Answer:
3
Step-by-step explanation:
Let's find the first term in terms of p.
So an arithmetic sequence is a linear relation.
That means it will have the same slope no matter the pair of points used. We are given the slope, the common difference, is 2. We are going to use point (9, 3+3p) and (1, t(1)) along with m=2 to find t(1).
[t(1)-(3+3p)]/[1-9]=2
Simplify denominator
[t(1)-(3+3p)]/[-8]=2
Multiply both sides by -8
t(1)-(3+3p)=-16
Add (3+3p) on both sides
t(1)=-16+3+3p
Combine like terms
t(1)=-13+3p
This means we can find the next term by adding 2 this.
t(2)=-11+3p
Let's find the next term by adding 2 this.
t(3)=-9+3p
Finally we can find the 4th term by adding 2 to this
t(4)=-7+3p
We are given the sum of the first 4 terms is 2p-10. So we can write:
-13+3p+-11+3p+-9+3p+-7+3p=2p-10
Combine like terms on left
12p-40=2p-10
Subtract 2p on both sides
10p-40=-10
Add 40 on both sides
10p=30
Divide both sides by 10
p=3.
-----------------
Checking:
t(1)=-13+3p=-13+3(3)=-13+9=-4
t(2)=-11+3p=-11+3(3)=-11+9=-2
t(3)=-9+3p=-9+3(3)=-9+9=0
t(4)=-7+3p=-7+3(3)=-7+9=2
----sum of the first 4 is -4
And 2p-10 at p=3 gives 2(3)-10=6-10=-4
So this part checks out
In general, the pattern that those 4 terms I wrote out follow t(n)=(-13-2)+2n+3p. I know this because the 0th term would have been (-13-2)+3p and this part goes up by 2 each time. The plus 3p part doesn't change.
Anyways t(9)=(-13-2)+2(9)+3p=-15+18+3p=3+3p. And this part looks good too.
