Let $a _ { 1 } , a _ { 3 } , \ldots$ 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
If $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots .$ are in harmonic progression, then $\frac { 1 } { a _ { 1 } } , \frac { 1 } { a _ { 2 } } , \frac { 1 } { a _ { 3 } } \ldots \ldots$ are in AP.
First term of $\mathrm { AP } , \frac { 1 } { a _ { 1 } } = \frac { 1 } { 5 }$
20th term of $\mathrm { AP } , \frac { 1 } { a _ { 20 } } = \frac { 1 } { 25 }$
$\Rightarrow \frac { 1 } { 5 } + 19 d = \frac { 1 } { 25 }$
$\Rightarrow d = \frac { - 4 } { 19 \times 25 }$
We have to find the least positive integer n for which $a _ { n } < 0$
$\Rightarrow \frac { 1 } { 5 } + ( n - 1 ) d < 0$
$\Rightarrow \frac { 1 } { 5 } + ( n - 1 ) d < 0$
$\Rightarrow \frac { 1 } { 5 } + ( n - 1 ) > \frac { 95 } { 4 }$
$\Rightarrow n > 24.75$
The sum of the first 20 term of the sequence $0.7,0.77,0.777 , \ldots ,$ is
[1] $\frac { 7 } { 81 } \left( 179 - 10 ^ { - 20 } \right)$
[2] $\frac { 7 } { 9 } \left( 99 - 10 ^ { - 20 } \right)$
[3] $\frac { 7 } { 81 } \left( 179 + 10 ^ { - 20 } \right)$
[4] $\frac { 7 } { 9 } \left( 99 + 10 ^ { - 20 } \right)$
We have,
$0.7 + 0.77 + 0.777 + \ldots$ upto 20 terms
$\Rightarrow 7\text{ }\left( \frac{1}{10}+\frac{11}{{{10}^{2}}}+\frac{111}{{{10}^{3}}}+\ldots upto\ 20\ terms \right)$
$\Rightarrow \frac{7}{9}\text{ }\left( \frac{9}{10}+\frac{99}{{{10}^{2}}}+\frac{999}{{{10}^{3}}}+\ldots upto\ 20\ terms \right)$
$\Rightarrow \frac{7}{9}\text{ }\left( \left( 1-\frac{1}{10} \right)+\left( 1-\frac{1}{100} \right)+\left( 1-\frac{1}{1000} \right)+\ldots upto\ 20\ terms \right)$ $\Rightarrow \frac{7}{9}\text{ }\left( 20-\text{ }\left( \frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\ldots upto\ 20\ terms \right) \right)$
$\Rightarrow \frac{7}{9}\left( 20-\left( \frac{1}{10}\left( \frac{1-{{\left( \frac{1}{10} \right)}^{20}}}{1-\frac{1}{10}} \right) \right) \right)$
$\Rightarrow \frac { 7 } { 9 } \left( 20 - \frac { 1 } { 9 } \left( 1 - \left( \frac { 1 } { 10 } \right) ^ { 20 } \right) \right)$
$\Rightarrow \frac { 7 } { 81 } \left( 179 + \left( \frac { 1 } { 10 } \right) ^ { 20 } \right)$
If $( 10 ) ^ { 9 } + 2 ( 11 ) ^ { 1 } ( 10 ) ^ { 8 } + 3 ( 11 ) ^ { 2 } ( 10 ) ^ { 7 } + \ldots + 10 ( 11 ) ^ { 9 } = k ( 10 ) ^ { 9 }$, then k is equal to
[1] $\frac { 121 } { 10 }$
[2] $\frac { 441 } { 100 }$
[3] 100
[4] 110
We have,
$( 10 ) ^ { 9 } + 2 ( 11 ) ^ { 1 } ( 10 ) ^ { 5 } + 3 ( 11 ) ^ { 2 } ( 10 ) ^ { 7 } + \ldots + 10 ( 11 ) ^ { 9 } = k ( 10 ) ^ { 9 }$
Dividing by $( 10 ) ^ { 9 }$ on both sides, we get
$1 + 2 \left( \frac { 11 } { 10 } \right) + 3 \left( \frac { 11 } { 10 } \right) ^ { 2 } + \ldots + 10 \left( \frac { 11 } { 10 } \right) ^ { 9 } = k \cdots(i)$
Multiplying $\left( \frac { 11 } { 10 } \right)$ on both side of above equation we get,
$\left( \frac { 11 } { 10 } \right) + 2 \left( \frac { 11 } { 10 } \right) ^ { 2 } + 3 \left( \frac { 11 } { 10 } \right) ^ { 3 } + \ldots + 10 \left( \frac { 11 } { 10 } \right) ^ { 10 } = \left( \frac { 11 } { 10 } \right) k \quad \ldots$ (ii)
Subtracting (ii) from (i), we get
$k - \left( \frac { 11 } { 10 } \right) k = 1 + \left( \frac { 11 } { 10 } \right) + \left( \frac { 11 } { 10 } \right) ^ { 2 } + \ldots + \left( \frac { 11 } { 10 } \right) ^ { 9 } - 10 \left( \frac { 11 } { 10 } \right) ^ { 10 }$
$\Rightarrow k \left( 1 - \left( \frac { 11 } { 10 } \right) \right) = \frac { 1 \left( \left( \frac { 11 } { 10 } \right) ^ { 10 } - 1 \right) } { \left( \frac { 11 } { 10 } \right) - 1 } - 10 \left( \frac { 11 } { 10 } \right) ^ { 10 }$
$\Rightarrow k \left( 1 - \left( \frac { 11 } { 10 } \right) \right) = - 10$
$\Rightarrow k = 100$
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] $2 + \sqrt { 3 }$
Let $a , a r , a r ^ { 2 }$ are in GP.
