Notes: `a_4 = 16, a_10 = 46` Write the first 5 terms of the arithmetic sequence.

Friday, August 29, 2014

$\displaystyle{a}_{{4}}={16},{a}_{{10}}={46}$ Write the first 5 terms of the arithmetic sequence.

$\displaystyle{a}_{{4}}={16}$

$\displaystyle{a}_{{10}}={46}$

To determine the first five terms of this arithmetic sequence, consider its nth term formula which is:

$\displaystyle{a}_{{n}}={a}_{{1}}+{\left({n}-{1}\right)}{d}$

To apply this, plug-in the given nth terms.

Plugging in a_4=16, the formula becomes:

$\displaystyle{16}={a}_{{1}}+{\left({4}-{1}\right)}{d}$

$\displaystyle{16}={a}_{{1}}+{3}{d}$ (Let this be EQ1.)

Also, substituting a_10=46, the formula becomes:

$\displaystyle{46}={a}_{{1}}+{\left({10}-{1}\right)}{d}$

$\displaystyle{46}={a}_{{1}}+{9}{d}$ (Let this be EQ2.)

Then, use these two equations to solve for the...

$\displaystyle{a}_{{4}}={16}$

$\displaystyle{a}_{{10}}={46}$

To determine the first five terms of this arithmetic sequence, consider its nth term formula which is:

$\displaystyle{a}_{{n}}={a}_{{1}}+{\left({n}-{1}\right)}{d}$

To apply this, plug-in the given nth terms.

Plugging in a_4=16, the formula becomes:

$\displaystyle{16}={a}_{{1}}+{\left({4}-{1}\right)}{d}$

$\displaystyle{16}={a}_{{1}}+{3}{d}$ (Let this be EQ1.)

Also, substituting a_10=46, the formula becomes:

$\displaystyle{46}={a}_{{1}}+{\left({10}-{1}\right)}{d}$

$\displaystyle{46}={a}_{{1}}+{9}{d}$ (Let this be EQ2.)

Then, use these two equations to solve for the values of a_1 and d. To do so, isolate a_1 in the first equation.

$\displaystyle{16}={a}_{{1}}+{3}{d}$

$\displaystyle{16}-{3}{d}={a}_{{1}}$

Plug-in this to the second equation.

$\displaystyle{46}={a}_{{1}}+{9}{d}$

$\displaystyle{46}={16}-{3}{d}+{9}{d}$

$\displaystyle{46}={16}+{6}{d}$

$\displaystyle{30}={6}{d}$

$\displaystyle{5}={d}$

Then, solve for a_1. To do so, plug-in d=5 to the first equation.

$\displaystyle{16}={a}_{{1}}+{3}{d}$

$\displaystyle{16}={a}_{{1}}+{3}{\left({5}\right)}$

$\displaystyle{16}={a}_{{1}}+{15}$

$\displaystyle{1}={a}_{{1}}$

Then, plug-in these two values a_1=1 and d=5 to the formula of nth terms of arithmetic sequence.

$\displaystyle{a}_{{n}}={a}_{{1}}+{\left({n}-{1}\right)}{d}$

$\displaystyle{a}_{{n}}={1}+{\left({n}-{1}\right)}{\left({5}\right)}$

$\displaystyle{a}_{{n}}={1}+{5}{n}-{5}$

$\displaystyle{a}_{{n}}={5}{n}-{4}$

Now that the formula of a_n is known, use this to solve for the values of a_2, a_3 and a_5. (Take note that the values of a_1 and a_4 are already known.)

1st term: $\displaystyle{a}_{{1}}={1}$

2nd term: $\displaystyle{a}_{{2}}={5}{\left({2}\right)}-{4}={6}$

3rd term: $\displaystyle{a}_{{3}}={5}{\left({3}\right)}-{4}={11}$

4th term: $\displaystyle{a}_{{4}}={16}$

5th term: $\displaystyle{a}_{{5}}={5}{\left({5}\right)}-{4}={21}$

Therefore, the first five terms of the arithmetic sequence are {1, 6, 11, 16, 21,...}.

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