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What is the 24th term of the arithmetic sequence where a1 = 8 and a9 = 56 ?
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Well first of all you need the equation for the arithmetic sequence\[a _{n}=a _{1}+(n-1)d\] where \[d = a _{k}-a _{k-1}\] or simply the difference of the the term and the term before. Now applying this formula you have that:\[a_1=8\] and \[a_9 = 8+ (9-1)d = 8+8d\] And since we already have a value for \[a_9 = 56\] we have that\[56 = 8+8d\] which turns to\[56-8=8d \rightarrow48=8d \rightarrow6=d\] Now that we have the difference we just plug it in:\[a_{24}=8+(24-1)(6)=8+23*6= 146\]. Peace, Love and Happiness from Puerto Rico
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