Let the first term of the geometric series be a and the common ratio be r. Then we have the equations ar = 3 and a + ar + ar^2 + ar^3 + ar^4 = a(1-r^5)/(1-r) = ?

<p><img src="https://scontent.fktm9-2.fna.fbcdn.net/v/t1.15752-9/329241671_1325068365059723_8598123340993255931_n.jpg?_nc_cat=103&amp;ccb=1-7&amp;_nc_sid=ae9488&amp;_nc_ohc=08yRNyAT6C8AX-eJWAz&amp;_nc_ht=scontent.fktm9-2.fna&amp;oh=03_AdRneT7y13-dxjJ_UGSSEevOt8vDfzeMxqe_ql16ulIatg&amp;oe=643FA376" alt="https://scontent.fktm9-2.fna.fbcdn.net/v/t1.15752-9/329241671_1325068365059723_8598123340993255931_n.jpg?_nc_cat=103&amp;ccb=1-7&amp;_nc_sid=ae9488&amp;_nc_ohc=08yRNyAT6C8AX-eJWAz&amp;_nc_ht=scontent.fktm9-2.fna&amp;oh=03_AdRneT7y13-dxjJ_UGSSEevOt8vDfzeMxqe_ql16ulIatg&amp;oe=643FA376" width="317" height="627"></p>

<p>Let's denote the first term of the geometric series as 'a' and the common ratio as 'r'.</p><p>The formula for the nth term of a geometric series is given by:</p><p>tn = a * r^(n-1)</p><p>Given that the third term (t3) is equal to 3, we can substitute the values into the formula:</p><p>t3 = a * r^(3-1)</p><p>3 = a * r^2</p><p>Now, we need to find the product of the first five terms (P5):</p><p>P5 = a * (a * r) * (a * r^2) * (a * r^3) * (a * r^4)</p><p>= a^5 * r^(1 + 2 + 3 + 4)</p><p>= a^5 * r^10</p><p>We know that t3 = 3, so we can rewrite it as:</p><p>3 = a * r^2</p><p>From this equation, we can solve for 'a' in terms of 'r':</p><p>a = 3 / r^2</p><p>Substituting this value of 'a' back into the expression for P5, we have:</p><p>P5 = (3 / r^2)^5 * r^10</p><p>= 3^5 / r^(2*5) * r^10</p><p>= 243 / r^10 * r^10</p><p>= 243</p><p>Therefore, the product of the first five terms of the geometric series is 243.</p>