Question

In an arithmetic sequence, a17 = -40 and

Answer 1

$a_{17}$= -40

a + 16d = -40 ...... -(i)

$a_{28}$= -73

a + 27d = -73 .......(ii)

Subtracting (i) and (ii)

a + 16 d - a - 27 d = - 40 + 73

-11d = 33

d = -3

a = -40 - 16d = -40 - 16(-3) = 8

$a_n$ = 8 + (n-1)-3

= 8 -3n +3

= -3n + 11

Hence recursive formula :

$a_n$= -3n + 11

Answer 2

Answer:

$a_n=-3n+11$

Step-by-step explanation:

We are given the values of two terms in this arithmetic sequence: $a_{17}=-40$ and $a_{28}=-73$. We want to find the recursive formula of this sequence, which will be in the form $a_n=a_1+d(n-1)$, where $a_1$ is the first term and d is the common difference.

Here, we can pretend that $a_{17}$ will replace the $a_1$ term, while $a_{28}$ replaces the $a_n$ term. This way, n becomes 28 and 1 becomes 17. Now, we can write:

$a_n=a_1+d(n-1)$

$a_{28}=a_{17}+d(28-17)$

Substitute in the values we know:

$a_{28}=a_{17}+d(28-17)$

$-73=-40+d(28-17)$

Solve for d:

$-73=-40+d(28-17)$

$-73=-40+d(11)$

11d = -33

d = -3

Now, we need to find our first term. We can do this by replacing $a_n$ with $a_{28}$ again, but this time, we're actually going to use $a_1$:

$a_{28}=a_1+d(28-1)$

Plug in the values we know:

$a_{28}=a_1+d(28-1)$

$-73=a_1-3(27)$

Solve for $a_1$:

-73 = $a_1$ - 81

$a_1$ = 8

Put these altogether:

$a_n=8-3(n-1)=-3n+11$

Thus, the recursive formula is $a_n=-3n+11$.