Class 8-12 Mathematics Practice Questions, Solutions | ICSE ISC CBSE Math Mathematics Made Easy for School Students

$\displaystyle \textbf{Question 1. }\text{If the fifth term of a G.P. is }2,\text{ then write the product of its }9\text{ terms.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In a G.P., the product of }9\text{ terms is }(T_5)^9.$
$\displaystyle T_5=2$
$\displaystyle \therefore \text{Product of }9\text{ terms}=2^9=512$
$\\$

$\displaystyle \textbf{Question 2. }\text{If }(p+q)\text{th and }(p-q)\text{th terms of a G.P. are }m\text{ and }n\text{ respectively,}$
$\displaystyle \text{then write its }p\text{th term.}$
$\displaystyle \text{Answer:}$
$\displaystyle T_{p+q}=m,\qquad T_{p-q}=n$
$\displaystyle \text{In a G.P., }T_p^2=T_{p+q}\cdot T_{p-q}$
$\displaystyle \therefore T_p^2=mn$
$\displaystyle \therefore T_p=\sqrt{mn}$
$\\$

$\displaystyle \textbf{Question 3. }\text{If }\log_x a,\ a^{x/2}\text{ and }\log_b x\text{ are in G.P., then write the value of }x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the terms are in G.P.,}$
$\displaystyle \left(a^{x/2}\right)^2=(\log_x a)(\log_a x)$
$\displaystyle a^x=1$
$\displaystyle \therefore x=0$
$\\$

$\displaystyle \textbf{Question 4. }\text{If the sum of an infinite decreasing G.P. is }3\text{ and the sum of the squares}$
$\displaystyle \text{of its terms is }\frac{9}{2}, \text{then write its first term and common ratio.}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{a}{1-r}=3$
$\displaystyle \frac{a^2}{1-r^2}=\frac{9}{2}$
$\displaystyle a=3(1-r)$
$\displaystyle \frac{9(1-r)^2}{(1-r)(1+r)}=\frac{9}{2}$
$\displaystyle \frac{1-r}{1+r}=\frac{1}{2}$
$\displaystyle 2-2r=1+r$
$\displaystyle r=\frac{1}{3}$
$\displaystyle a=3\left(1-\frac{1}{3}\right)=2$
$\displaystyle \therefore \text{First term}=2,\quad \text{common ratio}=\frac{1}{3}$
$\\$

$\displaystyle \textbf{Question 5. }\text{If }p\text{th, }q\text{th and }r\text{th terms of a G.P. are }x,y,z\text{ respectively,} \\ \text{then write the value of }x^{q-r}y^{r-p}z^{p-q}.$
$\displaystyle \text{Answer:}$
$\displaystyle x=ar^{p-1},\qquad y=ar^{q-1},\qquad z=ar^{r-1}$
$\displaystyle x^{q-r}y^{r-p}z^{p-q}$
$\displaystyle =(ar^{p-1})^{q-r}(ar^{q-1})^{r-p}(ar^{r-1})^{p-q}$
$\displaystyle =a^{q-r+r-p+p-q}r^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}$
$\displaystyle =a^0r^0=1$
$\\$

$\displaystyle \textbf{Question 6. }\text{If }A_1,A_2\text{ be two A.M.'s and }G_1,G_2\text{ be two G.M.'s between }a\text{ and }b,$
$\displaystyle \text{then find the value of }\frac{A_1+A_2}{G_1G_2}.$
$\displaystyle \text{Answer:}$
$\displaystyle A_1=\frac{2a+b}{3},\qquad A_2=\frac{a+2b}{3}$
$\displaystyle \therefore A_1+A_2=a+b$
$\displaystyle \text{Also, }G_1G_2=ab$
$\displaystyle \therefore \frac{A_1+A_2}{G_1G_2}=\frac{a+b}{ab}$
$\\$

$\displaystyle \textbf{Question 7. }\text{If second, third and sixth terms of an A.P. are consecutive terms of a G.P.,}$
$\displaystyle \text{write the common ratio of the G.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle T_2=a+d,\quad T_3=a+2d,\quad T_6=a+5d$
$\displaystyle (a+2d)^2=(a+d)(a+5d)$
$\displaystyle a^2+4ad+4d^2=a^2+6ad+5d^2$
$\displaystyle d(2a+d)=0$
$\displaystyle \therefore d=-2a$
$\displaystyle \text{Common ratio}=\frac{a+2d}{a+d}=3$
$\\$

$\displaystyle \textbf{Question 8. }\text{Write the quadratic equation the arithmetic and geometric means}$
$\displaystyle \text{of whose roots are } A\text{ and }G\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{\alpha+\beta}{2}=A,\qquad \sqrt{\alpha\beta}=G$
$\displaystyle \therefore \alpha+\beta=2A,\qquad \alpha\beta=G^2$
$\displaystyle \therefore \text{Required equation is }x^2-2Ax+G^2=0$
$\\$

$\displaystyle \textbf{Question 9. }\text{Write the product of }n\text{ geometric means between two numbers }a\text{ and }b.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Product of }n\text{ geometric means between }a\text{ and }b=(ab)^{n/2}$
$\\$

$\displaystyle \textbf{Question 10. }\text{If }a=1+b+b^2+b^3+\cdots\text{ to }\infty,\text{ then write }b\text{ in terms of } \\ a\text{ given that }|b|<1.$
$\displaystyle \text{Answer:}$
$\displaystyle a=1+b+b^2+b^3+\cdots$
$\displaystyle a=\frac{1}{1-b}$
$\displaystyle 1-b=\frac{1}{a}$
$\displaystyle b=1-\frac{1}{a}=\frac{a-1}{a}$
$\\$