# 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.

## Question 7

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: $-\frac{7}{8}.$

## Question 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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