If the middle term of the GP is doubled, then new numbers are in AP.
$\Rightarrow a , 2 a r , a r ^ { 2 }$ are in AP.
$\Rightarrow 4 a r = a + a r ^ { 2 }$
$\Rightarrow r ^ { 2 } - 4 r + 1 = 0$
$\Rightarrow r = 2 \pm \sqrt { 3 }$
since the series is an increasing GP so,
$\Rightarrow r = 2 + \sqrt { 3 }$
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
Let $\mathrm { S } _ { 3 } , \mathrm { S } _ { 5 } , \mathrm { S } _ { 7 } , \mathrm { S } _ { 15 } , \mathrm { S } _ { 35 } , \mathrm { S } _ { 21 }$ be the sum of integers divisible by $3,5,7,15,35$ and 21 respectively.
Here,
$S _ { 3 } = 3 + 6 + 9 + \ldots + 99$
$S _ { 5 } = 5 + 10 + 15 + \ldots + 100$
$S _ { 7 } = 7 + 14 + 21 + \ldots + 98$
$S _ { 15 } = 15 + 30 + 45 + \ldots + 90$
$S _ { 15 } = 35 + 70$
$S _ { 31 } = 21 + 42 + 63 + 84$
So, the required sum is:
$= S _ { 3 } + S _ { 5 } + S _ { 7 } - S _ { 15 } - S _ { 35 } - S _ { 21 }$
$= \frac { 33 } { 2 } ( 3 + 99 ) + \frac { 20 } { 2 } ( 5 + 100 ) + \frac { 14 } { 2 } ( 7 + 98 ) - \frac { 6 } { 2 } ( 15 + 90 ) - ( 35 + 70 ) - \frac { 4 } { 2 } ( 21 + 84 )$
$= 1683 + 1050 + 735 - 315 - 105 - 210$
$= 2838$
The maximum value of the sum of the AP $50,48,46,44 , \ldots$ is
[1] 650
[2] 450
[3] 558
[4] 648
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., $\quad 50 + ( n - 1 ) ( - 2 ) = 0 \Rightarrow \quad n = 26$
Therefore, $S _ { 26 } = \frac { 26 } { 2 } ( 50 + 0 ) = 26 \times 25 = 650$
If $a _ { i } > 0 , i = 1,2,3 , \ldots , 50$ and $a _ { 1 } + a _ { 2 } + a _ { 3 } + \ldots + a _ { 50 } = 50 ,$ then the minimum value of $\frac { 1 } { a _ { 1 } } + \frac { 1 } { a _ { 2 } } + \frac { 1 } { a _ { 3 } } + \ldots \frac { 1 } { a _ { 50 } }$ is equal to
[1] 150
[2] 100
[3] 50
[4] 200
We know that in any series
$A M \geq H M$
So,
AM of first 50 terms $= \frac { a _ { 1 } + a _ { 2 } + a _ { 3 } + \ldots + a _ { \mathrm { so } } } { 50 }$
HM of the first 50 terms $=\frac{50}{\left( \frac{1}{{{a}_{1}}}+\frac{1}{{{a}_{2}}}+\frac{1}{{{a}_{3}}}+\ldots +\frac{1}{{{a}_{50}}} \right)}$
Therefore, $\frac { a _ { 1 } + a _ { 2 } + a _ { 3 } + \ldots + a _ { \mathrm { so } } } { 50 } \geq \frac { 50 } { \left( \frac { 1 } { a _ { 1 } } + \frac { 1 } { a _ { 2 } } + \frac { 1 } { a _ { 3 } } + \ldots + \frac { 1 } { a _ { \mathrm { 50 } } } \right) }$
