A geometric series is the sum of terms from a geometric sequence, where each term is found by multiplying the previous term by a common ratio. For a finite geometric series, we can use \(S_n=a_1\frac{1-r^n}{1-r}\) when \(r\neq1\). These problems include finding finite sums, using sigma notation, solving for missing values, and identifying the first term or common ratio.
Notes
Practice Problems
\(\textbf{1)}\) \(\displaystyle\sum_{i=1}^{5}3(2)^{i}\)
The answer is \(186\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,a_1=3(2)^1=6\)
\(\,\,\,\,\,\,r=2\)
\(\,\,\,\,\,\,n=5\)
\(\,\,\,\,\,\,S_5=6\frac{1-2^5}{1-2}\)
\(\,\,\,\,\,\,S_5=6\frac{1-32}{-1}\)
\(\,\,\,\,\,\,S_5=6(31)\)
\(\,\,\,\,\,\,S_5=186\)
\(\textbf{2)}\) \(a_1=3, \, r=5,\,\) solve for \(S_8\).
The answer is \(S_8=292,968\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_8=3\frac{1-5^8}{1-5}\)
\(\,\,\,\,\,\,S_8=3\frac{1-390625}{-4}\)
\(\,\,\,\,\,\,S_8=3\frac{-390624}{-4}\)
\(\,\,\,\,\,\,S_8=3(97656)\)
\(\,\,\,\,\,\,S_8=292968\)
\(\textbf{3)}\) \(a_1=3,\, a_5=48,\,\) solve for r and \(S_8\).
The answer is \(r=2\) and \(S_8=765\)
\(\,\,\,\,\,\,a_n=a_1r^{n-1}\)
\(\,\,\,\,\,\,48=3r^{5-1}\)
\(\,\,\,\,\,\,48=3r^4\)
\(\,\,\,\,\,\,16=r^4\)
\(\,\,\,\,\,\,r=2\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_8=3\frac{1-2^8}{1-2}\)
\(\,\,\,\,\,\,S_8=3\frac{1-256}{-1}\)
\(\,\,\,\,\,\,S_8=3(255)\)
\(\,\,\,\,\,\,S_8=765\)
\(\textbf{4)}\) \(r=\frac{1}{2},\, S_6=1,260,\,\) solve for \(a_1\).
The answer is \(a_1=640\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,1260=a_1\frac{1-\left(\frac{1}{2}\right)^6}{1-\frac{1}{2}}\)
\(\,\,\,\,\,\,1260=a_1\frac{1-\frac{1}{64}}{\frac{1}{2}}\)
\(\,\,\,\,\,\,1260=a_1\frac{\frac{63}{64}}{\frac{1}{2}}\)
\(\,\,\,\,\,\,1260=a_1\left(\frac{63}{32}\right)\)
\(\,\,\,\,\,\,a_1=1260\cdot\frac{32}{63}\)
\(\,\,\,\,\,\,a_1=640\)
\(\textbf{5)}\) \(\frac{1}{9}+1+9+ \cdots +729\)
The answer is \(820 \frac{1}{9}\,\) or \( \,\frac{7381}{9}\)
\(\,\,\,\,\,\,a_1=\frac{1}{9}\)
\(\,\,\,\,\,\,r=\frac{1}{\frac{1}{9}}=9\)
\(\,\,\,\,\,\,\frac{1}{9},\,1,\,9,\,81,\,729\)
\(\,\,\,\,\,\,n=5\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_5=\frac{1}{9}\cdot\frac{1-9^5}{1-9}\)
\(\,\,\,\,\,\,S_5=\frac{1}{9}\cdot\frac{1-59049}{-8}\)
\(\,\,\,\,\,\,S_5=\frac{1}{9}\cdot\frac{-59048}{-8}\)
\(\,\,\,\,\,\,S_5=\frac{7381}{9}\)
\(\,\,\,\,\,\,S_5=820\frac{1}{9}\)
\(\textbf{6)}\) Find the sum of the first 6 terms of \(4+12+36+\cdots\)
The answer is \(1456\)
\(\,\,\,\,\,\,a_1=4\)
\(\,\,\,\,\,\,r=\frac{12}{4}=3\)
\(\,\,\,\,\,\,n=6\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_6=4\frac{1-3^6}{1-3}\)
\(\,\,\,\,\,\,S_6=4\frac{1-729}{-2}\)
\(\,\,\,\,\,\,S_6=4(364)\)
\(\,\,\,\,\,\,S_6=1456\)
\(\textbf{7)}\) Find \(S_7\) if \(a_1=5\) and \(r=2\).
