3 Marks Question

Question 513 Marks

If $S_p$ denotes the sum of the series $1+\text{r}^{\text{p}}+\text{r}^{2\text{r}}+\ \dots\text{ to }\infty$ and $S_p$ the sum of the series $1-\text{r}^{\text{p}}+\text{r}^{\text{2p}}-\ \dots\text{ to }\infty,$ prove that $\text{S}_\text{p} + \text{S}_\text{p} = 2 \text{S}_{2\text{p}}.$

Answer

$\text{S}_\text{p}=1+\text{r}^{\text{p}}+\text{r}^{2\text{p}}+\ \dots\ +\infty$
$\text{S}_\text{p}=\frac{1}{1-\text{r}^{\text{r}}}$
$\text{S}_\text{p}=1-\text{r}^\text{p}+\text{r}^{2\text{p}}+\ \dots\ +\infty$
$\text{S}_\text{p}=\frac{1}{1+\text{r}^\text{p}}$
Now,
$\text{S}_\text{p}+\text{S}_\text{p}=\frac{1}{1-\text{r}^\text{p}}+\frac{1}{1+\text{r}^{\text{p}}}$
$=\frac{2}{1-\text{r}^{2\text{p}}}$
$\text{S}_\text{p}+\text{S}_\text{p}=2\times\text{S}_{2\text{p}}$

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Question 523 Marks

If $a, b, c$ are in A.P. and $a, x, b$ and $b, y, c $are in G.P., show that $x2, b2, y2$ are in A.P.

Answer

$a, b, c$ are in A.P.
$\Rightarrow2\text{b}=\text{a}+\text{c }\cdots(\text{i})$
$a, x, b$ are in G.P.
$\Rightarrow\text{x}^2=\text{ab }\cdots{\text{(ii}})$
$b, y, c$ are in G.P.
$\Rightarrow\text{y}^2=\text{bc}\cdots(\text{iii})$
Now, putting the value of $a$ and $c$:
$\Rightarrow2\text{b}=\frac{\text{x}^2}{\text{b}}+\frac{\text{y}^2}{\text{b}}$
$\Rightarrow2\text{b}^2=\text{x}^2+\text{y}^2$
Therefore, $x^2, b^2$ and $y^2$ are also in A.P.

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Question 533 Marks

The sum of terms of the G.P. 3, 6, 12, ... is 381. Find the value of n.

Answer

3, 6, 12, ... n 381
$\text{a}={3},\text{r}=\frac{6}{{3}}={2},\text{n}=?,\text{S}_\text{n}=381$
We know that
$\text{S}_\text{n}=\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{r}}$
$381=\frac{3(2)^\text{n}-1}{2-1}$
$\frac{381}{3}=2^{\text{n}}-1$
$127=2^\text{n}-1$
$128=2^\text{n}$
$2^7=2^{\text{n}}$
$\Rightarrow\text{n}=7$

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Question 543 Marks

Find the two numbers whose A.M. is 25 and G.M. is 20.

Answer

Given,
A.M. = 25
G.M. = 20
Now,
$\text{A}.\text{M}.=\frac{\text{a}+\text{b}}{2}=25$
And, $\text{G}.\text{M}=\sqrt{\text{ab}}=20$
$\text{a}+\text{b}=50,\text{ab}=400$
$(\text{a}-\text{b})=\sqrt{(\text{a}+\text{b})^2-4\text{ab}}$
$=\sqrt{(50)^2-16000}$
$=\sqrt{2500-1600}$
$=\pm30$
$\text{a}-\text{b}=\pm30\\ {\text{a}+\text{b}=50}\\ \overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \ \ \text{2a} = \ \ \ 80$
$\text{a}=40$
Also, $-2\text{b}=-20$
$\text{b}=10$
$\therefore$ The numbers are 40, 10.

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Question 553 Marks

The ratio of the sum of first three term is to that of first $6$ terms of a G.P. is $125 : 152$. Find the common ratio.

