Question

find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule A(n)=-6+(n-1)(1/5)

Answer 1

Answer with explanation:

The formula that represents the nth term of a arithmetic sequence is given by:

`A(n)=-6+(n-1)((1)/(5))`

Now, we are asked to find the first, fourth, and tenth terms of the arithmetic sequence.

i.e. we are asked to find the value of A(n) when n=1 ,4 and 10

• when n=1 we have:

`A(1)=-6+(1-1)((1)/(5))\\\\i.e.\\\\A(1)=-6+0\\\\i.e.\\\\A(1)=-6`

• now when n=4 we have:

`A(4)=-6+(4-1)* ((1)/(5))\\\\i.e.\\\\A(4)=-6+3* (1)/(5)\\\\i.e.\\\\A(4)=-6+(3)/(5)\\\\i.e.\\\\A(4)=(-6* 5+3)/(5)\\\\i.e.\\\\A(4)=(-30+3)/(5)\\\\i.e.\\\\A(4)=(-27)/(5)`

• when n=10 we have:

`A(10)=-6+(10-1)* ((1)/(5))\\\\i.e.\\\\A(10)=-6+9* (1)/(5)\\\\i.e.\\\\A(10)=-6+(9)/(5)\\\\i.e.\\\\A(10)=(-6* 5+9)/(5)\\\\i.e.\\\\A(10)=(-30+9)/(5)\\\\i.e.\\\\A(10)=(-21)/(5)`

Answer 2

A(1) = -6 +(1 -1)*(1/5) = -6

A(4) = -6 +(4 -1)*(1/5) = -5 2/5

A(10) = -6 +(10 -1)*(1/5) = -4 1/5