Arithmetic and geometric sequences and their summations Examples


  • Examples
    (a) Find the sum of all terms of the arithmetic sequence 1, 4, 7, ..., 100.
    (b) Find the sum of all terms of the geometric sequence \(x{\rm{ }},{\rm{ }}{x^4},{\rm{ }}{x^7},{\rm{ }} \cdots {\rm{ }},{\rm{ }}{x^{100}}\).
    (c) It is given that \(x \ne 0\) and \(x \ne 1\). Solve the equation
    \((x - 1) + ({x^4} - 4) + ({x^7} - 7) + \cdots + ({x^{100}} - 100) = \frac{{15{\rm{ }}453 - x + {x^{103}}}}{{{x^3} - 1}}\).
  • Solutions
    (a) Let n be the number of terms of the sequence.
    \(\begin{array}{1}100 = 1 + (n - 1) \times 3\\3(n - 1) = {99^{}}\\n - 1 = {33^{}}\\n = {34^{}}\end{array}\)
    ∴ \(\begin{array}{1}1 + 4 + 7 + \cdots + 100 = \frac{{34}}{2}(1 + 100)\\ = {\underline{\underline {1{\rm{ }}717}} ^{}}\end{array}\)

    \(\begin{array}{1}(b)x + {x^4} + {x^7} + \cdots + {x^{100}} = \frac{{x[{{({x^3})}^{34}} - 1]}}{{{x^3} - 1}}\\ = {\underline{\underline {\frac{{x({x^{102}} - 1)}}{{{x^3} - 1}}}} ^{}}\end{array}\)

    \(\begin{array}{1}(c)(x - 1) + ({x^4} - 4) + ({x^7} - 7) + \cdots + ({x^{100}} - 100) = \frac{{15{\rm{ }}453 - x + {x^{103}}}}{{{x^3} - 1}}\\(x + {x^4} + {x^7} + \cdots + {x^{100}}) - (1 + 4 + 7 + \cdots + 100) = {\frac{{15{\rm{ }}453 - x + {x^{103}}}}{{{x^3} - 1}}^{}}\\\frac{{x({x^{102}} - 1)}}{{{x^3} - 1}} - 1{\rm{ }}717 = {\frac{{15{\rm{ }}453 - x + {x^{103}}}}{{{x^3} - 1}}^{}}\\\frac{{{x^{103}} - x - 1{\rm{ }}717{x^3} + 1{\rm{ }}717}}{{{x^3} - 1}} = {\frac{{15{\rm{ }}453 - x + {x^{103}}}}{{{x^3} - 1}}^{}}\\{x^{103}} - x - 1{\rm{ }}717{x^3} + 1{\rm{ }}717 = 15{\rm{ }}453 - x + {x^{103}}^{}\\ - {\rm{ }}1{\rm{ }}717{x^3} = 13{\rm{ }}{736^{^{}}}\\{x^3} = - {8^{^{}}}\\{x^3} = {( - 2)^3}^{}\\x = {\underline{\underline { - {\rm{ }}2}} ^{}}\end{array}\)


  • Examples
    The first term of a geometric sequence is \(12x + 3\), the common ratio is \(\frac{1}{2} - 3x\) and the sum to infinity is \(5 - x\). Find the value of x.
  • Solutions
    \(\begin{array}{1}\frac{{12x + 3}}{{1 - ({\textstyle{1 \over 2}} - 3x)}} = 5 - x\\\frac{{12x + 3}}{{3x + {\textstyle{1 \over 2}}}} = 5 - {x^{^{^{^{^{^{}}}}}}}\\12x + 3 = \frac{5}{2} + \frac{{29}}{2}x - 3{x^2}^{^{^{}}}\\3{x^2} - \frac{5}{2}x + \frac{1}{2} = {0^{^{^{^{}}}}}\\6{x^2} - 5x + 1 = {0^{^{^{}}}}\\(3x - 1)(2x - 1) = {0^{^{^{}}}}\end{array}\)
    \(x = \frac{1}{3}\) or \(x = \frac{1}{2}\)
    ∵ \( - {\rm{ }}1 < \frac{1}{2} - 3x < 1\) and \(\frac{1}{2} - 3x \ne 0\)
    ∴ \( - {\rm{ }}\frac{3}{2} < - 3x < \frac{1}{2}\) and \(3x \ne \frac{1}{2}\)
    \(\frac{1}{2} > x > - \frac{1}{6}\) and \(x \ne \frac{1}{6}\)
    i.e. \(x = \underline{\underline {\frac{1}{3}}} \)


