Chapter 8: Sequences and Series

Ans:

The deposited amount to the bank is Rs. $500$.

The amount obtained after the end of the first year 

$=$Rs. $500\left( 1+\dfrac{1}{10} \right)=Rs.\,500\left( 1.1 \right)$.

The amount obtained after the end of the second year,

$=Rs.\,500\left( 1.1 \right)\left( 1.1 \right)$.

The amount obtained after the end of third year,

$=Rs.\,500\left( 1.1 \right)\left( 1.1 \right)\left( 1.1 \right)$ and …so on

Therefore, after the end of $10$ years, the amount $=Rs.\,\,500\left( 1.1 \right)\left( 1.1 \right)...\left( 10\,\,\text{times} \right)$

$=Rs.\,\,500{{\left( 1.1 \right)}^{10}}$.

Ans:

First suppose that $a$ and $b$ are the roots of the quadratic equation.

Then, A.M. and G.M. of the roots $a$ and $b$ are such that

A.M.$=\dfrac{a+b}{2}$ and

G.M.$=\sqrt{ab}$.

Therefore, by the condition given to us,

$\dfrac{a+b}{2}=8$

$\Rightarrow a+b=16$                                                   …… (i)

and $\sqrt{ab}=5$

$\Rightarrow ab=25$                                                      …… (ii)

Now, we know that, in a quadratic equation of $x$,

${{x}^{2}}-x\left( \text{sum of roots} \right)+\left( \text{product of the roots} \right)=0$

$\Rightarrow {{x}^{2}}-\left( a+b \right)x+\left( ab \right)=0$

$\Rightarrow {{x}^{2}}-16x+25=0$, substituting the values of $a+b$ and $ab$.

Hence, the needed quadratic equation is ${{x}^{2}}-16x+25=0$.