Answer

Let Sum of first three terms $= a + ar + ar^2$
The ratio $=\frac{\text{a}+\text{ar}+\text{ar}^2}{\text{a}+\text{ar}+\text{ar}^2+\text{ar}^3+\text{ar}^4+\text{ar}^5}$
$=\frac{1+\text{r}+\text{r}^2}{1+\text{r}+\text{r}^2+\text{r}^3(1+\text{r}+\text{r}^2)}\cdots(1)$
Let $\text{A}=1+\text{r}+\text{r}^2\cdots(2)$
$\text{Ratio}=\frac{\text{A}}{\text{A}+\text{r}^3\text{A}}=\frac{125}{152}$
$\frac{1}{1+\text{r}^3}=\frac{125}{152}$
$152+125+125\text{r}^3$
$\text{r}^3=\frac{27}{125}$
$\text{r}=\frac35$

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Question 563 Marks

Show that the ratio of the sum of the first n terms of a G.P. to the sum of terms from $(\text{n}+1)^\text{th}\text{ to }(2\text{n})^\text{th}\text{ terms is }\frac{1}{\text{r}^\text{n}}.$

Answer

Sum of first n term of G.P.
$=\text{a}+\text{a}_2+\text{a}_3+\ ...\ +\text{a}_\text{n}$
$=\text{a}+\text{ar}+\text{ar}^2+\ ...\ +\text{ar}^{\text{n}-1}$ $[\therefore\text{t}_\text{n}=\text{ar}^{\text{n}-1}]\cdots(\text{i})$
Also sum of term from
$(\text{n}+1)^{\text{th}}\text{ to }(2\text{n})^{\text{th}}\text{ term is}$
$=\text{a}_{\text{n}+1}+\text{a}_{\text{n}+1}+\dots\ +\text{a}_{2\text{n}}$
$=\text{ar}^{\text{n}}+\text{ar}^{\text{n}-1}+\ ...\ +\text{ar}^{2\text{n}-1}\cdots{(\text{i})}$
Ratio of (i) and (ii) is:
$=\frac{\text{a}+\text{ar}+\text{ar}^2+\ ...\ \text{ar}^{\text{n}-1}}{\text{ar}^{\text{n}}+\text{ar}^{\text{n}-1}+\ ...\ +\text{ar}^{2\text{n}-1}}$ $\Big[\because\text{S}_\text{n}=\frac{\text{a}(1-\text{r}^\text{n})}{1-\text{r}}\Big]$
$=\frac{\frac{\text{a}(1-\text{r}^{\text{n}})}{1-\text{r}}}{\frac{\text{ar}^\text{n}(1-\text{r}^\text{n})}{1-\text{r}}}$
$=\frac{1}{\text{r}^\text{n}}$

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Question 573 Marks

Find the sum: $\sum\limits_{\text{n}=1}^{10}\bigg\{\Big(\frac12\Big)^{\text{n}-1}+\Big(\frac15\Big)^{\text{n}+1}\bigg\}.$

Answer

$\sum\limits_{\text{n}=1}^{10}\bigg\{\Big(\frac12\Big)^{\text{n}-1}+\Big(\frac15\Big)^{\text{n}+1}\bigg\}$
$=\sum\limits_{\text{n}=1}^{10}\Big(\frac12\Big)^{\text{n}-1}+\sum\limits_{\text{n}=1}^{10}\Big(\frac15\Big)^{\text{n}+1}$
$=1+\frac12+\frac{1}{2^2}+\ \cdots\ +\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+\ \cdots$
$=\frac{\Big(1-\frac{1}{2^{10}}\Big)}{1-\frac12}+\frac{\frac15\Big(1-\frac{1}{5^{10}}\Big)}{1-\frac15}$
$=\frac{2^{10}-1}{2^9}+\frac{5^{10}-1}{5^{11}}$

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Question 583 Marks

How many terms of G.P. $3,\frac32,\frac34\cdots$ are needed to give the sum $\frac{3069}{512}?$

Answer

$\text{Sum}=\frac{3069}{512}=\frac{3\big(1-\frac{1}{2^{\text{n}}}\big)}{\frac12}$
$1-\frac{1}{2^\text{n}}=\frac{3069}{512\times6}=\frac{1023}{512\times2}$
$1-\frac{1023}{1024}=\frac{1}{2^\text{n}}$
$\frac{1}{2^\text{n}}=\frac{1}{1024}$
$\text{n}=10$

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Maths 3 Marks Question