PROGRESSION

 Question 1:

Let $a_1,a_3,\dots$ be in harmonic progression with $a_1=5$ and $a_{20}=25.$ The least positive integer $n$ for which $a_n<0$ is

[1] 22

[2] 23

[3] 24

[4] 25

Answer & Solution

Option # 4

If $a_1,a_2,a_3,\dots .$ are in harmonic progression, then $\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3}\dots \dots$ are in AP.

First term of $AP,\frac{1}{a_1}=\frac{1}{5}$

20th term of $AP,\frac{1}{a_{20}}=\frac{1}{25}$

$\Rightarrow \frac{1}{5}+19d=\frac{1}{25}$

$\Rightarrow d=\frac{-4}{19\times 25}$

We have to find the least positive integer n for which $a_n<0$

$\Rightarrow \frac{1}{5}+(n-1)d<0$

$\Rightarrow \frac{1}{5}+(n-1)d<0$

$\Rightarrow \frac{1}{5}+(n-1)>\frac{95}{4}$

$\Rightarrow n>24.75$

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Question 2:
The sum of the first 20 term of the sequence $0.7,0.77,0.777,\dots ,$ is

[1] $\frac{7}{81}(179-{10}^{-20})$

[2] $\frac{7}{9}(99-{10}^{-20})$

[3] $\frac{7}{81}(179+{10}^{-20})$

[4] $\frac{7}{9}(99+{10}^{-20})$

Answer & Solution

Option # 3

We have,

$0.7+0.77+0.777+\dots$ upto 20 terms

$\Rightarrow 7(\frac{1}{10}+\frac{11}{{10}^2}+\frac{111}{{10}^3}+\dots upto20terms)$

$\Rightarrow \frac{7}{9}(\frac{9}{10}+\frac{99}{{10}^2}+\frac{999}{{10}^3}+\dots upto20terms)$

$\Rightarrow \frac{7}{9}((1-\frac{1}{10})+(1-\frac{1}{100})+(1-\frac{1}{1000})+\dots upto20terms)$ $\Rightarrow \frac{7}{9}(20-(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\dots upto20terms))$

$\Rightarrow \frac{7}{9}(20-\left(\frac{1}{10}\left(\frac{1-{\left(\frac{1}{10}\right)}^{20}}{1-\frac{1}{10}}\right)\right))$

$\Rightarrow \frac{7}{9}(20-\frac{1}{9}(1-{\left(\frac{1}{10}\right)}^{20}))$

$\Rightarrow \frac{7}{81}(179+{\left(\frac{1}{10}\right)}^{20})$

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Question 3:
If $(10{)}^9+2(11{)}^1(10{)}^8+3(11{)}^2(10{)}^7+\dots +10(11{)}^9=k(10{)}^9$, then k is equal to

[1] $\frac{121}{10}$

[2] $\frac{441}{100}$

[3] 100

[4] 110

Answer & Solution

Option # 3

We have,

$(10{)}^9+2(11{)}^1(10{)}^5+3(11{)}^2(10{)}^7+\dots +10(11{)}^9=k(10{)}^9$

Dividing by $(10{)}^9$ on both sides, we get

$1+2\left(\frac{11}{10}\right)+3{\left(\frac{11}{10}\right)}^2+\dots +10{\left(\frac{11}{10}\right)}^9=k\cdots \left(i\right)$

Multiplying $\left(\frac{11}{10}\right)$ on both side of above equation we get,

$\left(\frac{11}{10}\right)+2{\left(\frac{11}{10}\right)}^2+3{\left(\frac{11}{10}\right)}^3+\dots +10{\left(\frac{11}{10}\right)}^{10}=\left(\frac{11}{10}\right)k\dots$ (ii)

Subtracting (ii) from (i), we get

$k-\left(\frac{11}{10}\right)k=1+\left(\frac{11}{10}\right)+{\left(\frac{11}{10}\right)}^2+\dots +{\left(\frac{11}{10}\right)}^9-10{\left(\frac{11}{10}\right)}^{10}$

$\Rightarrow k(1-\left(\frac{11}{10}\right))=\frac{1({\left(\frac{11}{10}\right)}^{10}-1)}{\left(\frac{11}{10}\right)-1}-10{\left(\frac{11}{10}\right)}^{10}$

$\Rightarrow k(1-\left(\frac{11}{10}\right))=-10$

$\Rightarrow k=100$

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Question 4:
Three positive numbers form an increasing GP. If the middle term of the GP is doubled, then new numbers are in AP. Then, the common ratio of the GP is

[1] $\sqrt{2}+\sqrt{3}$

[2] $3+\sqrt{3}$

[3] $2-\sqrt{3}$

[4] $2+\sqrt{3}$

Answer & Solution

Option # 4

Let $a,ar,a{r}^2$ are in GP.

If the middle term of the GP is doubled, then new numbers are in AP.

$\Rightarrow a,2ar,a{r}^2$ are in AP.

