Solve the following Question.(1 Marks) - Maths STD 11 Questions - Vidyadip

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Solve the following Question.(1 Marks)

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18 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 Mark

If $A_1, A_2$ be two $AM's$ and $G_1, G_2$ be two $GM's$ between $a$ and $b$, then find the value of $\frac{\text{A}_1+\text{A}_2}{\text{G}_1\text{G}_2}.$

Answer

$a, A_1, A_2, b$ are in A.P.
$A_1 - a = A_2 - A_1 = b - A_2 $____$(1)$
And $a, G_1, G_2, b$ are in G.P.
$\text{G}_1=\sqrt{\text{a}\text{G}_2},\text{G}_2=\sqrt{\text{G}_1\text{b}}$
$\Rightarrow\text{G}_1\text{G}_2=\sqrt{\text{ab}\text{G}_1\text{G}_2}$
$\Rightarrow\sqrt{\text{G}_1\text{G}_2}=\sqrt{\text{ab}}$
$\Rightarrow\text{G}_1\text{G}_2=\text{ab}\cdots(2)$
From equation (1),
$2\text{A}_1 = \text{A}_2 + \text{a}$
$2\text{A}_2 = \text{A}_1 + \text{b}$
Adding these two,
$2(\text{A}_1+\text{A}_2)=(\text{A}_1+\text{A}_2)+\text{a}+\text{b}$
$\text{A}_1+\text{A}_2=\text{a}+\text{b}$
$\Rightarrow\frac{\text{A}_1+\text{A}_2}{\text{G}_1\text{G}_2}=\frac{\text{a}+\text{b}}{\text{ab}}$

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Question 21 Mark

If $\text{a} = 1 + \text{b} + \text{b}_2 + \text{b}_3 + ...\text{to }\infty,$ then write b in terms of a given that |b|<1.

Answer

$\text{a} = 1 + \text{b} + \text{b}_2 + \text{b}_3 + ...\text{to }\infty,$$\text{a}=\frac{1}{1-\text{b}}$ $\Big[\text{Since},\text{S}_\infty=\frac{\text{a}}{1-\text{r}}\Big]$
$\text{a}(1-\text{b})=1$
$\text{a}-\text{ab}=1$
$\text{ab}=\text{a}-1$
$\text{b}=\frac{\text{a}-1}{\text{a}}$

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Question 31 Mark

If second, third and sixth term of an A.P. are consecutive terms of a G.P., write the common ratio of G.P.

Answer

$a_2, a_3, a_6$ of A.P. are in G.P.
$\text{a}_2 = \text{a} + \text{d}\cdots(\text{i})$
$\text{a}_3=\text{a}+2\text{d}\cdots(\text{ii})$
$\text{a}_6=\text{a}+5\text{d}\cdots(\text{iii})$
$\frac{\text{a}_3}{\text{a}_2}=\frac{\text{a}_6}{\text{a}_3}$
$\frac{\text{a}+2\text{d}}{\text{a}+\text{d}}=\frac{\text{a}+5\text{d}}{\text{a}+2\text{d}}$
$\text{a}^2+4\text{d}^2+4\text{ad}=\text{a}^2+5\text{ad}+\text{ad}+5\text{d}^2$
$4\text{d}^2-5\text{d}^2=6\text{ad}-4\text{ad}$
$-\text{d}^2=2\text{ad}$
$-\text{d}=2\text{a}$
$2\text{a}=-\text{d}$
$\text{a}_2=-\frac{\text{d}}{2}+\text{d}=\frac{\text{d}}{2}$
$\text{a}_3=-\frac{\text{d}}{2}+2\text{d}=\frac{3\text{d}}{2}$
$\text{a}_{62}=-\frac{\text{d}}{2}+5\text{d}=\frac{9\text{d}}{2}$
$\text{r}=\frac{\text{a}_3}{\text{a}_2}=\frac{\text{a}_6}{\text{a}_3}$
$=\frac{\frac{3\text{d}}{2}}{\frac{\text{d}}{2}}=\frac{\frac{9\text{d}}{2}}{\frac{3\text{d}}{2}}$
$\text{r}=3$

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Question 41 Mark

If the fifth term of a G.P. is $2$, then write the product of its $9$ terms.

