Question

What is the 25th term of the arithmetic sequence where a1 = 8 and a9 = 48? A 113B 118C 123D 128

Answer 1

To solve for the 25th term of the arithmetic sequence:

$\begin{gathered} a_1=8 \\ a_9=48 \end{gathered}$

For an arithmetic sequence

$\begin{gathered} T_n=a+(n-1_{})d \\ a=a_1=8 \\ a_9=a+(9-1)d \\ a_9=a+8d \\ 48=8+8d \\ 48-8=8d \\ 40=8d \\ \text{divide both side by 8} \\ \frac{40}{8}=\frac{8d}{8} \\ d=4 \end{gathered}$

25th term = T_25

$\begin{gathered} T_{25}=a+(25-1)d \\ T_{25}=a+24d \\ T_{25}=8+24(5) \\ T_{25}=8+120 \\ T_{25}=128 \end{gathered}$

Therefore the 25th term of the arithmetic sequence = 128

Hence the correct answer is Option D