An example of an arithmetic sequence is<\/p>\n
10, 14, 18, 22, 26, 30, 34….<\/p>\n
You add the same amount each time to get the next term. \u00a0In the case above, of course, you add 4 each time. \u00a0We call 4 the common difference, while t1<\/sub> is 10. t5<\/sub> is 26.<\/p>\n Example 1<\/strong>: In the sequence above, what is t101<\/sub>?<\/p>\n Solution: We use the formula tn<\/sub>=t1<\/sub>+d(n-1)<\/p>\n n is the term number (or the index) We plug in the values we know:<\/p>\n t101<\/sub>=10+4(101-1)<\/p>\n Now, we simplify the right side. Remember BEDMAS, the order of operations: we must first simplify inside the brackets, then multiply, and finally add:<\/p>\n t101<\/sub>=10 + 4(100)=10+400=410<\/p>\n Apparently, t101<\/sub>=410.<\/p>\n Example 2<\/strong>: In the above sequence, which term is 202?<\/p>\n Solution: Again, we start with tn<\/sub>=t1<\/sub>+d(n-1).<\/p>\n This time, we know tn<\/sub> is 202; we just don’t know what n is. In other words, we need to know where 202 is in the sequence: Is it the 30th term? The 61st? We need to find out.<\/p>\n We plug in the known values:<\/p>\n 202=10+4(n-1)<\/p>\n Next, we subtract 10 from both sides:<\/p>\n 192=4(n-1)<\/p>\n We divide both sides by 4:<\/p>\n 192\/4=n-1<\/p>\n We simplify:<\/p>\n 48=n-1<\/p>\n Now, we add 1 to both sides:<\/p>\n 49=n<\/p>\n Apparently, n=49 when tn<\/sub>=202: t49<\/sub>=202. Therefore, in the arithmetic sequence 10,14,18,22,26,30,34…, the 49th term is 202. Or, the index of 202 is 49.<\/p>\n Arithmetic sequences are not always intuitive, but sticking with the formulas and suggested methods can make the questions surprisingly straightforward. I’ll be talking more about them in future posts:)<\/p>\n Source:<\/p>\n Travers, Kenneth et al. Using Advanced Algebra<\/em>. Toronto: Doubleday Canada Limited, \u00a0\u00a0\u00a01977.<\/p>\n
\nd is the common difference
\nt1<\/sub> is the first term<\/p>\n