The fifth term of an arithmetic series is 9, and the sum of the first 16 terms is 480. How do you find the first three terms of the sequence?

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The fifth term of an arithmetic series is 9, and the sum of the first 16 terms is 480. How do you find the first three terms of the sequence?

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Answer 1 · 2016-01-24T14:14:39

Solve to find the first term and common difference of the sequence and hence the first three terms:

$\frac{15}{7}$ $15/7$ , $\frac{27}{7}$ $27/7$ , $\frac{39}{7}$ $39/7$

Explanation:

The general term of an arithmetic sequence is given by the formula:

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

where $a$ $a$ is the initial term and $d$ $d$ is the common difference.

The sum of $N$ $N$ consecutive terms of an arithmetic sequence is $N$ $N$ times the average of the first and last terms, so:

$16 \sum n = 1 a_{n} = 16 \frac{a_{1} + a_{16}}{2} = 8 (a + (a + 15 d)) = 16 a + 120 d$ $sum_(n=1)^16 a_n = 16 (a_1 + a_16)/2 = 8(a+(a+15d)) = 16a+120d$

From the conditions of the question we have:

$240 = 16 a + 120 d$ $240 = 16a + 120d$

$9 = a_{5} = a + 4 d$ $9 = a_5 = a+4d$

Hence:

$96 = 240 - 16 \cdot 9 = (16 a + 120 d) - 16 (a + 4 d)$ $96 = 240 - 16*9 = (16a+120d) - 16(a+4d)$

$= color(red)(cancel(color(black)(16a)))+120d - color(red)(cancel(color(black)(16a)))-64d = 56d$

So $d = 96/56 = 12/7$

and $a = 9 - 4d = 9-48/7 = 15/7$

Hence the first three terms of the series are:

$15/7$ , $27/7$ , $39/7$

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The fifth term of an arithmetic series is 9, and the sum of the first 16 terms is 480. How do you find the first three terms of the sequence?

1 Answer

Explanation:

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