Question

What is the 22nd term of the arithmetic sequence where a1 = 8 and a9 = 56 ?

Answer 1

a 9 = a 1 + 8 d
56 = 8 + 8 d
8 d = 48 
as d is equal to 6

so a22=a1+21d
so we get by putting value 8+126
which is 134

Answer 2

Answer:

The 22nd term of the arithmetic sequence is 134.

Step-by-step explanation:

Given:   The arithmetic sequence where $a_1=8$ and  $a_9=56$

We have to find the 22 term of the arithmetic sequence.

Consider the given sequence with $a_1=8$ and $a_9=56$

We know , For a given sequence in an Arithmetic sequence with first term $a_1$ and common difference d , we have,

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

We first find the common difference "d".

$a_9=56$

$a_9=a_1+(9-1)d$

$a_1=8$ , we have,

$56=8+8d$

Solve for d , we have,

48 = 8d

d = 6

Thus, 22nd term is $a_{22}=a_1+(22-1)d$

$a_{22}=8+21\cdot 6$

$a_{22}=134$

Thus, The 22nd term of the arithmetic sequence is 134.