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)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,...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 = 1a 2 = −12

Answer:

a_9_6=-1,234

we know that

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

a_n=a_1+d(n-1)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,...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 = 1a 2 = −12a 3 = −25

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

2−a 1 = −12−1 = −13

Find the 96th term of the arithmetic sequence

a_n=a_1+d(n-1)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=-1,234a 3 = −25

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

2−a 1 = −12−1 = −13

Find the 96th term of the arithmetic sequence

a_n=a_1+d(n-1) 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

$thank \: you$