Part A
This is a question on Arithmetic Progression (A.P)
The nth term of an A.P is given as;
[tex]\begin{gathered} A_n=a+(n-1)d \\ \text{where A}_n\text{ is the nth term} \\ n\text{ is the number of terms in the sequence} \\ d=\text{common difference} \end{gathered}[/tex][tex]\begin{gathered} when=10,A_{10}=15 \\ A_{10}=a+(10-1)d \\ 15=\text{ a+9d} \end{gathered}[/tex][tex]\begin{gathered} \text{when n = 14, A}_{14}=35 \\ A_{14}=a+(14-1)d \\ 35=a+13d \end{gathered}[/tex]Hence, we have two linear equations to be solved simultaneously
a + 9d = 15........................ equation 1
a + 13d = 35 ..................... equation 2
Subtracting equation 1 from equation 2, we have;
13d - 9d = 35 - 15
4d = 20
[tex]\begin{gathered} d=\frac{20}{4}=5 \\ d=5 \end{gathered}[/tex]The common difference is 5
Part B
Substituting d = 5 into equation 1, we have;
a + 9(5) = 15
a + 45 = 15
a = 15 - 45
a = -30
The first term is -30
Part C
The general equation from the equation of the nth term of an A.P putting the values of a and d gotten will be;
[tex]\begin{gathered} A_n=-30+(n-1)5 \\ A_n=-30_{}+5n-5 \\ A_n=-35+5n \end{gathered}[/tex]Part D
From the general equation of the sequence gotten above, we can deduce that
[tex]A_n=5n-35[/tex]Which is similar to equation of a straight line in the form y = mx + C.
The graph to be plotted will be of the form y = 5x - 35
Part E
[tex]\begin{gathered} A_n=-30+(n-1)5 \\ A_n=-30+5n-5 \\ A_n=-35+5n \\ A_n=5n-35 \\ y=mx+b \\ \text{This can be related to y = mx + b which is the equation of a straight line} \end{gathered}[/tex]
