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

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?

  1. $4.10$
  2. $5.12$
  3. $29.52$
  4. $43.92$
  5. $49.04$
Reveal answer Fill a bubble to check yourself
E Correct answer
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 $2 \times 4/5 \times 4/5 \times 4/5 \times 10 +38.8 = 256/25 + 38.8 =49.04$

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

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

  1. $83667$
  2. $54954$
  3. $99994$
  4. $79894$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Odd integers divisible by 3 are: $3,9,15,21.....999$
Let $a=3$   and   $d=6$
$\therefore T _n=999=3+(n-1)6$
$\Rightarrow n-1=\dfrac{996}{6}=166$
$\therefore n=167$
$\therefore S _n=\dfrac{167}{2}[2\times 3+(167-1)6]=83667$
Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

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

  1. $\cfrac { 89 }{ 128 } $
  2. $\cfrac { 57 }{ 384 } $
  3. $\cfrac { 85 }{ 128 } $

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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

  1. $8$
  2. $9$
  3. $10$
  4. $11$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $i$
  2. $-i$
  3. $0$
  4. $i-1$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $= a$
$S(1-a)=\dfrac{1(a^{n}-1)}{a-1}-na^{n}$
$S=\dfrac{1-(a^{n})}{(a-1)^{2}}-\dfrac{na^{n}}{1-a}$

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

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: 

  1. only 1

  2. 1 and 8

  3. only 8

  4. 8 and 49

  5. 1,8, and 49
Reveal answer Fill a bubble to check yourself
E Correct answer
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.
Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

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

  1. $3$ or $1\dfrac {1}{2}$
  2. $1$
  3. $2\dfrac {1}{2}$
  4. $6$
  5. $9$ or $3$
Reveal answer Fill a bubble to check yourself
E Correct answer
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 $9$.