use \(a + 10d = 38\):

\[a + 10 \cdot 7 = 38\]

\[a + 70 = 38\]

\[a = 38 - 70 = -32\]

Step 5: Find the 31st term.

\[a_{31} = a + 30d\]

Substitute \(a = -32\) and \(d = 7\):

\[a_{31} = -32 + 30 \cdot 7 = -32 + 210 = 178\]

Conclusion: The 31st term of the AP is 178.

NCERT Solutions – Chapter-wise Questions & Answers