Question

Find the 96th term of the arithmetic sequence 1, -12, -25, ...1,−12,−25,

Answer 1

Answer:

$a_9_6=-1,234$

Step-by-step explanation:

we know that

The rule to calculate the a_n term in an arithmetic sequence is

$a_n=a_1+d(n-1)$

where

d is the common difference

a_1 is the first term

n is the number of terms

we have

$1,-12,-25,...$

Remember that In an Arithmetic Sequence the difference between one term and the next is a constant. and this constant is called the common difference

so

$a_1=1\\a_2=-12\\a_3=-25$

$d=a_2-a_1=-12-1=-13$

Find the 96th term of the arithmetic sequence

$a_n=a_1+d(n-1)$

we have

$a_1=1\\d=-13\\n=96$

substitute

$a_9_6=1+(-13)(96-1)$

$a_9_6=1+(-13)(95)$

$a_9_6=1-1,235$

$a_9_6=-1,234$