Question 825215
a geometric sequence has a sixth term of 24 and a common ratio of 2/3. find the first three terms

<pre>
GP formula: {{{a[n] = a[1]r^(n - 1)}}}
{{{a[6] = a[1](2/3)^(6 - 1)}}}___Substituting 6 for n, and {{{2/3}}} for r
{{{24 = a[1](2/3)^5}}}_____Substituting 24 for {{{a[n]}}}
{{{24 = a[1](32/243)}}}
{{{a[1] = 24}}}÷{{{32/243}}}
{{{a[1] = 24(243/32)}}}
{{{a[1] = 3cross(24)(243/4cross(32))}}}
{{{a[1] = 729/4)}}}, or {{{highlight_green(a[1] = 182.25)}}}

The first term, or {{{a[1]}}}(182.25), and the common ratio, or r ({{{2/3}}}), should be substituted into the GP formula,
{{{a[n] = a[1]r^(n - 1)}}} to determine the 2nd term ({{{a[2]}}}), and 3rd term ({{{a[3]}}}).

You'll then have all 3 terms.

You can do the check!! 

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