Tag: summing geometric series

Questions Related to summing geometric series

Given $A=2^{65}$ and $B=(2^{64}+2^{63}+2^{62}+....+2^0)$

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

  1. B is $2^{64}$ larger than A

  2. A and B are equal

  3. B is larger than A by $1$

  4. A is larger than B by $1$

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Correct Option: D
Explanation:

B is in G.P. with $a=2^0, r=2, n=65$
$\therefore S _n=\frac {a(r^n-1)}{r-1}=\frac {2^0(2^{65}-1)}{2-1}$
$\therefore B=2^{65}-1$
$\Rightarrow B=A-1$
$\therefore$ A is larger than B by $1$.

The sum of the geometric sequence is given as $S=\cfrac{a(1-r^n)}{1-r}$, where $r$ is the

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

  1. constant

  2. term

  3. common difference

  4. common ratio

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Correct Option: D
Explanation:
We know $S=\dfrac {a(1-r^n)}{1-r}$
In a geometric sequence, the ratio of terms is constant and is known as the common ratio $(r)$.
So, in the formula for summation, $r$ is common ratio.

If $a _1,\, a _2,\, a _3,\dots,a _n$ are in geometric progression. Then the given geometric progression is a

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

  1. finite geometric progression

  2. finite harmonic progression

  3. infinite geometric progression

  4. finite arithmetic progression

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Correct Option: A
Explanation:
Given sequence is $a _1,a _2,a _3,....a _n$
Since the last term is given, so the given progression is finite and it is given that progression is geometric progression.
So, the given progression is a finite geometric progression.

$x, 2x, 4x, . . .$
The first term in the sequence above is $x$, and each term thereafter is equal to twice the previous term. Find the sum of the first five terms of this sequence.

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

  1. $10x$

  2. $15x$

  3. $30x$

  4. $31x$

  5. $32x$

Show answer
Correct Option: D
Explanation:

The first term in given Geometric series, $a _1=x$
The common ratio $r$ $=\dfrac{4x}{2x}=2$
No. of terms, $n$ $=5$
Applying sum of GP formula,
$S _n=\dfrac{a(1-r^n)}{1-r}$
      $=\dfrac{x(1-2^5)}{1-2}$
      $=\dfrac{x(1-32)}{-1}=31x$
Hence,option D is correct.

Find the sum of the following G.P. to $n$ terms $0.5 + 0.55 + 0.555 + 0.5555 + .....$

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

  1. $\dfrac {5}{9}\left[9n-1+\dfrac {1}{10^n}\right]$

  2. $\dfrac {5}{81}\left[5n-1-\dfrac {1}{10^n}\right]$

  3. $\dfrac {5}{81}\left[9n-1+\dfrac {1}{10^n}\right]$

  4. $-\dfrac {5}{9}\left[9n-1+\dfrac {1}{10^n}\right]$

Show answer
Correct Option: C

Let $n > 1$ be the positive integer. The largest positive integer $m$, such that $n^m + 1$ divides $1 + n + n^2 ..... n^{125}$ is

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

  1. $60$

  2. $62$

  3. $63$

  4. $64$

Show answer
Correct Option: C
Explanation:

We have,

 $1 + n + n^2 ..... n^{125}$
$=\dfrac{n^{126}-1}{n-1}$
$=\dfrac{(n^{63})^2-1}{n-1}$
$=\dfrac{(n^{63}-1)(n^{63}+1)}{n-1}$
This will be divisible by $n^m+1$ for largest $m$ when $m=63$.

The sum of the first three terms of an increasing G.P. is $13$ and their product is $27$. The sum of the first $5$ terms is,

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

  1. $323$

  2. $363$

  3. $109$

  4. $254$

Show answer
Correct Option: B
Explanation:

Let the G.P be $\dfrac{a}{r}, a, ar$.


So,
$\dfrac{a}{r}+ a+ ar=13$

And
$\dfrac{a}{r}\times  a\times  ar=27$

$a^3=27$
$a=3$

Therefore,
$\dfrac{3}{r}+ 3+ 3r=13$

$3+ 3r+ 3r^2=13r$

$3r^2-10r+3=0$

$3r^2-9r-r+3=0$

$3r(r-3)-1(r-3)=0$

$(3r-1)(r-3)=0$

$r=\dfrac{1}{3}, 3$

So, $r=3$

So, the G.P is $1, 3, 9$.

Now, the sum of first five terms
$=\dfrac{3(3^5-1)}{3-1}$

$=\dfrac{3(243-1)}{2}$

$=\dfrac{3(242)}{2}$

$=3(121)=363$

Hence, this is the answer.

If $i^{2}=-1$, then sum $i+i^{2}+i^{3}+.......$ to $1000$ terms is equal to

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

  1. $1$

  2. $-1$

  3. $i$

  4. $0$

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Correct Option: D

The sum of sequence $0.15,0.015,0.0015,.....$ upto 20 term is ?

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

  1. $\dfrac{1}{6}[1-(0.1)^{20}]$

  2. $\dfrac{1}{6}[1+(0.1)^{20}]$

  3. $\dfrac{1}{3}[1-(0.1)^{20}]$

  4. None of these

Show answer
Correct Option: A
Explanation:

Since the sequence in Geometric progression 

where, $a=0.15;\ r=\dfrac { 0.015 }{ 0.15 } =0.1;\ n=20\ \therefore { S } _{ n }=\dfrac { a({ r }^{ n }-1) }{ (r-1) } \ =\dfrac { 0.15({ \left( 0.1 \right)  }^{ 20 }-1) }{ 0.1-1 } \ =\dfrac { 1 }{ 6 } \left( 1-{ (0.1) }^{ 20 } \right) $

The sum of $10$ terms of GP $\frac { 1 } { 2 } + \frac { 1 } { 4 } + \frac { 1 } { 8 } + \ldots$ is-

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

  1. $\frac { 2 ^ { 10 } - 1 } { 2 ^ { 10 } }$

  2. $\frac { 2 ^ { 9 } - 1 } { 2 ^ { 9 } }$

  3. $\frac { 2 ^ { 10 } - 1 } { 2 ^ { 9 } }$

  4. $\frac { 2 ^ { 9 } - 1 } { 2 ^ { 10 } }$

Show answer
Correct Option: A
Explanation:
$S=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...$ is in G.P
where $a=\dfrac{1}{2}$ and $r=\dfrac{\dfrac{1}{4}}{\dfrac{1}{2}}=\dfrac{1}{2}$
${S} _{n}=\dfrac{a\left({r}^{n}-1\right)}{r-1}$
${S} _{10}=\dfrac{\dfrac{1}{2}\left({\left(\dfrac{1}{2}\right)}^{10}-1\right)}{\dfrac{1}{2}-1}$
${S} _{10}=\dfrac{\dfrac{1}{2}\left({\left(\dfrac{1}{2}\right)}^{10}-1\right)}{\dfrac{-1}{2}}$

$=1-\dfrac{1}{{2}^{10}}$
$=\dfrac{{2}^{10}-1}{{2}^{10}}$