Answer

Here, $a_5 = 2$
Nine terms of a G.P. are,
$\frac{\text{a}}{\text{r}^4}\times\frac{\text{a}}{\text{r}^3}\times\frac{\text{a}}{\text{r}^2}\times\frac{\text{a}}{\text{r}}\times\text{a}\times\text{ar}\times\text{ar}^2\times\text{ar}^3\times\text{ar}^4$
$=(\text{a})^9$
$=(2)^9$
$=512$

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Question 51 Mark

If $(p + q)^{th}$ and $(p - q)^{th}$ terms of a G.P. are m and respectively, then write its pth term.

Answer

$(p+q)^{\text {th }} \text { term }=a r^{p+q-1}=m$
$(p-q)^{\text {th }} \text { term }=a r^{p-q-1}=n$
$m n=a^2 r^{3 p-2}$
$\sqrt{m n}=a r^{p-1}$
$=p^{\text {th }} \text { term }$

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Question 61 Mark

If $\log_\text{x}\text{a},\text{a}^{\frac{\text{x}}{2}}$ and (p - q)th are in G.P., then write the value of x.

Answer

$\log_\text{x}\text{a},\text{a}^{\frac{\text{x}}{2}}$ and $\log_\text{b}\text{x}$ are in G.P.
$\therefore\Big(\text{a}^{\frac{\text{x}}{2}}\Big)^2=\log_\text{x}\text{a}\times\log_\text{b}\text{x}$
$\Rightarrow\text{a}^\text{x}=\frac{\log_\text{b}\text{a}}{\log_\text{b}\text{x}}\times\log_\text{b}\text{x}$
$\Rightarrow\text{a}^\text{x}=\log_\text{b}\text{a}$
Now, by taking $\log_\text{a}$ on both the sides:
$\Rightarrow\text{x}=\log_\text{a}(\log_\text{b}\text{a})$

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Question 71 Mark

Write the product of n geometric means between two number a and b.

Answer

Let $\text{x}_1, \text{x}_2, \text{x}_3 \dots\text{x}_\text{n}$ be n G.M. between a and b then
$\text{a},\text{x}_1, \text{x}_2, \text{x}_3\dots \text{x}_\text{n},\text{b}$
$\text{b}=\text{a}_{\text{x}+2}=\text{ar}^{\text{n}+1}$
$\text{r}=\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{1}{\text{n}+1}}\cdots(1)$
$\text{x}_1=\text{a}_2=\text{ar}$
$\text{x}_2=\text{a}_3=\text{ar}^2$
$\text{x}_\text{n}=\text{a}_{\text{n}+1}=\text{ar}^{\text{n}}$
Product of n GM
$=\text{x}_1,\text{x}_2\dots\text{x}_\text{n}$
$=(\text{ar})\big(\text{ar}^2\big)\dots\big(\text{ar}^\text{n}\big)$
$=\text{a}^{\text{n}}\text{r}^{1+2+\dots+\text{n}}$
$=\text{a}^\text{n}\text{r}\frac{\text{n}(\text{n}+1)}{2}$
$=\text{a}^\text{n}\Big(\frac{\text{b}}{\text{a}}\Big)^{\frac{\text{n}}{2}}$ [Using equation (1)]
$=\text{a}^{\text{n}}\frac{\text{b}^{\frac{\text{n}}{2}}}{\text{a}^{\frac{\text{n}}{2}}}$
$=\text{a}^{\frac{\text{n}}{2}}\text{b}^{\frac{\text{n}}{2}}$
$=(\text{ab})^{\frac{\text{n}}{2}}$

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Question 81 Mark

If sum of an infinite decreasing G.P. is $3$ and the sum of the squares its term is $\frac{9}{2},$ then write its first term and common differance.

