Question

In the arithmetic sequence t1, t2, t3, . . . , tn, . . . , t1 = 23 and tn = tn – 1 – 3 for each n > 1. What is the value of n when tn = –4 ?

Answer 1

An arithmetic sequence is a string of numbers that is produced by repeatedly applying a certain arithmetic operation, in this case the string of numbers $t_{1} , t_{2}, t_{3}, ..., t_{n}, ...$ is thus determined, the first term $t_{1}$ is equivalent to number 23 and each next term, $t_{n}$ with $n>1$ (i.e. $t_{2} , t_{3}, t_{4}, ...$) is calculated using the given formula $t_{n}= t_{n-1} -3$, so the next term in the sequence, $t_{2}$  will be:

$t_{2}= t_{1}-3\\ t_{2}= 23-3 = 20$

in the same way the following terms are calculated

$t_{3} = t_{2}-3\\t_{3} = 20-3 = 17\\t_{4} = t_{3}-3\\t_{4} = 17-3 = 14\\...\\t_{9} = t_{8}-3\\t_{9} = 2-3 = -1\\t_{10} = t_{9}-3\\t_{10} = -1-3 = -4$

Another way to achieve this is to use the formula to calculate the terms $t_{n}$, note that:

$t_{2} = t_{1}-3\\t_{3} = t_{2}-3 = (t_{1}-3) -3 = t_{1}-6\\t_{4} = t_{3}-3 = (t_{1}-6) -3 = t_{1}-9=t_{1}-3(3)\\t_{5} = t_{4}-3 = (t_{1}-9) -3 = t_{1}-12=t_{1}-3(4)$

So, as you see each tn becomes $t_{1}$ minus a multiple of -3, or what is the same:

$t_{n}= t_{1} -3 (n-1)\\t_{n}= 23-3 (n-1)$

Replacing the desired value

$23-3 (n-1)=-4\\23 + 4 = 3 (n-1)\\27 = 3 (n-1)\\27/3 = (n-1)9 = n-1\\n = 10$

Answer

the value of n when $t_{n} =-4$ is $n=10$, $t_{10}$