Answer:
Step-by-step explanation:
<h3>Arithmetic sequence:</h3>
6th term is 14 ⇒
a + (6 - 1)d = 14
a + 5d = 14 --------------(I)
14th term is 6 ⇒
a + (14-1)d = 6
a + 13d = 6 ----------------(II)
Subtract equation (II) from equation(I)
(I) a + 5d = 14
(II) a + 13d = 6
<u>- - -</u>
-8d = 8
d = 8 ÷(-8)
Plugin d = -1 in equation (I)
a + 5(-1) = 14
a -5 = 14
a = 14 + 5
20th term:
= 19 - 19
Answer: 560
Step-by-step explanation:
There are [20, 22, 24, 26, 28]: 5 even integers for each ten numbers.
The 20's set can be written as 5*20 + (2+4 + 6 + 8) = 5*20 + (20) = 120
The 30's can be written: 5*30 + 20 = 170
40's: 5*40+20 = 220
50's: 5*50+20 = 270
60's: 5*60+20 = 320
Each set is incremented by 50
We want the sets for 20, 30, and 40, plus the number 50.
(120+170+220+50) = 560
You can also do this in Excel (attached). Set the first cell to 20, the next cell below equal to the cell above plus 2. Then draw the second cell down until you've reached 50. Then sum the cells to arrive at 560.
1) Let's write out both expressions subtracting 4m²+2mn+8n² from 2m²+6mn+2n²
2) Note that when we subtract 4m^2 + 2mn + 8n^2 from 2m^2 + 6mn + 2n^2 we need to swap the sign by placing -1 outside the parentheses and then combine like terms adding those terms algebraically.