Tag: maths

Questions Related to maths

A rubber ball is dropped from a height of $10$ meters. If the ball always rebounds $\dfrac {4}{5}$ the distance it has fallen, calculate, how far, in meters, will the ball have travelled at the moment it hits the ground for the fourth time?

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $4.10$

  2. $5.12$

  3. $29.52$

  4. $43.92$

  5. $49.04$

Show answer
Correct Option: E
Explanation:
  • for the first time it hits the ground , it travels $10$ m
  • for the second time it hits the ground , it travels $2 \times 4/5 \times 10 = 16+10 =26$
  • for the third time it hits the ground , it travels $ 2\times 4/5 \times 4/5 \times 10 +26 =64/5+26 = 38.8$
  • for the fourth time it hits the ground , it travels $2 \times 4/5 \times 4/5 \times 4/5 \times 10 +38.8 = 256/25 + 38.8 =49.04$

Find the sum of odd integers between $1$ and $1000$ which are divisible by $3$.

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $83667$

  2. $54954$

  3. $99994$

  4. $79894$

Show answer
Correct Option: A
Explanation:
Odd integers divisible by 3 are: $3,9,15,21.....999$
Let $a=3$   and   $d=6$
$\therefore T _n=999=3+(n-1)6$
$\Rightarrow n-1=\dfrac{996}{6}=166$
$\therefore n=167$
$\therefore S _n=\dfrac{167}{2}[2\times 3+(167-1)6]=83667$

How many terms of the series $1+3+9+ ...$sum to $121$?

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $5$

  2. $6$

  3. $4$

  4. $3$

Show answer
Correct Option: A
Explanation:

The given series is a G.P
Sum of n terms of a G.P is $ a* {(\frac{(r^n\;-\;1)}{(r-1)})} $
Here a=1 and r=3
Substituting it in the equation and finding n we get n=5

What is the sum of first eight terms of the series $1-\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } -\cfrac { 1 }{ 8 } +.....$?

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $\cfrac { 89 }{ 128 } $

  2. $\cfrac { 57 }{ 384 } $

  3. $\cfrac { 85 }{ 128 } $

  4. None of the above

Show answer
Correct Option: C
Explanation:

Given series is a sum of terms of GP with common ratio $-\dfrac{1}{2}$
Sum of $n$ terms of GP with common ratio $r$ and first term $a$ is $\dfrac { a(1-r^{ n }) }{ 1-r } $
Putting $a=1,n=8$ and $r=-\dfrac { 1 }{ 2 } $ in above equation, we have 

Sum $=\dfrac { 1(1-(-\frac { 1 }{ 2 } )^{ 8 }) }{ 1-(-\frac { 1 }{ 2 } ) } =\dfrac { 1-\frac { 1 }{ 256 }  }{ \frac { 3 }{ 2 }  } =\dfrac { 255 }{ 128\times 3 } =\dfrac { 85}{128} $
Hence, option C is correct

What is the greatest value of the positive integer n satisfying the condition $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +  ...... + \dfrac{1}{2^{n - 1}} < 2 - \dfrac{1}{1000}$?

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $8$

  2. $9$

  3. $10$

  4. $11$

Show answer
Correct Option: C
Explanation:

Given : $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+......+\dfrac{1}{2^{n−1}} < 2−\dfrac{1}{1000}$

Left side forms a sum of a finite geometric series with first term $1$ and common ratio $\dfrac{1}{2}$ with $n$ terms.
 Sum $ =\dfrac{a(1-r^{ n })}{(1-r)} = \dfrac{1(1-(0.5)^{ n })}{0.5} = 2-{ 2 }^{ 1-n }$
 So, $2-{ 2 }^{ 1-n } < 2-\dfrac { 1 }{ 1000 }$  
We have,
${ 2 }^{ n-1 } < 1000$ we get the max value of $n = 10$.
Hence, C is correct.

