How do you write the nth term rule for the arithmetic sequence with $a_{7} = 34$ $a_7=34$ and $a_{18} = 122$ $a_18=122$ ?

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How do you write the nth term rule for the arithmetic sequence with $a_{7} = 34$ $a_7=34$ and $a_{18} = 122$ $a_18=122$ ?

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Answer 1 · 2016-07-31T06:33:26

$n^{t h}$ $n^(th)$ term of the arithmetic sequence is $8 n - 22$ $8n-22$ .

Explanation:

$n^{t h}$ $n^(th)$ term of an arithmetic sequence whose first term is $a_{1}$ $a_1$ and common difference is $d$ $d$ is $a_{1} + (n - 1) d$ $a_1+(n-1)d$ .

Hence $a_{7} = a_{1} + (7 - 1) \times d = 34$ $a_7=a_1+(7-1)xxd=34$ i.e. $a_{1} + 6 d = 34$ $a_1+6d=34$

and $a_{18} = a_{1} + (18 - 1) \times d = 122$ $a_18=a_1+(18-1)xxd=122$ i.e. $a_{1} + 17 d = 122$ $a_1+17d=122$

Subtracting firt equation from second equation, we get

$11 d = 122 - 34 = 88$ $11d=122-34=88$ or $d = \frac{88}{11} = 8$ $d=88/11=8$

Hence $a_{1} + 6 \times 8 = 34$ $a_1+6xx8=34$ or $a_{1} = 34 - 48 = - 14$ $a_1=34-48=-14$

Hence $n^{t h}$ $n^(th)$ term of the arithmetic sequence is $- 14 + (n - 1) \times 8$ $-14+(n-1)xx8$ or $- 14 + 8 n - 8 = 8 n - 22$ $-14+8n-8=8n-22$ .

Answer 2 · 2016-07-31T06:37:04

$a_{n} = 8 n - 22$ $color(blue)(a_n=8n-22)$

Explanation:

The given data are

$a_{7} = 34$ $a_7=34$ and $a_{18} = 122$ $a_18=122$

We can set up 2 equations

$a_{n} = a_{1} + (n - 1) \cdot d$ $a_n=a_1+(n-1)*d$

$a_{7} = a_{1} + (7 - 1) \cdot d$ $a_7=a_1+(7-1)*d$

$34 = a_{1} + 6 \cdot d$ $34=a_1+6*d" "$ first equation
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$a_{n} = a_{1} + (n - 1) \cdot d$ $a_n=a_1+(n-1)*d$

$a_{18} = a_{1} + (18 - 1) \cdot d$ $a_18=a_1+(18-1)*d$

$122 = a_{1} + 17 \cdot d$ $122=a_1+17*d" "$ second equation

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

By method of elimination using subtraction, let us use first and second equations

$34 = a_{1} + 6 \cdot d$ $34=a_1+6*d" "$ first equation

$122 = a_{1} + 17 \cdot d$ $122=a_1+17*d" "$ second equation

By subtraction, we have the result

$88 = 0 + 11 d$ $88=0+11d$

$d = \frac{88}{11} = 8$ $d=88/11=8$

Solving now for $a_{1}$ $a_1$ using the first equation and $d = 8$ $d=8$

$34 = a_{1} + 6 \cdot d$ $34=a_1+6*d" "$ first equation

$34 = a_{1} + 6 \cdot 8$ $34=a_1+6*8" "$

$34 = a_{1} + 48$ $34=a_1+48$

$a_{1} = - 14$ $a_1=-14$

We can write the $n t h$ $nth$ term rule now

#a_n=-14+8*(n-1)

$a_{n} = - 14 - 8 + 8 n$ $a_n=-14-8+8n$

$a_{n} = 8 n - 22$ $color(blue)(a_n=8n-22)$

God bless....I hope the explanation is useful.

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How do you write the nth term rule for the arithmetic sequence with $a_{7} = 34$ $a_7=34$ and $a_{18} = 122$ $a_18=122$ ?

2 Answers

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How do you write the nth term rule for the arithmetic sequence with a7=34a_7=34 and a18=122a_18=122?

2 Answers

Explanation:

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How do you write the nth term rule for the arithmetic sequence with $a_{7} = 34$ $a_7=34$ and $a_{18} = 122$ $a_18=122$ ?