Question

Given the first term and the common difference of an arithmetic sequence find the recursive formula and the three terms in the sequence after the last one given.show work please!

Answer 1

An arithmetic sequence can be expressed explicitly or recursively

The recursive formula of an arithmetic sequence is calculated as:

$a_{n+1} =a_n + d$

(19) a1 = 3/5 and d = -1/3

The recursive formula of the arithmetic sequence isL

$a_{n+1} =a_n -\frac 13$

The second term of the arithmetic sequence is:

$a_{1+1} =a_1 -\frac 13$

$a_2 =\frac 35 - \frac 13$

$a_2 =\frac 4{15}$

The third term of the arithmetic sequence is:

$a_{3} =a_2 -\frac 13$

$a_3 =\frac 4{15} - \frac 13$

$a_3 =-\frac 1{15}$

Hence, the first three terms of the sequence are 3/5, 4/15 and -1/15

(20) a1 = 39 and d = -5

The recursive formula of the arithmetic sequence is:

$a_{n+1} =a_n -5$

The second term of the arithmetic sequence is:

$a_2 =a_1 -5$

$a_2 =39 - 5$

$a_2 =34$

The third term of the arithmetic sequence is:

$a_3 =a_2 -5$

$a_3 =34-5$

$a_3 =29$

Hence, the first three terms of the sequence are 39, 34 and 29

(21) a1 = -26 and d = 200

The recursive formula of the arithmetic sequence is:

$a_{n+1} =a_n +200$

The second term of the arithmetic sequence is:

$a_2 =a_1 +200$

$a_2 =-26 +200$

$a_2 =174$

The third term of the arithmetic sequence is:

$a_3 =a_2 +200$

$a_3 =174 +200$

$a_3 =374$

(22) a1 = -9,2 and d = 0.9

The recursive formula of the arithmetic sequence is:

$a_{n+1} =a_n +0.9$

The second term of the arithmetic sequence is:

$a_2 =a_1 +0.9$

$a_2 =-9.2 +0.9$

$a_2 =-8.3$

The third term of the arithmetic sequence is:

$a_3 =a_2 +0.9$

$a_3 =-8.3+0.9$

$a_3 =-7.4$

Hence, the first three terms of the sequence are -9.2, -8.3 and -7.4

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