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

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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.

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:

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

The 31st term of an AP whose 11th term is 38 and 16th term is 73 is 178. a, a + d, a + 2d, a + 3d…. is the general form of arithmetic progression. Hence, the formula to find the nth term is: an = a + (n – 1) d.

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