What is the first term of an arithmetic sequence in which the seventh term is 3 and the sixteenth term is -15?

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Answer 1 · 2015-07-17T22:16:33

The first term of the sequence is $15$ $color(red)(15)$ .

We know that $a_{7} = 3$ $a_7 = 3$ and $a_{16} = - 15$ $a_16 = -15$ .

We also know that $a_{n} = a_{1} + (n - 1) d$ $a_n = a_1 + (n-1)d$ .

$a_{16} = a_{1} + (16 - 1) d$ $a_16 = a_1 + (16-1)d$

Equation (1): $- 15 = a_{1} + 15 d$ $-15 = a_1 + 15d$

and

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

Equation (2): $3 = a_{1} + 6 d$ $3 = a_1 + 6d$

Subtract Equation (2) from Equation (1).

$- 15 - 3 = 15 d - 6 d$ $-15-3 = 15d-6d$

$- 18 = 9 d$ $-18 = 9d$

$d = - \frac{18}{9}$ $d = -18/9$

Equation (3): $d = - 2$ $d= -2$

Substitute Equation (3) in Equation (1).

$- 15 = a_{1} + 15 d$ $-15 = a_1 + 15d$

$- 15 = a_{1} + 15 (- 2) = a_{1} - 30$ $-15 = a_1 + 15(-2) = a_1 -30$

$a_{1} = 30 - 15$ $a_1 = 30-15$

$a_{1} = 15$ $a_1 = 15$