Question

the sum of 5 consecutive even numbers is 200. what is the 5th number in this sequence?

Answer 1

Answer:

n+(n+1)+(n+2)+(n+3)+(n+4)= 200

n+n+n+n+n+1+2+3+4= 200

5n+10= 200

5n= 200-10

5n= 190

n= 190/5

n=38

Now substitute n by 38.

38+(38+1)+(38+2)+(38+3)+(38+4)= 200

38+39+40+41+42= 200

77+40+83= 200

160+40= 200

200= 200

Step-by-step explanation:

Hope it helps.

Answer 2

Answer:

$tn = a + n - 1 \times d \\ 5 = 5 + 200 - 1 \times 5 \\ 5 = 205 - 1 \times 5 \\ 5 = 204 \times 5 \\ 5 = 1020 \\ tn = 1020 \div 5 \\ tn = 25$

is a answer