  • Examples
    At the beginning of a month, Mr. Lee borrowed $P from a bank at an interest rate of r% p.a. The interest is compounded monthly at the end of each month. From the beginning of the second month, he repays $x at the beginning of each month until the loan is fully repaid. The last repayment may be less than $x.
    (a) Express the balance owed in the following periods in terms of P, r% and x.
        (i) At the end of the first month
        (ii) At the end of the second month
        (iii) At the end of the third month
    (b) Prove that the balance owed at the end of the nth month is \(\$ \{ P{(1 + \frac{{r\% }}{{12}})^n} - x(1 + \frac{{r\% }}{{12}})[\frac{{{{(1 + {\textstyle{{r\% } \over {12}}})}^{n - 1}} - 1}}{{{\textstyle{{r\% } \over {12}}}}}]\} \). (
    c) Let \(P = 100{\rm{ }}000\) and \(r = 12\).
        (i) If Mr. Lee repays the loan in 28 instalments, find the least value of x. (Give your answer correct to 2 decimal places.)
        (ii) If Mr. Lee repays $5 000 each month, when will he fully repay the loan?
  • Solutions
    (a) (i) Balance owed at the end of the first month\( = \underline{\underline {\$ P(1 + \frac{{r\% }}{{12}})}} \)
        (ii) Balance owed at the end of the second month\(\begin{array}{l} = \$ [P(1 + \frac{{r\% }}{{12}}) - x](1 + \frac{{r\% }}{{12}})\\ = \underline{\underline {\$ [P{{(1 + \frac{{r\% }}{{12}})}^2} - x(1 + \frac{{r\% }}{{12}})]}} \end{array}\)
        (iii) Balance owed at the end of the third month\(\begin{array}{l} = \$ [P{(1 + \frac{{r\% }}{{12}})^2} - x(1 + \frac{{r\% }}{{12}}) - x](1 + \frac{{r\% }}{{12}})\\ = \underline{\underline {\$ [P{{(1 + \frac{{r\% }}{{12}})}^3} - x{{(1 + \frac{{r\% }}{{12}})}^2} - x(1 + \frac{{r\% }}{{12}})]}} \end{array}\)

    (b) Balance owed at the end of the nth month
    \(\begin{array}{l} = \$ [P{(1 + \frac{{r\% }}{{12}})^n} - x{(1 + \frac{{r\% }}{{12}})^{n - 1}} - x{(1 + \frac{{r\% }}{{12}})^{n - 2}} - {\rm{ }} \cdots {\rm{ }} - x(1 + \frac{{r\% }}{{12}})]\\ = \$ \{ P{(1 + \frac{{r\% }}{{12}})^n} - x(1 + \frac{{r\% }}{{12}})[1 + (1 + \frac{{r\% }}{{12}}) + {(1 + \frac{{r\% }}{{12}})^2} + {\rm{ }} \cdots {\rm{ }} + {(1 + \frac{{r\% }}{{12}})^{n - 2}}]\} \\ = \$ {\{ P{(1 + \frac{{r\% }}{{12}})^n} - x(1 + \frac{{r\% }}{{12}})[\frac{{{{(1 + {\textstyle{{r\% } \over {12}}})}^{n - 1}} - 1}}{{1 + {\textstyle{{r\% } \over {12}}} - 1}}]\} ^{^{^{^{^{}}}}}}\\ = \$ {\{ P{(1 + \frac{{r\% }}{{12}})^n} - x(1 + \frac{{r\% }}{{12}})[\frac{{{{(1 + {\textstyle{{r\% } \over {12}}})}^{n - 1}} - 1}}{{{\textstyle{{r\% } \over {12}}}}}]\} ^{^{^{^{^{^{}}}}}}}\end{array}\)

    \(\begin{array}{1}(c) (i)100{\rm{ }}000{(1 + \frac{{12\% }}{{12}})^{29}} - x(1 + \frac{{12\% }}{{12}})[\frac{{{{(1 + {\textstyle{{12\% } \over {12}}})}^{29 - 1}} - 1}}{{{\textstyle{{12\% } \over {12}}}}}] \le 0\\100{\rm{ }}000{(1.01)^{29}} \le x(1.01)(\frac{{{{1.01}^{28}} - 1}}{{0.01}})\\x \ge {\frac{{100{\rm{ }}000{{(1.01)}^{28}}(0.01)}}{{{{1.01}^{28}} - 1}}^{^{}}}\end{array}\)
    \(x \ge 4{\rm{ }}112.44\) (corr. to 2 d.p.)
    ∴ The least value of x is 4 112.44.
    \(\begin{array}{1}    (ii) 100{\rm{ }}000{(1 + \frac{{12\% }}{{12}})^n} - 5{\rm{ }}000(1 + \frac{{12\% }}{{12}})[\frac{{{{(1 + {\textstyle{{12\% } \over {12}}})}^{n - 1}} - 1}}{{{\textstyle{{12\% } \over {12}}}}}] \le 0\\100{\rm{ }}000{(1.01)^n} \le 5{\rm{ }}000(1.01)(\frac{{{{1.01}^{n - 1}} - 1}}{{0.01}})\\20{(1.01)^{n - 1}}(0.01) \le {1.01^{n - 1}} - 1\\0.2{(1.01)^{n - 1}} - {1.01^{n - 1}} \le - {1^{^{}}}\\0.8{(1.01)^{n - 1}} \ge {1^{^{}}}\\{1.01^{n - 1}} \ge {1.25^{^{}}}\\(n - 1)\log 1.01 \ge \log {1.25^{^{^{}}}}\\n - 1 \ge {\frac{{\log 1.25}}{{\log 1.01}}^{}}\\n \ge \frac{{\log 1.25}}{{\log 1.01}} + {1^{^{^{^{^{}}}}}}\end{array}\)
    \(n \ge 23.43\) (corr. to 2 d.p.)
    ∴ Nothing outstanding at the end of the 24th month.
    i.e. Mr. Lee will fully repay the loan at the beginning of the 24th month.