5 Marks Questions

Question 515 Marks

Let $a_n $ be the nth term of the G.P. of positive numbers. Let $\sum\limits_{\text{n}=1}^{100}\text{a}_{2\text{n}}=\alpha\text{ and}\sum\limits_{\text{n}=1}^{10}\text{a}_{2\text{n}-1}=\beta,$ such that $\alpha\neq\beta.$ Prove that the common ratio of the G.P. is $\frac{\alpha}{\beta}$

Answer

Given: $\sum\limits_{\text{n}=1}^{100}\text{a}_{2\text{n}}=\alpha$ $\Rightarrow\text{a}_2+\text{a}_4+\text{a}_6+\ ...\ +\text{a}_{200}=\alpha\cdots(\text{i})$
Also, $\sum\limits_{\text{n}=1}^{10}\text{a}_{2\text{n}-1}=\beta,$
$\Rightarrow\text{a}_1+\text{a}_3+\text{a}_5+\ ...\ +\text{a}_{199}=\beta\cdots(\text{ii})$
Sum of G.P, $\text{S}_\text{n}=\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{n}}$
$=\text{a}=\text{a}_2,\text{r}=\text{r}^2,\text{n}=100$
$\text{ar}+\text{ar}^3+\text{ar}^5+\ ...\ +\text{ar}^{199}=\alpha$
$\text{ar}\frac{\Big(1-\big(\text{r}^2\big)^{100}\Big)}{1-\text{r}^2}=\alpha\cdots(\text{iii})$
$\text{a}+\text{ar}^2+\text{ar}^4+\ ...\ +\text{ar}^{198}=\beta$
$\frac{\text{a}\Big(1-\big(\text{r}^2\big)^{100}\Big)}{1-\text{r}^{2}}=\beta\cdots(\text{iv})$
$\text{r}(\beta)=\alpha\cdots(\text{v})$
$\text{r}=\frac{\alpha}{\beta}$ [From (iv) and (v)]

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Question 525 Marks

If a, b, c, are in G.P., prove that:
$\frac{1}{\text{a}^2+\text{b}^2},\frac{1}{\text{b}^2+\text{c}^2},\frac{1}{\text{c}^2+\text{d}^2}\text{ are in G.P.}$

Answer

a, b, c, d are in G.P. $\therefore\text{b}^2=\text{ac}$ $\text{ad}=\text{bc}$ $\text{c}^2=\text{bd}\cdots(1)$ $\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\Big(\frac{1}{\text{b}^2}\Big)^2+\frac{2}{\text{b}^2\text{c}^2}+\Big(\frac{1}{\text{c}^2}\Big)^2$ $\Rightarrow\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\Big(\frac{1}{\text{ac}}\Big)^2+\frac{1}{\text{b}^2\text{c}^2}+\frac{1}{\text{b}^2\text{c}^2}+\Big(\frac{1}{\text{bd}}\Big)^2$ [Using (1)] $\Rightarrow\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\frac{1}{\text{a}^2\text{c}^2}+\frac{1}{\text{a}^2\text{d}^2}+\frac{1}{\text{b}^2\text{c}^2}+\frac{1}{\text{b}^2\text{d}^2}$ [Using (1)]$\Rightarrow\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\frac{1}{\text{a}^2}\Big(\frac{1}{\text{c}^2}+\frac{1}{\text{d}^2}\Big)+\frac{1}{\text{b}^2}\Big(\frac{1}{\text{c}^2}+\frac{1}{\text{d}^2}\Big)$
$\Rightarrow\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big)^2=\Big(\frac{1}{\text{a}^2+\text{b}^2}\Big)\Big(\frac{1}{\text{c}^2}+\frac{1}{\text{d}^2}\Big)$
$\therefore\Big(\frac{1}{\text{b}^2+\text{c}^2}\Big),\Big(\frac{1}{\text{c}^2+\text{d}^2}\Big)\text{ and are also in G.P.}$

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Question 535 Marks

The product of three numbers in G.P. is $125$ and the sum of their products taken in pairs is $87\frac{1}{2}.$ Find them.

