The explicit formula for arithmetic sequence is:
an=a+(n-1)d
where:
a=first term
d=common difference
given:
a3=22
a(17)=-20
substituting this in our equation we get:
22=a+(3-1)d
22=a+2d
a=22-2d........i

also:
-20=a+(17-1)d
-20=a+16d.....ii
but substituting i in ii we get:
-20=22-2d+16d
-20-22=14d
-42=14d
d=-3
but:
a=22-2d
a=22-2(-3)
a=28
thus the formula will be:
an=28-3(n-1)
thus the first term will be 28
the 2nd term will be:
a2=28-3(2-1)
a2=25

the 3rd term will be:
a3=28-3(3-1)
a3=28-6
a3=22

a4=28-3(4-1)
a4=28-9
a4=15

a5=28-3(5-1)
a5=28-3(4)
a5=28-12
a5=15

5 0

2 years ago

Solve this equation 7x+3=5(x-3)+2x

zaharov [31]

Answer: nonsense

Step-by-step explanation:

7x+3=5(x-3)+2x

7x+3=5x-15+2x

7x+3-5x+15-2x=0

18=0

Non-sense

8 0

2 years ago

Simplify the expression below.<br> 4+2^3 -|-4|

pantera1 [17]

Answer:

8

Explanation:

Too lazy to type it all

5 0

3 years ago

Read 2 more answers

(1 ÷ x- 1)+(2÷ x+2)=(3÷2) solve and check for extraneous solutions

Sliva [168]

$\frac{1}{x-1}+ \frac{2}{x+2}= \frac{3}{2}$
$\frac{1(x+2)}{(x-1)(x+2)}+ \frac{2(x-1)}{(x+2)(x-1)}= \frac{3(x-1)(x+2)}{2}$
$\frac{x+2+2(x-1)}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}$
$\frac{x+2+2x-1}{(x-1)(x+2)}= \frac{3(x-1)(x+2)}{2}$
$\frac{3x+1}{(x-1)(x+2)}= \frac{3(x^2+x-2)}{2}$
$\frac{}{(x-1)(x+2)}= \frac{3x^2+3x-6)}{2}$
$\frac{3x+1}{(x^2+x-2)}= \frac{3x^2+3x-6)}{2}$
$\frac{2(3x+1)}{2(x^2+x-2)}= \frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}$
$\frac{2(3x+1)}{2(x^2+x-2)}-\frac{3x^2+3x-6)(x^2+x-2)}{2(x^2+x-2)}=0$
$\frac{2(3x+1)-3x^2+3x-6}{2(x^2+x-2)}=0$

this will be continued

5 0

3 years ago