Question

What is the first term of the quadratic sequence:x; 16; 41; 71; ...

Answer 1

To calculate the first term of the quadratic sequence given;

`x,16,41,71`

Take note that the nth term of a quadratic sequence is given by the formula;

`an^2+bn+c`

We shall use the terms provided to find the values of a, b and c, after which we shall use these to find the value of the first term.

Let us first of all confirm its a quadratic sequence by determining the first difference and the second difference.

`\begin{gathered} 1st\text{ Difference;} \\ 25,30 \end{gathered}`

We find the common difference for the first difference and this will serve as our 2nd difference.

`\begin{gathered} 2nd\text{ Difference;} \\ 5,5 \end{gathered}`

We now divide the 2nd difference by 2 to get the value of a (in the equation for the nth term).

`\begin{gathered} a=(5)/(2) \\ a=2.5 \end{gathered}`

If a = 2.5, the the first term in the nth term is;

`2.5n^2`

Next we substitute the numbers 1 to 4 into 2.5n^2 and we'll have;

`\begin{gathered} 1,2,3,4 \\ 2.5(1)^2=2.5 \\ 2.5(2)^2=10 \\ 2.5(3)^2=22.5 \\ 2.5(4)^2=40 \end{gathered}`

We now have a sequence which is;

`2.5,10,22.5,40`

And the differences are;

7.5, 12.5, 17.5

We shall take the nth term of this sequence of differences as follows;

`\begin{gathered} U_n=a_1+(n-1)d \\ Where\text{ a}=7.5,d=5 \\ U_n=7.5+(n-1)5 \\ U_n=7.5+(5n-5) \\ U_n=7.5+5n-5 \\ U_n=5n+2.5 \end{gathered}`

With this result, we now have the values of b and c (to be used in the equation of the nth term).

`\begin{gathered} b=5n \\ c=2.5 \end{gathered}`

Where the equation is;

`\begin{gathered} an^2+bn+c \\ a=2.5n^2,b=5n,c=2.5 \\ \end{gathered}`

The first term becomes;

`\begin{gathered} a_1=2.5(1)^2+5(1)+2.5 \\ a_1=2.5+5+2.5 \\ a_1=10 \end{gathered}`

Withn this formula for the nth term you can go ahead and find any indicated term in this quadratic sequence.

ANSWER:

The first term is 10