Question 7 & 8, Exercise 4.6

Solutions of Question 7 & 8 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 14 th term of H.P. $\frac{1}{4}, \frac{1}{7}, \frac{1}{10}, \frac{1}{13}, \ldots$

Solution.

Given the $ \frac{1}{4}, \frac{1}{7}, \frac{1}{10}, \frac{1}{13}, \ldots $ is an arithmetic sequence.

Here, $ a_1 = \frac{1}{4} $, $d = \frac{1}{7} - \frac{1}{4} = -\frac{3}{28},$ $ n = 14$

The general term of the A.P. is given as: $$a_n = a_1 + (n-1)d.$$ Thus, \begin{align*} a_{14} &= \frac{1}{4} + (13)\left(-\frac{3}{28}\right) \\ &= \frac{1}{4} + 13\left(-\frac{3}{28}\right) \\ &= \frac{1}{4} - \frac{39}{28}\\ &= -\frac{8}{7} \text{ is in A.P.} \end{align*} Thus, the 14th term in the harmonic progression is: $-\frac{7}{8}.$

$7,4,1, \ldots$ is arithmetic sequence, find the 17th term in H.P.

Solution.

Given $7,4,1, \ldots$ is arithmetic sequence.

Here $a_1=7$, $d=4-7=-3$, $n=17$.

The general term of A.P is given as $$ a_n=a_1+(n-1)d. $$ Thus \begin{align*} a_{17}&=7+(16)(-3) \\ &= 7-48 \\ &= -41 \text{ is in A.P.} \end{align*} Hence 17th term in H.P is $-\dfrac{1}{41}$.

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