$\Rightarrow \frac{50}{50}\ge \frac{50}{\frac{1}{{{a}_{1}}}+\frac{1}{{{a}_{2}}}+\frac{1}{{{a}_{3}}}+\ldots +\frac{1}{{{a}_{50}}}}$
$\Rightarrow \quad \frac{1}{{{a}_{1}}}+\frac{1}{{{a}_{2}}}+\frac{1}{{{a}_{3}}}+\ldots +\frac{1}{{{a}_{50}}}\ge 50$
Hence, minimum value of $\frac{1}{{{a}_{1}}}+\frac{1}{{{a}_{2}}}+\frac{1}{{{a}_{3}}}+\ldots +\frac{1}{{{a}_{50}}}$ is 50
The sum to infinity of the series $1 + \frac { 2 } { 3 } + \frac { 6 } { 3 ^ { 2 } } + \frac { 10 } { 3 ^ { 3 } } + \frac { 14 } { 3 ^ { 4 } } + \ldots \ldots \ldots \quad \ldots$ is
[1] 6
[2] 2
[3] 3
[4] 4
$S = 1 + \frac { 2 } { 3 } + \frac { 6 } { 3 ^ { 2 } } + \frac { 10 } { 3 ^ { 3 } } + \frac { 14 } { 3 ^ { 4 } } + \ldots \ldots .$ (i)
$\frac { S } { 3 } = 1 + \frac { 1 } { 3 } + \frac { 2 } { 3 ^ { 2 } } + \frac { 6 } { 3 ^ { 3 } } + \frac { 10 } { 3 ^ { 4 } } + \ldots \ldots .$ (ii)
Subtracting ( ii ) from (i) we get, $\frac { 2 } { 3 } = 1 + \frac { 1 } { 3 } + \frac { 4 } { 3 ^ { 2 } } + \frac { 4 } { 3 ^ { 3 } } \ldots$
$= \frac { 4 } { 3 } + \frac { 4 } { 3 ^ { 2 } } \left( \frac { 1 } { 1 - \frac { 1 } { 3 } } \right) = \frac { 4 } { 3 } + \frac { 4 } { 3 ^ { 2 } } \times \frac { 3 } { 2 } = 2$
$S = 3$
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] $> 2 b ^ { 3 }$
[2] $>2{{c}^{3}}$
[3] $< 2 b ^ { 3 }$
[4] $< 2 c ^ { 3 }$
Since $\quad x ^ { a } = y ^ { b } = z ^ { c } = \lambda$ (say)
Therefore, $\quad x = \lambda ^ { 1 / a } , y = \lambda ^ { 1 / b } , z = \lambda ^ { 1 / c }$
Now, because x , y , z are in GP
So, $\quad y ^ { 2 } = z x$
$\Rightarrow \lambda ^ { 2 / b } = \lambda ^ { 1 / c } \cdot \lambda ^ { 1 / a }$
$\Rightarrow {{\lambda }^{2/b}}={{\lambda }^{1/c}}+1/a$
$\Rightarrow \quad \frac { 2 } { b } = \frac { 1 } { a } + \frac { 1 } { c }$
Therefore, a, b, and c are in HP
Now, GM >HM
$\Rightarrow \sqrt{ac}>b...(i)$
Now, for three numbers $a ^ { 3 } , b ^ { 3 } , c ^ { 3 }$
$A M > G M$
$\Rightarrow \quad \frac { a ^ { 3 } + c ^ { 3 } } { 2 } > ( \sqrt { a c } ) ^ { 3 } > b ^ { 3 }$
Or, $a ^ { 3 } + c ^ { 3 } > 2 b ^ { 3 }$
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
Let a , ar, $a r ^ { 2 } , \ldots$
$a + a r = 12 \cdots (i)$
$a{{r}^{2}}+a{{r}^{3}}=48\cdots (ii)$
dividing (ii) by (i), we have
$\frac { \mathrm { ar } ^ { 2 } ( 1 + \mathrm { r } ) } { \mathrm { a } ( \mathrm { r } + 1 ) } = 4$
$\Rightarrow r ^ { 2 } = 4$ if $r \neq - 1$
Therefore, r=-2 and a=-12.
