Question

An arithmetic sequence has its 4th term equal to 28 and its 10th term is 70. Find its 55th term.

an =
n =

an =
n =
a1 =
d =
a55 =

Answer 1

Answer:

The 55th term of the sequence is 385.

Step-by-step explanation:

For an arithmetic sequence, the fourth term,

$a_{4}=28$

and the tenth term,

$a_{10}=70$

We know that the $n^{th}$ term of an arithmetic sequence is,

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

where $a$ is the first term and $d$ is the common difference of the sequence.

So using this we have the following equations.

$a_4=a+(4-1)d=28\\\\\implies a+3d=28$

and

$a_{10}=a+(10-1)d=70\\\\\implies a+9d=70$

Second equation - first equation yields

$9d-3d=70-28\\\\\implies 6d= 42\\\\\implies d= 7$

So the common difference is 7. Substitute the value of $d$ in one of the equations, we obtain,

$a+3d=28\\\\\implies a+3\times 7=28\\\\\implies a+21=28\\\\\implies a=28-21=7$

So the first term is 7. Hence the 55th term is,

$a+(55-1)d=7+54\times 7 =385$