Question

The first four terms of an arithmetic sequence are given.

27, 32, 37, 42, ...

What is the 60th term of the sequence?

Answer 1

Answer:

$a_6_0=322$

Step-by-step explanation:

we know that

The rule to calculate the an term in an arithmetic sequence is

$a_n=a_1+d(n-1)$

where

d is the common difference

a_1 is the first term

we have that

$a_1=27\\a_2=32\\a_3=37\\a_4=42$

$a_2-a_1=32-27=5$

$a_3-a_2=37-32=5$

so

The common difference is d=5

$a_4-a_3=42-37=5$

Find  60th term of the sequence

$a_n=a_1+d(n-1)$

we have

$a_1=27\\d=5\\n=60$

substitute

$a_6_0=27+5(60-1)$

$a_6_0=27+5(59)$

$a_6_0=322$

Answer 2

The 60th term of the sequence should be 322 when the first four terms should be given.

Calculation of the 60th term of the sequence:

Since

a1 = 27

a2 = 32

a3 = 37

And, a4 = 42

So,

= 27 + 5(60 - 1)

= 27 + 5(59)

= 322

hence, The 60th term of the sequence should be 322 when the first four terms should be given.

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