Answer

Let G.P. $be a, ar, ar^2$
$\text{S}'_\infty=\frac{\text{a}}{1-\text{r}}$
$3=\frac{\text{a}}{1-\text{r}}\cdots(\text{i})$
Square of terms
$a^2, a^2r^2, a^2r^4, ...$
$\text{S}"_\infty=\frac{\text{a}^2}{1-\text{r}^2}$
$\frac92=\frac{\text{a}^2}{1-\text{r}^2}\cdots(\text{ii})$
Using equation (i) and (ii),
$\frac{9}{2\times3}=\frac{\text{a}^2}{1-\text{r}^2}\times\frac{1-\text{r}}{\text{a}}$
$\frac{3}{2}=\frac{\text{a}}{1+\text{r}}\cdots(\text{iii})$
Using equation (iii) and (ii),
$\frac{3}{2\times3}=\frac{\text{a}}{1+\text{r}}\times\frac{1-\text{r}}{\text{a}}$
$\frac12=\frac{1-\text{r}}{1+\text{r}}$
$1+\text{r}=2-2\text{r}$
$3\text{r}=1$
$\text{r}=\frac{1}{3}$
From equation (i),
$3=\frac{\text{a}}{1-\frac13}$
$\text{a}=3\times\frac23$
$\text{a}=2$
$\text{r}=\frac13$

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Question 91 Mark

If $p^{th}, q^{th}$ and rth terms of a G.P. are $x, y, z$ respectively, then write the value of $x^{q-r} y^{r-p} z^{p-q}.$

Answer

$\text{a}_\text{p}=\text{x}=\text{ar}^{\text{p}-1}\cdots(\text{i})$
$\text{a}_\text{q}=\text{y}=\text{ar}^{\text{q}-1}\cdots(\text{ii})$
$\text{a}_\text{r}=\text{z}=\text{ar}^{\text{r}-1}\cdots(\text{iii})$
$\text{x}^{\text{q}-\text{r}}\text{y}^{\text{r}-\text{p}}\text{z}^{\text{p}-\text{q}}$
$=\big[\text{ar}^{\text{p}-1}\big]\big[\text{ar}^{\text{q}-1}\big]^{\text{r}-\text{p}}\big[\text{ar}^{\text{r}-1}\big]^{\text{p}-\text{q}}$
$=\text{a}^{\text{q}-\text{r}+\text{r}-\text{p}+\text{q}}\text{ r}^{\text{pq}-\text{pr}-\text{q}+\text{r}+\text{qr}-\text{r}+\text{p}+\text{rp}-\text{qr}-\text{p}+\text{q}}$
$=\text{a}^0\text{r}^0$
$=1$
$\text{x}^{\text{q}-\text{r}}\times\text{y}^{\text{r}-\text{p}}\times\text{z}^{\text{p}-\text{q}}=1$

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Question 101 Mark

Write the quadration the arithmetic and geometric means of whose roots are A and G respectively.

Answer

Let $\alpha,\beta, $ be roots of equation, so equation is
$\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta=0$
Here,
$\frac{\alpha+\beta}{2}=\text{A}$
$\alpha+\beta=2\text{A}$
And $\sqrt{\alpha\beta}=\text{G}$
$\alpha\beta=\text{G}^2$
So, Required equation is,
$\text{x}^2-2\text{Ax}+\text{G}^2=0$

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Question 111 Mark

Write the sum of the series $1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + ..... + (2n - 1)^2 - (2n)^2$

Answer

We have,
$1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + ..... + (2n - 1)^2 - (2n)^2$
$= (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + ...... + (2n - 1 - 2n)(2n - 1 + 2n)$
$= -1[1 + 2 + 3 + 4 + ..... + 2n - 1 + 2n]$
$=-1\Big[\frac{2\text{n}(2\text{n}+1)}{2}\Big]$ $\Big[\because\ 1 + 2\ ....\text{n} =\frac{\text{n}(\text{n}+1)}{2}\Big]$
$=-\text{n}(2\text{n}+1)$
Hence, sum of the series $= -n(2n + 1)$

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Question 121 Mark

If the sum of first $n$ even natural numbers is equal to $k$ times the sum of first $n$ odd natural numbers, then write the value of $k$.

