1 Answer

Chatina · Answer 1 · 2023-03-10T08:23:04+0000

To calculate the first term of the quadratic sequence given;

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

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;

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;

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

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