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The missing number in the arithmetic sequence [tex]$16, 28, \ldots, 52$[/tex] is

A. 42
B. 38
C. 39
D. 40
E. 41

Sagot :

Let's solve this arithmetic sequence problem step by step.

We are given an arithmetic sequence with specific known elements:
- The first term [tex]\( a_1 = 16 \)[/tex]
- The third term [tex]\( a_3 = 28 \)[/tex]
- The fifth term [tex]\( a_5 = 52 \)[/tex]

In an arithmetic sequence, the difference between consecutive terms is constant and is called the "common difference," denoted by [tex]\( d \)[/tex].

Firstly, to find [tex]\( d \)[/tex], we know that:
[tex]\[ a_5 = a_1 + 4d \][/tex]

Given [tex]\( a_5 = 52 \)[/tex] and [tex]\( a_1 = 16 \)[/tex]:
[tex]\[ 52 = 16 + 4d \][/tex]
Solving for [tex]\( d \)[/tex]:
[tex]\[ 52 - 16 = 4d \][/tex]
[tex]\[ 36 = 4d \][/tex]
[tex]\[ d = \frac{36}{4} \][/tex]
[tex]\[ d = 9 \][/tex]

Now that we have the common difference [tex]\( d = 9 \)[/tex], we next need to find the missing number which is the fourth term [tex]\( a_4 \)[/tex].

Using the value of [tex]\( d \)[/tex] and the fact that [tex]\( a_3 = 28 \)[/tex]:
[tex]\[ a_4 = a_3 + d \][/tex]
[tex]\[ a_4 = 28 + 9 \][/tex]
[tex]\[ a_4 = 37 \][/tex]

Therefore, the missing number in the given arithmetic sequence [tex]\( 16, 28, \_\_, 52 \)[/tex] is [tex]\( \boxed{37} \)[/tex].