Verified Answer :

meerkat18

To determine the first six terms of the arithmetic progression, add the common difference to the prior term

a1 = 3
a2 = 3 + -10 = -7
a3 = -7 + -10 = -17
a4 = -17 + - 10 = -27
a5 = -27 + -10 = -37
a6 = - 37 + -10 = -47
Plot these values in the number line.

isyllus

Answer:

3, -7, -17, -27, -37, -47

Step-by-step explanation:

Given:  First term: a= 3 and Common Difference, d = -10

We need to find the first six term of the sequence.

Formula: \(a_n=a+(n-1)d\)

where,

\(a_n\) nth term of series

a is first term

n is number of term

d is common difference

For Second term, n=2

\(a_2=3+(2-1)(-10) = 3-10=-7\)

For Third term, n=3

\(a_3=3+(3-1)(-10) = 3-20=-17\)

For Fourth term, n=4

\(a_4=3+(4-1)(-10) = 3-30=-27\)

For Fifth term, n=5

\(a_5=3+(5-1)(-10) = 3-40=-37\)

For Sixth term, n=6

\(a_6=3+(6-1)(-10) = 3-50=-47\)

Hence, The sequence is 3, -7, -17, -27, -37, -47

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