$\Rightarrow 4ar=a+a{r}^2$

$\Rightarrow r^2-4r+1=0$

$\Rightarrow r=2\pm \sqrt{3}$

since the series is an increasing GP so,

$\Rightarrow r=2+\sqrt{3}$

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Question 5:
The sum of the integers lying between 1 and 100 (both inclusive) and divisible by 3 or 5 or 7 is

[1] 818

[2] 1828

[3] 2838

[4] 3848

Answer & Solution

Option # 3

Let $S_3,S_5,S_7,S_{15},S_{35},S_{21}$ be the sum of integers divisible by $3,5,7,15,35$ and 21 respectively.

Here,

$S_3=3+6+9+\dots +99$

$S_5=5+10+15+\dots +100$

$S_7=7+14+21+\dots +98$

$S_{15}=15+30+45+\dots +90$

$S_{15}=35+70$

$S_{31}=21+42+63+84$

So, the required sum is:

$=S_3+S_5+S_7-S_{15}-S_{35}-S_{21}$

$=\frac{33}{2}(3+99)+\frac{20}{2}(5+100)+\frac{14}{2}(7+98)-\frac{6}{2}(15+90)-(35+70)-\frac{4}{2}(21+84)$

$=1683+1050+735-315-105-210$

$=2838$

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Question 6:
The maximum value of the sum of the AP $50,48,46,44,\dots$ is

[1] 650

[2] 450

[3] 558

[4] 648

Answer & Solution

Option # 1

For maximum value of the given sequence to n terms, when the nth term is either zero or the

smallest positive number of the sequence

ie., $50+(n-1)(-2)=0\Rightarrow n=26$

Therefore, $S_{26}=\frac{26}{2}(50+0)=26\times 25=650$

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Question 7:
If $a_i>0,i=1,2,3,\dots ,50$ and $a_1+a_2+a_3+\dots +a_{50}=50,$ then the minimum value of $\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\dots \frac{1}{a_{50}}$ is equal to

[1] 150

[2] 100

[3] 50

[4] 200

Answer & Solution

Option # 3

We know that in any series

$AM\ge HM$

So,

AM of first 50 terms $=\frac{a_1+a_2+a_3+\dots +a_{so}}{50}$

HM of the first 50 terms $=\frac{50}{(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\dots +\frac{1}{a_{50}})}$

Therefore, $\frac{a_1+a_2+a_3+\dots +a_{so}}{50}\ge \frac{50}{(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\dots +\frac{1}{a_{50}})}$

$\Rightarrow \frac{50}{50}\ge \frac{50}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\dots +\frac{1}{a_{50}}}$

$\Rightarrow \frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\dots +\frac{1}{a_{50}}\ge 50$

Hence, minimum value of $\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\dots +\frac{1}{a_{50}}$ is 50

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Question 8:
The sum to infinity of the series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\dots \dots \dots \dots$ is

[1] 6 

[2] 2

[3] 3

[4] 4

Answer & Solution

Option # 3

$S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\dots \dots .$ (i)

$\frac{S}{3}=1+\frac{1}{3}+\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+\dots \dots .$ (ii)

Subtracting ( ii ) from (i) we get, $\frac{2}{3}=1+\frac{1}{3}+\frac{4}{3^2}+\frac{4}{3^3}\dots$

$=\frac{4}{3}+\frac{4}{3^2}\left(\frac{1}{1-\frac{1}{3}}\right)=\frac{4}{3}+\frac{4}{3^2}\times \frac{3}{2}=2$

$S=3$

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Question 9:
If $x^a=y^b=z^c,$ where $a,b,c$ are unequal positive numbers and $x,y,z$ are in GP, then $a^3+c^3$ is

[1] $>2b^3$

[2] $>2c^3$

[3]  $<2b^3$

[4] $<2c^3$

Answer & Solution

Option # 1

Since $x^a=y^b=z^c=\lambda$ (say)

Therefore, $x={\lambda }^{1/a},y={\lambda }^{1/b},z={\lambda }^{1/c}$

Now, because x , y , z are in GP

So, $y^2=zx$

$\Rightarrow {\lambda }^{2/b}={\lambda }^{1/c}\cdot {\lambda }^{1/a}$

$\Rightarrow {\lambda }^{2/b}={\lambda }^{1/c}+1/a$

$\Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

Therefore, a, b, and c are in HP

Now, GM >HM

$\Rightarrow \sqrt{ac}>b...\left(i\right)$

Now, for three numbers $a^3,b^3,c^3$

$AM>GM$

$\Rightarrow \frac{a^3+c^3}{2}>(\sqrt{ac}{)}^3>b^3$

Or, $a^3+c^3>2b^3$

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Question 10:
The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is

[1] –4

[2] –12

[3] 12

[4] 4

Answer & Solution

Option # 2

Let a , ar, $a{r}^2,\dots$

$a+ar=12\cdots \left(i\right)$

$a{r}^2+a{r}^3=48\cdots \left(ii\right)$

dividing (ii) by (i), we have

$\frac{{ar}^2(1+r)}{a(r+1)}=4$

$\Rightarrow r^2=4$ if $r\ne -1$

Therefore, r=-2 and a=-12.