Sequence problem and find a and d [closed]

given an arithmetic progression that T(14)=-8 and T(20)=-20, find the 50th term of the progression.

closed as off-topic by JonMark Perry, Xander Henderson, user99914, Leucippus, Claude LeiboviciDec 22 '17 at 6:52

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• $6$ seconds later you find yourself $12$ steps down the stairs. Then after $36$ seconds $\cdots$ – dxiv Dec 22 '17 at 4:41

HINT: $T(50)=T(20)+5(T(20)-T(14))$. You do not need to find $a$ and $d$ for this part of your question.

Pointers:

• The $n^{\text{th}}$ term of an Arithmetic progression with starting term $a$ and common difference $d$ is: $$T_n = a + (n-1)d$$

• Given: $T_{14}$ and $T_{20}$, can you solve the two equations to get $a$ and $d$?

Since there is always a constant difference "d" in an arithmetic progression, it doesn't matter where you begin. To make things easier, let's imagine a sequence that goes from the 14th to the 20th term of your sequence. So a(1)=T(14). This means that T(20) would be a(7) (since 20 is 6 more than 14 and 7 is 6 more than 1). We can now rewrite your problem as:

a(1)=-8 and a(7)=-20. Find the common difference.

a(n)=a(1)+(n-1)d a(6)=a(1)+(7-1)d -20=-8+6d -12=6d d=-2

Since a(1)=T(14) then T(50)=a(37) (since 50 is 36 more than 14 and 37 is 36 more than 1).

a(n)=a(1)+(n-1)d a(37)=a(1)+(37-1)d a(37)=-8+(37-1)(-2) a(37)=-8+(36)(-2) a(37)=-8+-72 a(37)=-80

This is T(50)

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