Find the 32 nd term of each sequence. 101,105,109,113, ...........

An arithmetic progression (AP) in mathematics is a list and maybe sequence of numbers with each term is obtained by adding a sequence to the term before it.

With the exception of arithmetic progression, the fixed number is denoted by the letter 'd.'
Common difference => d = a2 - a1 = a3 - a2 = a4 - a3 =...... = a - an-1.
nth term of an AP: an = a + (n - 1) d
The sum of n terms of an AP's => Sn = n/2(2a+(n-1)d) = n/2(a + l), where l = last term of the AP.

Finally, the solution to the question is as follows:

The numbers in the displayed sequence are as follows:

101,105,109,113, ...........

We must explore the 32nd number in the series because it contains 32 terms.

The common difference with both two consecutive terms of an AP which are equal is denoted by 'd.'

Let's assume the first term is'a₁' = 101.

Let's assume 'a₂' = 105 is the second term.

Let's assume the third term is 'a₃' = 109

Substitute the values the preceding equation; the 32nd term is-

d = a₂ - a₁

Obtain the common difference;

d = 105 - 101

d = 4

Put the values in the preceding formula; the 32nd term is

The formula finds the nth term equation;

nth term of an AP: an = a + (n - 1) d
Total number of terms is; n = 32
Initial term is a = 101
Common difference d = 4

Put the given values in nth term formula;

a₃₂ = a + (n - 1) d

a₃₂ = 101 + (32 - 1)(4)

a₃₂ = 101 + 128 - 4

a₃₂ = 225

Thus, the 32nd term value of the arithmetic sequence is calculated to be 225.

More information on the arithmetic sequence can be found here.

https://brainly.com/question/16954227

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