Step-by-step explanation:
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the shape answers could be incorrect-
Answer:
- 9 student tickets and 8 general admission tickets
Step-by-step explanation:
Let the number of student tickets is s and general admission tickets - g.
<u>Set equations and solve:</u>
<u>Double the first equation and subtract from the second to eliminate s:</u>
- 2s + 3g - 2s - 2g = 42 - 2*17
- g = 8
<u>Find s:</u>
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General formula for n-th term of arithmetical progression is
a(n)=a(1)+d(n-1).
For 3d term we have
a(3)=a(1) +d(3-1), where a(3)=7
7=a(1)+2d
For 7th term we have
a(7)=a(1) +d(7-1)
a(7)=a(1) + 6d
Also, we have that the <span>seventh term is 2 more than 3 times the third term,
a(7)=3*a(3)+2= 3*7+2=21+2=23
So we have, </span>a(7)=a(1) + 6d and a(7)=23. We can write
23=a(1) + 6d.
Now we can write a system of equations
23=a(1) + 6d
<span> - (7=a(1)+2d)
</span>16 = 4d
d=4,
7=a(1)+2d
7=a(1)+2*4
a(1)=7-8=-1
a(1)= - 1
First term a(1)=-1, common difference d=4.
Sum of the 20 first terms is
S=20 * (a(1)+a(20))/2
a(1)=-1
a(n)=a(1)+d(n-1)
a(20) = -1+4(20-1)=-1+4*19=75
S=20 * (-1+75)/2=74*10=740
Sum of 20 first terms is 740.
Add 2 to 3 and get 5 add 3 and get 8 add 5 to get 13 add 8 to get 21
