Question

What is the 23rd term of the arithmetic sequence where a1 = 8 and a9 = 48?

Answer 1

Answer:

23rd term of the arithmetic sequence is 118.

Step-by-step explanation:

In this question we have been given first term a1 = 8 and 9th term a9 = 48

we have to find the 23rd term of this arithmetic sequence.

Since in an arithmetic sequence

$T_{n}=a+(n-1)d$

here a = first term

n = number of term

d = common difference

since 9th term a9 = 48

48 = 8 + (9-1)d

8d = 48 - 8 = 40

d = 40/8 = 5

Now $T_{23}= a + (n-1)d$

= 8 + (23 -1)5 = 8 + 22×5 = 8 + 110 = 118

Therefore 23rd term of the sequence is 118.