Tag: sum of n terms of an gp

Questions Related to sum of n terms of an gp

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. $3$

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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

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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. $i-1$

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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 $9$.

The sum of $2n$ terms of a series of which every even term is $'a'$ times the terms before it, and every odd term $'c'$ times the terms before it, the first term being unity, is

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

  1. $\dfrac { \left( 1-a \right) \left( { a }^{ n }{ c }^{ n }-1 \right) }{ ac-1 }$

  2. $\dfrac { \left( 1+a \right) \left( { a }^{ n }{ c }^{ n }-1 \right) }{ ac+1 }$

  3. $\dfrac { \left( 1+a \right) \left( { a }^{ n }{ c }^{ n }-1 \right) }{ ac-1 }$

  4. $None\ of\ these$

Show answer
Correct Option: C
Explanation:

$\begin{array}{l} { T _{ 1 } }=1 \ { T _{ 2 } }=a \ { T _{ 3 } }=Ca \ { T _{ 4 } }=C{ a^{ 2 } } \ { T _{ 2n } }=a\frac { { 2n } }{ 2 } \cdot C\frac { { 2n } }{ 2 } -1={ a^{ n } }{ C^{ n-1 } } \ { 5 _{ 2n } }=1+a+ca.....{ a^{ n } }{ C^{ n-1 } } \ =1+\left[ { a+c{ a^{ 2 } }+{ c^{ 2 } }{ a^{ 3 } }...{ a^{ n } }{ c^{ n-1 } } } \right]  \ +\left[ { ca+{ c^{ 2 } }{ a^{ 2 } }+{ c^{ 3 } }{ a^{ 3 } }.....{ c^{ n-1 } }{ a^{ n-1 } } } \right]  \ =1+\frac { { a\left( { { a^{ n } }{ c^{ n-1 } } } \right)  } }{ { ac-1 } } +\frac { { ac\left( { { a^{ n-1 } }{ c^{ n-1 } }-1 } \right)  } }{ { ac-1 } }  \ =\frac { { \left( { { a^{ n } }{ c^{ n } }-1 } \right) \left( { a+1 } \right)  } }{ { ac-1 } }  \end{array}$

The sum of $10$ terms of the series $0.7 + .77 + .777 + \ldots \ldots \ldots$ is

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

  1. $\dfrac { 7 } { 9 } \left( 89 + \dfrac { 1 } { 10 ^ { 10 } } \right)$

  2. $\dfrac { 7 } { 81 } \left( 89 + \dfrac { 1 } { 10 ^ { 10 } } \right)$

  3. $\dfrac { 7 } { 81 } \left( 89 + \dfrac { 1 } { 10 ^ { 9 } } \right)$

  4. $\dfrac { 7 } { 9 } \left( 89 + \dfrac { 1 } { 10 ^ { 9 } } \right)$

Show answer
Correct Option: B
Explanation:
$0.7+0.77+0.777+......$
$=7\left( 0.1+0.11+0.111+...... \right) $
$=\dfrac { 7 }{ 9 } \left( 0.9+0.99+0.999+...... \right) $
$=\dfrac { 7 }{ 9 } \left( 1-0.1+1-0.1+1-0.001+...... \right) $
$=\dfrac { 7 }{ 9 } \left( 10-\left( 0.1+0.01+0.001+...... \right)  \right) $
$=\dfrac { 7 }{ 9 } \left( 10-\dfrac { 0.1\left( 1-{ 10 }^{ -10 } \right)  }{ 1-0.1 }  \right) =\frac { 7 }{ 9 } \left( 10-\dfrac { 0.1\left( { 10 }^{ 10 }-1 \right)  }{ 0.9\times { 10 }^{ 10 } }  \right) =\dfrac { 7 }{ 9 } \left( 10-\dfrac { 1 }{ 9 } +\dfrac { 1 }{ 9\times { 10 }^{ 10 } }  \right) $
$=\dfrac { 7 }{ 9 } \left( \dfrac { 89 }{ 9 } +\dfrac { 1 }{ 9\times { 10 }^{ 10 } }  \right) $
$=\dfrac { 7 }{ 81 } \left( 89+\dfrac { 1 }{ { 10 }^{ 10 } }  \right) $      [B]