Answer

It is given that the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers.
$\therefore\ 2+4+6+ \ ...\ +2\text{n}=\text{k}(1+3+5+\ ....\ +\text{n})$
$\Rightarrow2(1+2+3+ \ ...\ +\text{n})=\text{k}(1+3+5+\ ....\ +\text{n})$
$\Rightarrow2\times\frac{\text{n}(\text{n}+1)}{2}=\text{k}(1+3+5+\ ...\ +\text{n})$
$\Rightarrow\text{n}(\text{n}+1)=\text{k}(1+3+5+\ ...\ +\text{n})$
$\Rightarrow\text{n}(\text{n}+1)=\text{k}\times\text{n}^2$ $\big[\because$ sum of first odd natural numbers is $n^2$$\big]$
$\Rightarrow\frac{(\text{n}+1)}{\text{n}}=\text{k}$
Hence, $\text{k}=\frac{\text{n}+1}{\text{n}}$

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Question 131 Mark

Let $S_n$ denote the sum of the cubes of first n natural numbers and $s_n$ denote the sum of first $n$ natural numbers.
Then, write the value of $\sum\limits^\text{n}_{\text{r}=1}\frac{\text{S}_\text{r}}{\text{S}_\text{r}}$

Answer

We know that,
$\text{S}_\text{r}=1^3+2^3+3^3+\ ....\ +\text{r}^3=\Big[\frac{\text{r}(\text{r}+1)}{2}\Big]^2$
And, $\text{S}_\text{r}=1+2+3+\ ....\ +\text{r}=\frac{\text{r}(\text{r}+1)}{2}$
As, $\frac{\text{S}_\text{r}}{\text{S}_\text{r}}=\frac{\big[\frac{\text{r}(\text{r}+1)}{2}\big]^2}{\big[\frac{\text{r}(\text{r}+1)}{2}\big]}$
$=\frac{\text{r}(\text{r}+1)}{2}=\frac{1}{2}(\text{r}^2+\text{r})$
Now,
$\sum\limits^{\text{n}}_{\text{r}=1}\frac{\text{S}_\text{r}}{\text{S}_\text{r}}=\sum\limits^{\text{n}}_{\text{r}=1}\frac{1}{2}(\text{r}^2+\text{r})$
$=\frac{1}{2}\Bigg(\sum\limits^{\text{n}}_{\text{r}=1}\text{r}^2+\sum\limits^{\text{n}}_{\text{r}=1}\text{r}\Bigg)$
$=\frac{1}{2}\Big[\frac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}+\frac{\text{n}(\text{n}+1)}{2}\Big]$
$=\frac{1}{2}\times\frac{\text{n}(\text{n}+1)}{2}\times\Big[\frac{(2\text{n}+1)}{3}+1\Big]$
$=\frac{\text{n}(\text{n}+1)}{4}\Big[\frac{2\text{n}+4}{3}\Big]$
$=\frac{\text{n}(\text{n}+1)}{4}\times\frac{2(\text{n}+2)}{3}$
$=\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$

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Question 141 Mark

If $\sum\limits^{\text{n}}_{\text{r}=1}\text{r}=55,$ find $\sum\limits^{\text{n}}_{\text{r}=1}\text{r}^3$

Answer

$\sum\limits^{\text{n}}_{\text{r}=1}\text{r}=55$
$\Rightarrow\frac{\text{n}(\text{n}+1)}{2}=55\ ...(\text{i})$
Now,
$\sum\limits^{\text{n}}_{\text{r}=1}\text{r}^3=\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]^2$
$=\big[55\big]^2$ [Using equation (i), we get]
$=3025$
Hence, $\sum\limits^{\text{n}}_{\text{r}=1}\text{r}^3=3025$

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Question 151 Mark

Write the sum of the series $2 + 4 + 6 + 8 + ... + 2n$.

