# Question 1 and 2, Exercise 4.6

Solutions of Question 1 and 2 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 1

Find the indicated term of the harmonic progression. $\frac{1}{9}, \frac{1}{12}, \frac{1}{15}, \cdots \quad 7$ th term.

** Solution. **

$$\frac{1}{9}, \frac{1}{12}, \frac{1}{15}, \cdots \text{ is in H.P.}$$ $$9, 12, 15, ... \text{ is in A.P.}$$ Here $a_1=9$, $d=12-9=3$, $a_7=?$.

Gneral term of A.P is given as $$ a_n=a_1+(n-1)d. $$ Thus \begin{align*} a_7&=9+(6)(3) \\ & = 27 \text{ is in A.P} \end{align*} Hence the 7th term in H.P is $\dfrac{1}{27}$.

## Question 2

Find the indicated term of the harmonic progression. $\frac{1}{11}, \frac{1}{9}, \frac{1}{7}, \ldots \quad 10$ th term.

** Solution. **

\begin{align*} &\frac{1}{11}, \frac{1}{9}, \frac{1}{7}, \ldots\quad \text{ is in H.P.}\\ &11, 9, 7, \ldots \quad \text{ is in A.P.}\end{align*} Here $a_1 = 11$, $d = 9 - 11 = -2.$ $a_{10}=?$

The general term of the A.P. is given as $$ a_n = a_1 + (n-1)d. $$ Thus, \begin{align*} a_{10} &= 11 + (10-1)(-2) \\ &= 11 + 9(-2) \\ &= 11 - 18 \\ &= -7. \end{align*} Hence, the 10th term in H.P. is $-\frac{1}{7}.$

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