Question

What is a21 of the arithmetic sequence for which a7=−19 and a10=−28? A. -35 B. 35 C. -58 D. -61

Answer 1

Answer:

a21 = -61

Step-by-step explanation:

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

$-19=a_{1}+(7-1)d$

$-28=a_{1}+(10-1)d$ (subtract to eliminate a₁)

9 = -3d

d = -3

-19 = a₁ + (6)(-3)

-1 = a

a21 = -1 + (21 - 1)(-3)

= -61

Answer 2

Answer:

-61 (Answer D)

Step-by-step explanation:

The general formula for an arithmetic sequence with common difference d and first term a(1) is

a(n) = a(1) + d(n - 1)

Therefore, a(7) = -19 = a(1) + d(7 - 1), or a(7) = a(1) + d(6) = -19

and a(10) = a(1) + d(10 - 1) = -28, or a(1) + d(10 - 1) = -28

Solving the first equation a(1) + d(6) = -19 for a(1) yields a(1) = -19 - 6d.  We substitute this result for a(1) in the second equation:

-19 - 6d + 9d = -28.  Grouping like terms together, we get:

3d = -9, and so d = -3.

Going back to an earlier result:  a(1) = -19 - 6d.

Here, a(1) = -19 - 6(-3), or a(1) = -1.

Then the formula specifically for this case is a(n) = -1 - 3(n - 1)

and so a(21) = -1 - 3(20) = -61 (Answer D)