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gavmur [86]
3 years ago
14

The 80th term of an arithmetic sequence is twice the 30th term. If the first term of the sequence is 7, what is the 40th term?

Mathematics
1 answer:
VikaD [51]3 years ago
8 0
It seems that "d" or the common difference is a number that makes the sequence go into the larger positive realm of numbers. Therefore, 2*(7+30d)=7+80d. d=35/100. Now just do 7 + 40(35/100) = 21. 21 is your answer!
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Base 11 has the digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A} where A is treated as a single digit number

Base 12 has the digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B}

Base 13 has the digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C}

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Step-by-step explanation:

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For the arithmetic sequence 42, 32, 22, 12... find the 18th term.
Stella [2.4K]

Answer:

The18th term of the given sequence is -128

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To find the 18th term of the sequence:

42, 32, 22, 12, ..., we need to find the nth term of the sequence first.

The nth term of a sequence is given be the formula:

T_n=a+(n-1)d

Where a is the first term, and d is the common difference.

Here, a = 42, d = 32 - 42 = -10

\begin{gathered} T_n=42+(n-1)(-10) \\ =42-10n+10 \\ T_n=52-10n \end{gathered}

To find the 18th terem, substitute n = 18 into the nth term

\begin{gathered} T_{18}=52-10(18) \\ =52-180 \\ =-128 \end{gathered}

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