Question 5 & 6, Exercise 4.6

Solutions of Question 5 & 6 of Exercise 4.6 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find the indicated term of the harmonic progression. $\dfrac{1}{27}, \dfrac{1}{20}, \dfrac{1}{13}, \ldots \quad$ nth term.

Solution.

\begin{align*} &\frac{1}{27}, \frac{1}{20}, \frac{1}{13}, \ldots \quad \text{ is in H.P.} \\ &27, 20, 13, \ldots \quad \text{ is in A.P.} \end{align*} Here $a_1 = 27$, $d = 20 - 27 = -7$, $a_n=?$

The general term of the A.P. is given as $$ a_n = a_1 + (n-1)d. $$ Thus, \begin{align*} a_n &= 27 + (n-1)(-7) \\ &= 27 - 7(n-1) \\ &= 27 - 7n + 7 \\ &= 34 - 7n. \end{align*}

Hence, the $n$th term in H.P. is $\frac{1}{34 - 7n}.$

Find the indicated term of the harmonic progression. $\frac{1}{2}, \frac{1}{2 \frac{1}{2}}, \frac{1}{3}, \frac{1}{3 \frac{1}{2}}, \ldots$ nth term.

Solution.

\begin{align*} &\frac{1}{2}, \frac{1}{2 \frac{1}{2}}, \frac{1}{3}, \frac{1}{3 \frac{1}{2}}, \ldots \quad \text{ is in H.P.} \\ &2, 2 \frac{1}{2}, 3, 3 \frac{1}{2}, \ldots \quad \text{ is in A.P.} \end{align*} Here $a_1 = 2$, $d = 2 \frac{1}{2} - 2 = \frac{1}{2}$, $a_n=?$

The general term of the A.P. is given as $$ a_n = a_1 + (n-1)d. $$ Thus, \begin{align*} a_n &= 2 + (n-1)\left(\frac{1}{2}\right) \\ &= 2 + \frac{n-1}{2} \\ &= \frac{4}{2} + \frac{n-1}{2} \\ &= \frac{4 + n - 1}{2} \\ &= \frac{n+3}{2}. \end{align*} Hence, the $n$th term in H.P. is $\dfrac{1}{\frac{n+3}{2}}$ or $\dfrac{2}{n+3}$.

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