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Simple high school math problems!
1. The sum of the first items of an in the arithmetic series is Sn. If S9=72, what is a2+a4+a9?

S9=9a+36d=72 a+4d=8

a2+a4+a9 = a+d+a+3d+a+8d = 3a+ 12d = 3(a+4d)= 3 * 8 = 24

Second, the common ratio in geometric series is 4, and the sum of the first three terms is 2 1. What is the general formula?

a 1 = a a2 = 4a a3 = 16a a 1+a2+a3 = 2 1a = 2 1a = 1

an=4^(n- 1)

3. It is known that abcd is a geometric series, the vertex coordinate of the function y=x squared -2x+8 is (b, c), and what is ad?

The vertex of y = x 2-2x+8 = (x- 1) 2+7 is calculated as (1, 7) b= 1 c=7. q=c/d=7, so a =1/7d =

ad=7

4. arithmetic progression a 1+a2+a3+a4+a5+a6=27, what is a3? (Why? )

a 1+a2+a3+a4+a5+a6 = a3-2d+a3-d+a3+a3+d+a3+2d+a3+3d = 6 a3+3d = 27

I can't figure it out

5. The sum of the first few terms in the geometric series is Sn, 4a 1, 2a2, a3 becomes arithmetic progression. If a 1= 1, then S4=?

sn=a 1( 1-q^n)/( 1-q)

a 1 = 1 4a 1 = 4 a2 = q 2 a2 = 2q a3=q^2

2q-4=q^2-2q q=2

s4=( 1-2^4)/( 1-2)= 15

6. arithmetic progression a3=7, a5+a7=26, and the sum of the preceding items of an is Sn. What about an and Sn?

a5=a3+2d a7=a3+4d

a5+a7=2a3+6d=26

6d=26-2a3=26-2*7= 12

D=2, so a 1=a3-2d=7-4=3.

an=3+(n- 1)*2=2n+ 1

Sn=(3+2n+ 1)n/2=n(n+2)