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

The n^{th} term of an Arithmetic progression is the total number of terms in the sequence.

**Solution:**

a_{n} = a + (n - 1)d is the n^{th} term of Arithmetic Progression.

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

a_{n} = a + (n - 1)d

a_{11} = 38

a + (11 - 1) d = 38

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

a_{16} = 73

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

By subtracting the equations (2) from (1) for the values of a and d.

5d = 35

d = 7

Putting the value of d in the equation (1).

a = 38 - 70

= - 32

31^{st} terms is, a_{31} = a + (31 - 1)d

= - 32 + 30 × 7

= - 32 + 210

= 178

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

**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 - Chapter 5 Exercise 5.2 Question 7:

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

The 31st term of an AP in which the 11th term is 38 and the 16th term is 73 is 178