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3 years ago

What is the 23rd term of the arithmetic sequence where a1 = 8 and a9 = 48?

Mathematics

1 answer:

joja [24]3 years ago

4 0

Answer:

23rd term of the arithmetic sequence is 118.

Step-by-step explanation:

In this question we have been given first term a1 = 8 and 9th term a9 = 48

we have to find the 23rd term of this arithmetic sequence.

Since in an arithmetic sequence

$T_{n}=a+(n-1)d$

here a = first term

n = number of term

d = common difference

since 9th term a9 = 48

48 = 8 + (9-1)d

8d = 48 - 8 = 40

d = 40/8 = 5

Now $T_{23}= a + (n-1)d$

= 8 + (23 -1)5 = 8 + 22×5 = 8 + 110 = 118

Therefore 23rd term of the sequence is 118.

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
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\\ \quad \\
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\\ \quad \\
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\\ \quad \\
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\\ \quad \\
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$\bf \begin{array}{llll}
% right side info
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\\\\

\end{array}\\$

$\bf \begin{array}{llll}
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\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}
$

now, notice the template above... now let's see your function $\bf y=x^2+4\implies 
\begin{array}{llllll}
y=&1(&1x&+&0)^2+&4\\
&A&B&&C&D
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2 years ago

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Answer:

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Combine like terms

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