Consider the arithmetic sequences 5, 13, 21What 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? About the author Quinn
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: [tex]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[/tex] Reply
Answer:
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:
[tex]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[/tex]