The 31st term of an AP whose 11th term is 38 and 16th term is 73 is – (a) 150 (b) 178

By BYJU'S Exam Prep

Updated on: September 25th, 2023

The 31st term of an AP whose 11th term is 38 and 16th term is 73 is 178. It is given that a11 = 38 and a16 = 73 where a11 is the 11th term and a16 is the 16th term of an AP. We know that an = a + (n – 1) d is the formula to find the nth term of arithmetic progression

By using this formula we can write as:

38 = a + (11 – 1) d

73 = a + (16 – 1) d

38 = a + 10d ….. (1)

73 = a + 15d ….. (2)

These equations have just two variables. Let’s use the substitution method to resolve these.

Table of content

1. The 31st term of an AP whose 11th term is 38 and 16th term is 73 is – (a) 150 (b) 178

We have:

38 = a + 10d

On rearranging we get:

a = 38 – 10d

Now substitute the value of an in equation (2), and we get:

73 = 38 – 10d + 15d

In simplification we get the:

35 = 5d

Hence the Common difference is d = 7.

Now by substituting the value of d in equation (1), we get:

38 = a + 70

By rearranging and simplifying we get:

a = -32

Therefore, the common difference is d = 7 and

First term a = -32

Now we can find the value of the nth term of arithmetic progression using the formula:

an = a + (n – 1) d

Substituting the value we get:

a31 = -32 + (31 – 1) (7)

= -32 + 210

= 178

hence, the 31st term of AP is 178.

Arithmetic Progression:

A progression of numbers known as an arithmetic sequence is one in which, for every pair of consecutive terms, the second number is obtained by adding a predetermined number to the first one.

Summary: