Question

Write the first ten terms of a sequence whose first term is -10 and whose common difference is -2.

Answer 1

A=first term
n=number of terms
d=common difference

Formula: a+(n-1)d
10th term;
a=-10
n=10
d=-2

-10+(10-1)-2
-10+(9)-2
-10+-18
-10-18
10th term= -28

A.P= -10,-12,-14,-16,-18,-20,-22,-24,-26,-28,...

Answer 2

Answer: The first ten terms are;

-10, -12, -14, -16, -18, -20, -22, -24, -26 and -28

Step-by-step explanation:

This sequence will be an arithmetic sequence since they have a "common difference".

The nth term of a sequence is

Tn = a+(n-1)d

a is the first term = -10

n is the number of terms

d is the common difference = -2

Second term is when n= 2

T2 = a+d

T2 = -10-2 = -13

When n= 3

T3 = a+2d

T3 = -10 + 2(-2)

T3 = -10 -4

T3 = -14

When n=4

T4 = a+3d

T4 = -10 +3(-2)

T4 = -10 - 6

T4 = -16

When n = 5

T5 = a+4d

T5 = -10 + 4(-2)

T5 = -10-8

T5 = -18

When n=6

T6 = a+5d

T6 = -10+5(-2)

T6 = -10-10

T6 = -20

When n= 7

T7 = a+6d

T7 = -10+6(-2)

T7 = -10-12

T7 = -22

When n= 8

T8 = a+7d

T8 = -10+7(-2)

T8 = -10-14

T8 = -24

When n= 9

T9 = a+8d

T9 = -10+8(-2)

T9 = -10-16

T9 = -26

When n= 10

T10 = a+9d

T10 = -10+9(-2)

T10 = -10 - 18

T10 = -28

Therefore the first ten terms are;

-10, -12, -14, -16, -18, -20, -22, -24, -26 and -28