Question

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Answer 1

Answer: 912

===================================

Work Shown:

The starting term is a1 = 3. The common difference is d = 5 (since we add 5 to each term to get the next term). The nth term formula is

an = a1+d(n-1)

an = 3+5(n-1)

an = 3+5n-5

an = 5n-2

Plug n = 19 into the formula to find the 19th term

an = 5n-2

a19 = 5*19-2

a19 = 95-2

a19 = 93

Add the first and nineteenth terms (a1 = 3 and a19 = 93) to get a1+a19 = 3+93 = 96

Multiply this by n/2 = 19/2 = 9.5 to get the final answer

96*9.5 = 912

I used the formula

Sn = (n/2)*(a1 + an)

where you add the first term (a1) to the nth term (an), then multiply by n/2

-----------------

As a check, here are the 19 terms listed out and added up. We get 912 like expected.

3+8+13+18   +23+28+33+38    +43+48+53+58    +63+68+73+78   +83+88+93 = 912

There are 19 values being added up in that equation above. I used spaces to help group the values (groups of four; except the last group which is 3 values) so it's a bit more readable.

Answer 2

Answer:  912

Step-by-step explanation:

The given sequence {3, 8, 13, 18, ... } provides the following information:

• the first term (a₁) = 3
• the difference (d) = 5

We can use the information above to find the explicit rule of the sequence:

$a_n=a_1+d(n-1)\\.\quad =3+5(n-1)\\.\quad =3+5n-5\\.\quad =5n-2$

We can use the explicit rule to find the 19th term (a₁₉)

$a_n=5n-2\\a_{19}=5(19)-2\\.\quad =95-2\\.\quad =93$

Next, we can input the first and last term of the sequence into the Sum formula:

$S_n=\dfrac{a_1+a_n}{2}\times n\\\\S_{19}=\dfrac{a_1+a_{19}}{2}\times 19\\\\.\quad =\dfrac{3+93}{2}\times 19\\\\.\quad =\dfrac{96}{2}\times 19\\\\.\quad =48\times 19\\\\.\quad =912$