# Find the 31^{st} term of an AP whose 11^{th} term is 38 and the 16^{th} term is 73

**Solution:**

The n^{th} term of an Arithmetic progression is given by,

aₙ = a + (n - 1)d

Here, a is the first term, d is the common difference and n is the number of terms.

aₙ = a + (n - 1)d

a₁₁ = 38

a + (11 - 1) d = 38

a + 10d = 38 ...(1)

a₁₆ = 73

a + 15d = 73 ...(2)

By subtracting equation (1) from equation (2) we get,

a + 15d - ( a + 10d ) = 73 - 38

5d = 35

d = 7

Putting the value of d in equation (1),

a + 10(7) = 38

a = 38 - 70

= - 32

31^{st} terms is, a₃₁ = a + (31 - 1)d

= - 32 + 30 × 7

= - 32 + 210

= 178

The 31^{st} term of the AP is 178.

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 5

**Video Solution:**

## Find the 31^{st} term of an AP whose 11^{th} term is 38 and the 16^{th} term is 73

NCERT Solutions Class 10 Maths Chapter 5 Exercise 5.2 Question 7

**Summary:**

The 31^{st} term of an AP whose 11^{th} term is 38 and the 16^{th} term is 73 is 178.

**☛ Related Questions:**

- An AP consists of 50 terms of which the 3rd term is 12 and the last term is 106. Find the 29th term.
- If the 3rd and the 9th terms of an AP are 4 and -8, respectively, which term of this AP is zero.
- 17th term of an AP exceeds its 10th term by 7. Find the common difference.
- Which term of the AP 3,15,27,39... will be 132 more than its 54th term?

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