Question

can you guys help me solve A(n) = 1 + (n –1)(–5.7) find the first, fourth, and tenth terms.....sorry fro not putting this before:(

Answer 1

Answer:

I think this is of the chapter Arithmetic Progressions. So, here a term n is found out by the formula a+(n-1)*d

where a is the first term and d is the common difference.

As you have provided,

n= 1+(n-1)(-5.7)

Here a is 1. So the first term is 1.

Now find the 4th term. n will be 4.

a+(n-1)*d

1+(4-1)*(-5.7)

1+3*(-5.7)

1-17.1

-16.1

Now find the 10th term. n will be 10.

a+(n-1)*d

1+(10-1)*(-5.7)

1+9*(-5.7)

1-51.3

-50.3

So the first term of the AP is 1, the fourth term is -16.1 and the tenth term is -50.3.

Hope this helps...

Answer 2

Answer:

1 ; - 16.1 ; - 50.3

Step-by-step explanation:

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

A(n) = $a_{n}$ = 1 + (n – 1)( – 5.7) ⇒ $a_{1}= 1$ and d = - 5.7

$a_{1}= 1$

$a_{2}$ = 1 + (2 - 1)(- 5.7) = - 4.7

$a_{4}$ = 1 + (4 - 1)(- 5.7) = - 16.1

$a_{10}$ = 1 + (10 - 1)( - 5.7) = - 50.3