Question

Consider the arithmetic sequences 5, 13, 21
What is the common difference?
b) Write the next two terms.
c) Can the difference of any two terms of this sequence be 120?
1) Check whether 2020 is a term of this sequences?
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Answer 1

Answer:

1. 29 and 37 is next sequence

Answer 2

Answer:

a) Common Difference is 8

b) Next two terms are 29 and 37

c) T16 - T1 = 125 - 5 = 120

d) 2020 is not a term of this sequence

Step-by-step explanation:

$Arithmetic \: \: Sequence \: \: is : \\ 5, \: \: 13, \: \: 21,... \\ \\ a) \: \: \: d = T_{2} - T_{1} = 13 - 5 = 8 \\ \: \: \: \: \: Common \: \: Difference = 8 \\ \\ b) \: \: \: 21 + 8 = 29 \\ \: \: \: \: \: 29 + 8 = 37 \\ \: \: \: \: \: The \: \: Next \: \: two \: \: terms \: \: are : 29 \: \: and \: \: 37 \\ \\ c) \: \: \: 8 \times 15 = 120 \\ \boxed{ a = 5} \\ \: \: \: \: \: T_{16} = a + 15d \\ \: \: \: \: \: T_{16} = 5 + 15(8) \\ \: \: \: \: \: T_{16} = 5 + 120 \\ \: \: \: \: \: \boxed{ T_{16} = 125} \\ \: \: \: \: \: ∴ \: \: T_{16} - T_{1} = 125 - 5 = 120 \\ \\ d) \: \: \: T_{n} = a + (n - 1)d \\ 2020 = 5 + (n - 1)8 \\ 2020 = 5 + 8n - 8 \\ 2020 = 8n - 3 \\ 2020 + 3 = 8n \\ 2023 = 8n \\ \frac{2023}{8} = n \\ \boxed{ \underline{ \underline{252 \frac{7}{8}}} = n} \\ ∴ \: \: 2020 \: \: is \: \: not \: \: a \: \: term \: \: of \: \: this \: \: sequence$