The value of the sum $\sum _{ n=1 }^{ 13 }{ \left( { i }^{ n }+{ i }^{ n+1 } \right)  } $ where $i=\sqrt { -1 } $ is:

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $i$

  2. $-i$

  3. $0$

  4. $i-1$

Show answer
Correct Option: D
Explanation:

$\displaystyle \sum _{ n=1 }^{ 13 }{ ({ i }^{ n }+{ i }^{ n+1 }) }$ 

$=\displaystyle (i+1)\sum _{ n=1 }^{ 13 }{ { i }^{ n } } $
$=(i+1)\dfrac { (i({ i }^{ 13 }-1)) }{ i-1 } $
$=(i+1)\dfrac { (i(i-1)) }{ i-1 } $
$=(i+1)(i)$
$=i-1$
Hence, the correct answer is D.

Sum $1 + 2a + 3a^{2} + 4a^{3} + ....$ to $n$ terms.

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $\dfrac{1+(a^{n})}{(a-1)^{2}}-\dfrac{na^{n}}{1+a}$

  2. $\dfrac{1-2(a^{n})}{(a-1)^{2}}+\dfrac{na^{n}}{1-2a}$

  3. $\dfrac{1-(a^{n})}{(a-1)^{2}}-\dfrac{na^{n}}{1-a}$

  4. none of these

Show answer
Correct Option: C
Explanation:

Let $S=1+2a+3a^{2}+4a^{3}+$   ..........$+na^{n-1}$
Multiply both sides by $a$, we get
$Sa=0+a+2a^{2}+3a^{3}$   .............$(n-1)a^{n-1}+na^{n}$
Subtract both equations,
$S(1-a)=1+a+a^{2}+a^{3}$    .............$a^{n-1}-na^{n}$
Clearly above series is G.P
Common ratio $= a$
$S(1-a)=\dfrac{1(a^{n}-1)}{a-1}-na^{n}$
$S=\dfrac{1-(a^{n})}{(a-1)^{2}}-\dfrac{na^{n}}{1-a}$

The geometric mean if the series $1, 2, 4,...., 2^n$, is

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $2^{n + (1/2)}$

  2. $2^{(n + 1)/2}$

  3. $2^{n - (1/2)}$

  4. $2^{n/2}$

Show answer
Correct Option: D
Explanation:

$1, 2, 4, ....., 2^n$


No. of terms $=n+1$

$\therefore G.M.= (1.2.4......2^n)^{\dfrac{1}{n+1}}$
$= (2^{1+2+3+...+n})^{\dfrac{1}{n+1}}$

$= (2)^{\dfrac{n(n+1)}{2}.\dfrac{1}{n+1}}$

$=2^{n/2}$

If the sum $1+2+3 +....+ K$ is a perfect square N$^{2}$ and if N is less than 100, then the possible values for K are: 

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. only 1

  2. 1 and 8

  3. only 8

  4. 8 and 49

  5. 1,8, and 49

Show answer
Correct Option: E
Explanation:
S= K(K + l)/2 = $N^{2}$. The possible values for N$^{2}$ are $1^{2}, 2^{2}, 3^{2}.... 99^{2}$;
For K to be integral, the discriminant 1 + 8N$^{2}$ of the equation $K^{2} + K- 2N$= 0 must be a perfect square. This fact reduces the possible values for $N^{2}$ to $1^{2}, 6^{2}, and 35^{2}$. Hence the values of K are 1, 8, and 49.
Note: There are ways of shortening the number of trials for N2 still further.but these involve a knowledge of number-theoretic theorems. The shortest way to do this problem is by testing the choices given.

The sum to infinity of the terms of an infinite geometric progression is $6$. The sum of the first two terms is $4\dfrac {1}{2}$. The first term of the progression is

maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

  1. $3$ or $1\dfrac {1}{2}$

  2. $1$

  3. $2\dfrac {1}{2}$

  4. $6$

  5. $9$ or $3$

Show answer
Correct Option: E
Explanation:

Let the first two terms be $a$ and $ar$ where $-1 < r < 1$.
$\therefore a(1 + r) = 4\dfrac {1}{2}$
Since $s = a/(1 - r) = 6, a = 6(1 - r)$
$\therefore 6(1 - r)(1 + r) = 4\dfrac {1}{2}; \therefore r = \pm \dfrac {1}{2}; \therefore a = 3$ or $9$.