(Each question 4 marks)

Question 514 Marks

If $S_1, S_2, S_3$, be respectively the sums of n, 2n, 3n terms of a G.P., then prove that $\text{S}^2_1+\text{S}^2_2=\text{S}_1(\text{S}_2+\text{S}_3).$

Answer

$S_1$ = sum of n terms, S1 = sum of 2n terms, S1 = sum of 3n terms. Then, $\text{S}_1^2+\text{S}_2^2$
$=(\text{S}_\text{n})^2+(\text{S}_\text{2n})^2$
$=\Big(\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{r}}\Big)^2+=\Big(\frac{\text{a}(1-\text{r}^\text{2n})}{1-\text{r}}\Big)^2$
$=\frac{\text{a}^2}{(1-\text{r})^2}\Big[\big(1-(\text{r})^\text{n})^2+(1-\text{r}^{2\text{n}})\Big]$
$=\frac{\text{a}^2}{(1-\text{r})^2}\Big[1+\text{r}^{2\text{n}}-2\text{r}^\text{n}+1+\text{r}^{4\text{n}}-2\text{r}^{2\text{n}}\Big]$
$=\frac{\text{a}^2}{(1-\text{r})^2}\Big[2-\text{r}^{2\text{n}}-2\text{r}^\text{n}+\text{r}^{4\text{n}}\Big]\cdots(\text{i})$ Also, $\text{S}_1(\text{S}_2+\text{S}_3)$
$=\frac{\text{a}^2}{(1-\text{r})^2}\Big(\frac{\text{a}(1-\text{r}^{2\text{n}})}{1-\text{r}}+\frac{\text{a}(1-\text{r}^{3\text{n}})}{1-\text{r}}\Big)$
$=\frac{\text{a}^2}{(1-\text{r})^2}\big[(1-\text{r})^\text{n}(1-\text{r}^{2\text{n}})+(1-\text{r}^{\text{n}})(1-\text{r}^{3\text{n}})\big]$
$=\frac{\text{a}^2}{(1-\text{r})^2}\big[1-\text{r}^{2\text{n}}-\text{r}^\text{n}+\text{r}^{3\text{n}}-\text{r}^{3\text{n}}-\text{r}^{\text{n}}+1+\text{r}^{4\text{n}}\big]$
$=\frac{\text{a}^2}{(1-\text{r})^2}\big[2-\text{r}^{2\text{n}}-2\text{r}^\text{n}+\text{r}^{4\text{n}}\big]\cdots(\text{ii})$
$(\text{i})=(\text{ii})\text{ Hence, }\text{S}^2_1+\text{S}^2_2=\text{S}_1(\text{S}_2+\text{S}_3)$

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Question 524 Marks

If a, b, c, are in G.P., prove that: $\big(\text{a}^2-\text{b}^2\big),\big(\text{b}^2-\text{c}^2\big),\big(\text{c}^2-\text{d}^2\big)\text{ are in G.P.}$

Answer

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

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Question 534 Marks

A Person has 2 parents, 4 grandparents, 8 great parents, and so on. Find the number of his ancestors during the generation preceding his own.

Answer

To find number of ancestors, we will find the sum of $2, 2^2, 2^3$, ... Number of ancestors $=\frac{2(2^{10}-1)}{2-1}$ $=2(1024-1)$ $=2\times1023$ $=2046$

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Question 544 Marks

If the $p^{th}$ and $q^{th}$ terms of a G.P. and q and p respectively, show that $(p + q)^{th}$ term is $\Big(\frac{\text{q}^\text{p}}{\text{p}^\text{q}}\Big)^\frac{1}{\text{p}-\text{q}}.$

Answer

$n^{th}$ term of $G.P. = ar^{x-1} p^{th}$ term =$q = a.r^{p-1} q^{th}$ term = $p = a.r^{q-1}$^ $\frac{\text{q}}{\text{p}}=\text{r}^{\text{r}-4}$
$\text{r}=\Big(\frac{\text{q}}{\text{p}}\Big)^\frac{1}{\text{r}-\text{q}}$
$\text{a}=\text{p}\Big(\frac{\text{p}}{\text{q}}\Big)^\frac{1-\text{q}}{\text{p}-\text{q}}$
$\text{p}+\text{q}^\text{th}\text{term}=\text{p}\Big(\frac{\text{q}}{\text{p}}\Big)^{\frac{1-\text{q}}{\text{p}-\text{q}}}\Big(\frac{\text{q}}{\text{p}}\Big)^{\frac{\text{p}+\text{q}-1}{\text{p}-\text{q}}}$
$=\text{p}\Big(\frac{\text{q}}{\text{p}}\Big)^{\frac{1-\text{q}+\text{p}+\text{q}-1}{\text{p}-\text{q}}}$
$=\text{p}\Big(\frac{\text{q}}{\text{p}}\Big)^{\frac{\text{p}}{\text{p}-\text{q}}}$
$=\frac{\text{q}^\frac{\text{p}}{\text{p}-\text{q}}}{\text{p}^{\frac{\text{q}}{\text{p}-\text{q}}-1}}$
$=\frac{\text{q}^\frac{\text{p}}{\text{p}-\text{q}}}{\text{p}^{\frac{\text{q}}{\text{p}-\text{q}}}}$
$=\Big(\frac{\text{q}^\text{p}}{\text{p}^\text{q}}\Big)^\frac{1}{\text{p}-\text{q}}$

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Question 554 Marks

If A.M. and G.M. of roots of a quadratic equation are 8 and 5 respectively, then obtain the quadratic equation.

Answer

Let the roots of the quadratic equation be a and b. AM = 8 $\therefore\ \frac{\text{a}+\text{b}}{2}=8$ $\Rightarrow\text{a}+\text{b}=16\ \cdots(\text{i})$ Also, G = 5 $\Rightarrow\sqrt{\text{ab}}=5$ $\Rightarrow\text{ab}=5^2$ $\Rightarrow\text{ab}=25\ \cdots(\text{ii})$ Now, the quadratic equition is given by $\text{x}^2-(\text{a}+\text{b})\text{x}+\text{ab}=0$ $\Rightarrow\text{x}^2-16\text{x}+25=0$

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Question 564 Marks

Show that in an infinite G.P. with common ratio $\text{r}\big(|\text{r}|<1\big),$ each terms bears a constant ratio to the sum of all terms that follow it.

Answer

Let a be first term and r be common ratio of G.P. Here, $\frac{\text{a}_\text{n}}{\big(\text{a}_{\text{n}+1}+\text{a}_{\text{n}+2}+\ \dots\infty\big)}=\frac{\text{ar}^{\text{n}-1}}{\text{ar}^{\text{n}}+\text{ar}^{\text{n}+1}+\ \dots}$ $=\frac{\text{ar}^{\text{n}-1}}{\text{ar}^{\text{n}}\big(1+\text{r}+\text{r}^2+\ \dots\infty\big)}$ $=\frac{\text{ar}^{\text{n}-1}}{\text{ar}^{\text{n}}\Big(\frac{1}{1-\text{r}}\Big)}$ $=\Big(\frac{1-\text{r}}{\text{r}}\Big)$ Since r is a constant, so $\Big(\frac{\text{a}_\text{n}}{\text{a}_{\text{n}+1}+\text{a}_{\text{n}+2}+\ \dots\infty}\Big)=\text{k}\ (\text{constant})$ Such that $\text{k}=\Big(\frac{1-\text{r}}{\text{r}}\Big)$

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MATHS (Each question 4 marks)