Question

An arithmetic sequence has a 2nd term equal to 3 and 10th term equal to -13.
Find the term of the sequence that has value -93?​

Answer 1

Answer:

n = 50

Step-by-step explanation:

Let a be the first term and d be the common difference.

An arithmetic sequence has a 2nd term equal to 3 and 10th term equal to -13.

$a_n=a+(n-1)d$

According to the given condition,

$a_2=3\\\\a_{10}=-13$

or

$a+(2-1)d=3\\\\a+d=3\ ...(1)\\\\a+(10-1)d=-13\\\\a+9d=-13\ ...(2)$

Subtract equation (1) from (2).

a+9d-(a+d) = -13-3

8d = -16

d = -2

Put the value of d in equation (1).

a+(-2) = 3

a = 3+2

a = 5

Now,

$a+(n-1)d = -93\\\\5+(n-1)(-2)=-93\\\\5-2n+2=-93\\\\7+93=2n\\\\2n=100\\\\n=50$

So, 50th term has the value of -93.