The answer is \(635\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_7=5\frac{1-2^7}{1-2}\)
\(\,\,\,\,\,\,S_7=5\frac{1-128}{-1}\)
\(\,\,\,\,\,\,S_7=5(127)\)
\(\,\,\,\,\,\,S_7=635\)
\(\textbf{8)}\) Find \(S_5\) if \(a_1=10\) and \(r=-2\).
The answer is \(110\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_5=10\frac{1-(-2)^5}{1-(-2)}\)
\(\,\,\,\,\,\,S_5=10\frac{1-(-32)}{3}\)
\(\,\,\,\,\,\,S_5=10\frac{33}{3}\)
\(\,\,\,\,\,\,S_5=10(11)\)
\(\,\,\,\,\,\,S_5=110\)
\(\textbf{9)}\) \(\displaystyle\sum_{k=0}^{4}2(3)^k\)
The answer is \(242\)
\(\,\,\,\,\,\,a_1=2(3)^0=2\)
\(\,\,\,\,\,\,r=3\)
\(\,\,\,\,\,\,n=4-0+1=5\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_5=2\frac{1-3^5}{1-3}\)
\(\,\,\,\,\,\,S_5=2\frac{1-243}{-2}\)
\(\,\,\,\,\,\,S_5=242\)
\(\textbf{10)}\) \(\displaystyle\sum_{n=1}^{6}7\left(\frac{1}{2}\right)^{n-1}\)
The answer is \(\frac{441}{32}\)
\(\,\,\,\,\,\,a_1=7\)
\(\,\,\,\,\,\,r=\frac{1}{2}\)
\(\,\,\,\,\,\,n=6\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_6=7\frac{1-\left(\frac{1}{2}\right)^6}{1-\frac{1}{2}}\)
\(\,\,\,\,\,\,S_6=7\frac{1-\frac{1}{64}}{\frac{1}{2}}\)
\(\,\,\,\,\,\,S_6=7\frac{\frac{63}{64}}{\frac{1}{2}}\)
\(\,\,\,\,\,\,S_6=7\cdot\frac{63}{32}\)
\(\,\,\,\,\,\,S_6=\frac{441}{32}\)
\(\textbf{11)}\) Find the sum of the first 8 terms of \(2,6,18,54,\ldots\)
The answer is \(6560\)
\(\,\,\,\,\,\,a_1=2\)
\(\,\,\,\,\,\,r=\frac{6}{2}=3\)
\(\,\,\,\,\,\,n=8\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_8=2\frac{1-3^8}{1-3}\)
\(\,\,\,\,\,\,S_8=2\frac{1-6561}{-2}\)
\(\,\,\,\,\,\,S_8=6560\)
\(\textbf{12)}\) Find the sum of the first 5 terms of \(81+27+9+\cdots\)
The answer is \(121\)
\(\,\,\,\,\,\,a_1=81\)
\(\,\,\,\,\,\,r=\frac{27}{81}=\frac{1}{3}\)
\(\,\,\,\,\,\,n=5\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_5=81\frac{1-\left(\frac{1}{3}\right)^5}{1-\frac{1}{3}}\)
\(\,\,\,\,\,\,S_5=81\frac{1-\frac{1}{243}}{\frac{2}{3}}\)
\(\,\,\,\,\,\,S_5=81\frac{\frac{242}{243}}{\frac{2}{3}}\)
\(\,\,\,\,\,\,S_5=121\)
\(\textbf{13)}\) \(a_1=4,\, a_4=108,\,\) solve for \(r\) and \(S_6\).
The answer is \(r=3\) and \(S_6=1456\)
\(\,\,\,\,\,\,a_n=a_1r^{n-1}\)
\(\,\,\,\,\,\,108=4r^{4-1}\)
\(\,\,\,\,\,\,108=4r^3\)
\(\,\,\,\,\,\,27=r^3\)
\(\,\,\,\,\,\,r=3\)
\(\,\,\,\,\,\,S_6=4\frac{1-3^6}{1-3}\)
\(\,\,\,\,\,\,S_6=4\frac{1-729}{-2}\)
\(\,\,\,\,\,\,S_6=1456\)
\(\textbf{14)}\) \(a_1=6,\, r=4,\,\) solve for \(S_5\).