Answer

Let $T_n$ be the term of the given series and $S_n$ be the sum of the given series.
$\therefore\ \text{T}_\text{n}=2\text{n}$
$\therefore\ \text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\text{T}_\text{k}=\sum\limits^{\text{n}}_{\text{k}=1}2\text{k}$
$=2\sum\limits^{\text{n}}_{\text{k}=1}\text{k}$
$=2\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]$
$=\text{n}(\text{n}+1)$
Hence, $\text{S}_\text{n}=\text{n}(\text{n}+1)$

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Question 161 Mark

Write the sum of $20$ terms of the series $1+ \frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+\ ...$

Answer

Let the $n ^{\text {th }}$ term of the given series is $T_n$ and $S_n$ be the sum of the given series.
$\therefore\ \text{T}_\text{n}=\frac{1}{\text{n}}\big[1+2+3+\ ...\ +\text{n}]$
$=\frac{1}{\text{n}}\Big[\frac{\text{n}}{2}\big(2\times1+(\text{n}-1)\times1\big)\Big]$
$=\frac{1}{\text{n}}\times\frac{\text{n}}{2}\big[2+\text{n}-1\big]$
$=\frac{1}{2}(\text{n}+1)$
$\therefore\ \text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{k}=1}\frac{1}{2}(\text{k}+1)$
$=\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}\text{k}+\frac{1}{2}\sum\limits^{\text{n}}_{\text{k}=1}1$
$=\frac{1}{2}\Big[\frac{\text{n}(\text{n}+1)}{2}\Big]+\frac{1}{2}\times\text{n}$
$=\frac{\text{n}(\text{n}+1)}{4}+\frac{\text{n}}{2}$
$=\frac{\text{n}(\text{n}+1)+2\text{n}}{4}$
$=\frac{\text{n}[\text{n}+1+2]}{4}$
$=\frac{\text{n}[\text{n}+3]}{4}$
Putting, $n = 20$ we get
$\text{S}_{20}=\frac{20[20+3]}{4}$
$=5\times23$
$=115$

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Question 171 Mark

Write the sum to $n$ terms of a series whose $r^{th}$ term is $r + 2^r$

Answer

Let $T_n$ be the $n^{th}$ term of the series and S_n be the sum to $n$ terms of a series.
$\therefore\ \text{T}_\text{r}=\text{r}+2^{\text{r}}$ [given]
$\therefore\ \text{S}_\text{n}=\sum\limits^{\text{n}}_{\text{r}=1}(\text{r}+2^{\text{r}})$
$=\sum\limits^{\text{n}}_{\text{r}=1}\text{r}+\sum\limits^{\text{n}}_{\text{r}=1}2^{\text{r}}$
$=\frac{\text{n}(\text{n}+1)}{2}+\big[2^1+2^2\ ...\ 2^{\text{n}}\big]$
$=\frac{\text{n}(\text{n}+1)}{2}+\frac{2(2^\text{n}-1)}{(2-1)}$
$=\frac{\text{n}(\text{n}+1)}{2}+\frac{2\times2^\text{n}-2}{1}$
$=\frac{\text{n}(\text{n}+1)}{2}+2^{\text{n}+1}-2$
$\therefore\ \text{S}_\text{n}=\frac{\text{n}(\text{n}+1)}{2}+2^{\text{n}+1}-2$

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Question 181 Mark

Write the $50^{th}$ term of the series $2 + 3 + 6 + 11 + 18 + ....$

Answer

We have,
$a_1 = 2,$
$a_2 = 3 = 2 + 1,$
$a_3 = 6 = 2 + 1 + 3,$
$a_4 = 11 = 2 + 1 + 3 + 5,$
$.............................$
$.............................$
$..............................$
$a_{50} = 2 + 1 + 3 + 5 + .... (50 terms)$
$=2+\frac{49}{2}\big[2\times1+(49-1)\times2\big]$ (As, the terms apart $2$ are in A.P. with $a = 1$ and $d = 2$)
$=2+\frac{49}{2}(2+48\times2)$
$=2+\frac{49}{2}\times98$
$=2+49^2$
$=2+2401$
$=2403$
So, the $50^{th}$ term of the given series is $2403$

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