Answer

Let the three number in G.P. be $\frac{\text{a}}{\text{r}},\text{a},\text{ar}$ then product of these numbers $\Big(\frac{\text{a}}{\text{r}}\Big)(\text{a})(\text{ar})$
$\Rightarrow\text{a}^3=125=5^3$
$\text{a}=5$
Also, sum of these product in pair
$\Big(\frac{\text{a}}{\text{r}}\Big)(\text{a})+(\text{a})(\text{ar})+\Big(\frac{\text{a}}{\text{r}}\Big)(\text{ar})$
$=87\frac{1}{2}=\frac{195}{2}$
$=(5)^2\Big(\frac{1+\text{r}^2+\text{r}}{\text{r}}\Big)=\frac{195}{2}$
$1+\text{r}^2+\text{r}=\Big(\frac{195}{2\times25}\Big)^\text{r}$
$2(1+\text{r}^2+\text{r})=\frac{39}{5}\text{r}$
$10+10\text{r}^2+10\text{r}=39\text{r}$
$10\text{r}^2-25\text{r}-4\text{r}+10=0$
$5\text{r}(2\text{r}-5)-2(2\text{r}-5)=0$
$\text{r}=\frac{5}{2},\frac{2}{5}$
$\therefore\text{G.P. is }\frac{\text{a}}{\text{r}},\text{a},\text{ar}$
$10, 5,\frac{5}{2}, \ ... \text{or}\frac{5}{2},5,10\ ...$

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Question 545 Marks

If $a, b, c$ are in G.P., prove that:
$\frac{\text{a}^2+\text{ab}+\text{b}^2}{\text{bc}+\text{ca}+\text{ab}}=\frac{\text{b}+\text{a}}{\text{c}+\text{b}}$

Answer

$a, b, c$ are in G.P.
$a, b = ar, c = ar^2$
$\frac{\text{a}^2+\text{ab}+\text{b}^2}{\text{bc}+\text{ca}+\text{ab}}=\frac{\text{b}+\text{a}}{\text{c}+\text{b}}$
$\frac{\text{a}^2+\text{a}(\text{ar})+\text{a}^2\text{r}^2}{(\text{ar})\big(\text{ar}^2\big)+\big(\text{ar}^2\big)\text{a}+\text{a}(\text{ar})}=\frac{\text{ar}+\text{a}}{\text{ar}^2+\text{ar}}$
$\frac{\text{a}^2\big(1+\text{r}+\text{r}^2\big)}{\text{a}^2(\text{r}^3+\text{r}^2+\text{r})}=\frac{1+\text{r}}{\text{r}(1+\text{r})}$
$\frac{1}{\text{r}}=\frac{1}{\text{r}}$
$\text{L.H.S}=\text{R.H.S}$
So,
$\frac{\text{a}^2+\text{ab}+\text{b}^2}{\text{bc}+\text{ca}+\text{ab}}=\frac{\text{b}+\text{a}}{\text{c}+\text{b}}$

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Question 555 Marks

If a and b are the roots of $\text{x}^2-3\text{x}+\text{p}=0$ and c, d are roots $\text{x}^2-12\text{x}+\text{q}=0,$ where a, b, c, d from a G.P. Prove that (q + p) : (q - p) = 17 : 15.

Answer

Given,
a, b are roots of the equation $\text{x}^2-3\text{x}+\text{p}=0$
$\Rightarrow\text{a}+\text{b}=3,\text{ab}=\text{p}$
And c, d are roots of the equation $\text{x}^2-12\text{x}+\text{q}=0$
$\Rightarrow\text{c}+\text{d}=12, \text{cd}=\text{q}$
Let b = ar, c = ar2 and d = ar3, then a + b = 3 and c + d = 12
$\text{a}(1+\text{r})=3\text{ and ar}^2(1+\text{r})=12$
$\Rightarrow\frac{\text{ar}^2(1+\text{r})}{\text{a}(1+\text{r})}=\frac{12}{3}$
$\Rightarrow\text{r}=2$
And $\text{a}(\text{r}+2)=3$
$\Rightarrow\text{a}=1$
$\text{p}=\text{ab}$
$\text{p}=\text{a}\times\text{ar}$
$\text{p}=2$
$\text{q}=\text{cd}$
$=\text{ar}^2\times\text{ar}^3$
$\text{a}=32$
$\frac{\text{q}+\text{p}}{\text{q}-\text{p}}=\frac{32+2}{32-2}$
$=\frac{34}{30}$
$(\text{q}+\text{p}):(\text{q}-\text{p})=17:15$

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Question 565 Marks

If a, b, c, d are in G.P., prove that:
$(\text{b}+\text{c})(\text{b}+\text{d})=(\text{c}+\text{a})(\text{c}+\text{d})$

Answer

a, b and c are in G.P.$\therefore\text{b}^2=\text{ac }\cdots(1)$
$\text{L.H.S}=({\text{b}+\text{a})(\text{b}+\text{d})}$ $=\text{b}^2+\text{bd}+\text{bc}+\text{cd}$ $=\text{ac}+\text{c}^2+\text{ad}+\text{cd}$ $[\text{Using (1)}]$ $=\text{c}(\text{a}+\text{c})+\text{d}(\text{a}+\text{c})$ $=(\text{c}+\text{a})(\text{c}+\text{d})$ $=\text{R.H.S}$$\therefore\text{R.H.S}=\text{L.H.S}$

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Maths 5 Marks Questions