The answer is \(2046\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_5=6\frac{1-4^5}{1-4}\)
\(\,\,\,\,\,\,S_5=6\frac{1-1024}{-3}\)
\(\,\,\,\,\,\,S_5=6\frac{-1023}{-3}\)
\(\,\,\,\,\,\,S_5=6(341)\)
\(\,\,\,\,\,\,S_5=2046\)
\(\textbf{15)}\) Find the sum of \(5+10+20+40+80+160\).
The answer is \(315\)
\(\,\,\,\,\,\,a_1=5\)
\(\,\,\,\,\,\,r=2\)
\(\,\,\,\,\,\,n=6\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_6=5\frac{1-2^6}{1-2}\)
\(\,\,\,\,\,\,S_6=5\frac{1-64}{-1}\)
\(\,\,\,\,\,\,S_6=5(63)\)
\(\,\,\,\,\,\,S_6=315\)
\(\textbf{16)}\) Find \(S_4\) if \(a_1=-3\) and \(r=5\).
The answer is \(-468\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_4=-3\frac{1-5^4}{1-5}\)
\(\,\,\,\,\,\,S_4=-3\frac{1-625}{-4}\)
\(\,\,\,\,\,\,S_4=-3\frac{-624}{-4}\)
\(\,\,\,\,\,\,S_4=-3(156)\)
\(\,\,\,\,\,\,S_4=-468\)
\(\textbf{17)}\) \(a_1=12,\, r=\frac{1}{2},\,\) solve for \(S_5\).
The answer is \(\frac{93}{4}\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_5=12\frac{1-\left(\frac{1}{2}\right)^5}{1-\frac{1}{2}}\)
\(\,\,\,\,\,\,S_5=12\frac{1-\frac{1}{32}}{\frac{1}{2}}\)
\(\,\,\,\,\,\,S_5=12\frac{\frac{31}{32}}{\frac{1}{2}}\)
\(\,\,\,\,\,\,S_5=12\cdot\frac{31}{16}\)
\(\,\,\,\,\,\,S_5=\frac{93}{4}\)
\(\textbf{18)}\) \(r=3,\, S_4=160,\,\) solve for \(a_1\).
The answer is \(a_1=4\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,160=a_1\frac{1-3^4}{1-3}\)
\(\,\,\,\,\,\,160=a_1\frac{1-81}{-2}\)
\(\,\,\,\,\,\,160=a_1\frac{-80}{-2}\)
\(\,\,\,\,\,\,160=40a_1\)
\(\,\,\,\,\,\,a_1=4\)
\(\textbf{19)}\) \(\displaystyle\sum_{i=2}^{6}4(2)^i\)
The answer is \(496\)
\(\,\,\,\,\,\,a_1=4(2)^2=16\)
\(\,\,\,\,\,\,r=2\)
\(\,\,\,\,\,\,n=6-2+1=5\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_5=16\frac{1-2^5}{1-2}\)
\(\,\,\,\,\,\,S_5=16\frac{1-32}{-1}\)
\(\,\,\,\,\,\,S_5=16(31)\)
\(\,\,\,\,\,\,S_5=496\)
\(\textbf{20)}\) \(a_1=9,\, a_6=288,\,\) solve for \(r\) and \(S_6\).
The answer is \(r=2\) and \(S_6=567\)
\(\,\,\,\,\,\,a_n=a_1r^{n-1}\)
\(\,\,\,\,\,\,288=9r^{6-1}\)
\(\,\,\,\,\,\,288=9r^5\)
\(\,\,\,\,\,\,32=r^5\)
\(\,\,\,\,\,\,r=2\)
\(\,\,\,\,\,\,S_n=a_1\frac{1-r^n}{1-r}\)
\(\,\,\,\,\,\,S_6=9\frac{1-2^6}{1-2}\)
\(\,\,\,\,\,\,S_6=9\frac{1-64}{-1}\)
\(\,\,\,\,\,\,S_6=9(63)\)
\(\,\,\,\,\,\,S_6=567\)
See Related Pages
