Question

Given the Arithmetic sequence A1,A2,A3,A4 53, 62, 71, 80 What is the value of A38?

Answer 1

Answer:

$A_{38}=386$

Step-by-step explanation:

We have been given an arithmetic sequence gas $A_1,A_2,A_3,A_4$ as :53,62,71,80. We are asked to find $A _{38}$.

We know that an arithmetic sequence is in format $a_n=a_1+(n-1)d$, where,

$a_n$ = nth term,

$a_1$ = 1st term of sequence,

n = Number of terms,

d = Common difference.

We have been given that 1st term of our given sequence is 53.

Now, we will find d by subtracting 71 from 80 as:

$d=80-71=9$

$A_{38}=53+(38-1)9$

$A_{38}=53+(37)9$

$A_{38}=53+333$

$A_{38}=386$

Therefore, $A_{38}=386$.

Answer 2

Answer:

$A_{38} = 350$

Step-by-step explanation:

The 5th term of the arithmetic sequence is 53. We can write the equation:

$a + 4d = 53...(1)$

The 6th term of the arithmetic sequence is 62. We can write the equation:

$a + 5d = 62...(2)$

Subtract the first equation from the second one to get:

$5d - 4d = 62 - 53$

$d = 9$

The first term is

$a + 4(9) = 53$

$a + 36 = 53$

$a = 53 - 36$

$a = 17$

The 38th term of the sequence is given by:

$A_{38} = a + 37d$

$A_{38} = 17+ 37(9)$

$A_{38} = 350$