Tag: summing geometric series

Questions Related to summing geometric series

The sum of first 4 term of GP with $a=2,r=3$ is 

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

  1. $80$

  2. $26$

  3. $127$

  4. $8$

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Correct Option: A
Explanation:

$a=2,r=3$

Sum of first $4$ terms is $\dfrac{a(r^n-1)}{r-1}\\dfrac{2(3^4-1)}2=81-1=80$

Three numbers whose sum is $45$ are in A.P. If $5$ is subtracted from the first number and $25$ is added to third number, the numbers are in G.P. Then numbers can be

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

  1. $10,\ 15,\ 20$

  2. $8,\ 15,\ 22$

  3. $5,\ 15,\ 25$

  4. $12,\ 15,\ 18$

Show answer
Correct Option: A
Explanation:

Let the numbers be $a-d,a,a+d$

Their sum is $45\a-d+d+a+d=45\3a=45\a=15$
The changed numbers are $15-d-5,15,15+d+25\10-d,15,40+d$
Condition to be in GP is $b^2=ac\15^2=(10-d)(40+d)\225=400-30d-d^2\d^2+30d-175=0\d^2+35d-5d-175=0\d(d+35)-5(d+35)=0\(d-5)(d+35)=0\d=5,-35$
$a=15,d=5$
So the series is $10,15,20$

If $S$ is the sum to infinity of a $G.P.$ whose first terms is $a$ then the sum of the first $n$ terms is 

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

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

  2. $S\left[1-\left(1-\dfrac{a}{S}\right)\right]^{n}$

  3. $a\left[1-\left(1-\dfrac{a}{S}\right)\right]^{n}$

  4. $S\left[1-\left(1-\dfrac{S}{a}\right)\right]^{n}$

Show answer
Correct Option: A

For first $n$ natural numbers we have the following results with usual notations $ \displaystyle \sum _{r=1}^{n}r =\frac{n(n+1)}{2}, \sum _{r=1}^{n}r^{2} =\frac{n(n+1)(2n+1)}{6},\sum _{r=1}^{n}r^{3}=\left ( \sum _{r=1}^{n}r \right )^{2}$ If $\displaystyle a _{1}a _{2}....a _{n} \in A.P $ then sum to $n$ terms of the sequence $\displaystyle \frac{1}{a _{1}a _{2}},\frac{1}{a _{2}a _{3}},...\frac{1}{a _{n-1}a _{n}}$ is equal to $\displaystyle \frac{n-1}{a _{1}a _{n}}$
 and the sum to $ n$ terms of a $G.P$ with first term '$a$' & common ratio '$r$' is given by  $\displaystyle S _{n}= \frac{lr-a}{r-1}$ for $ r \neq 1 $ for $ r =1 $ sum to $n$ terms of same $G.P.$ is $n$ $a$, where the sum to infinite terms of$G.P.$ is the limiting value of
 $\displaystyle \frac{lr-a}{r-1} $ when $\displaystyle n \rightarrow \infty ,\left |  r \right | < l $ where $l$ is the last term of $G.P.$  On the basis of above data answer the following questionsThe sum of the series $\displaystyle 2+6+18+...+486 $ equals?

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

  1. 2184

  2. 1358

  3. 1456

  4. 728

Show answer
Correct Option: D
Explanation:

Let ${ S } _{ n }=2+6+18+...+486$

$\Rightarrow { S } _{ n }=2\left( 1+{ 3+3 }^{ 2 }+...+{ 3 }^{ 5 } \right) $
$\Rightarrow { S } _{ n }=2\left( \dfrac { { 3 }^{ 6 }-1 }{ 3-1 }  \right) =729-1$     ...[ sum of G.P series ]

$\Rightarrow { S } _{ n }=728$

Ans: D

$\displaystyle x(x+y)+x^{2}(x^{2}+y^{2})+x^{3}(x^{3}+y^{3})+$.......to n terms.

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

  1. $\displaystyle x^{2}\frac{(1-x^{2n})}{1-x^{2}}+xy\frac{(1-x^{n}y^{n})}{1-xy}$

  2. $\displaystyle x^{2}\frac{(1+x^{2n})}{1-x^{2}}+xy\frac{(1-x^{n}y^{n})}{1-xy}$

  3. $\displaystyle x^{2}\frac{(1+x^{2n})}{1+x^{2}}+xy\frac{(1+x^{n}y^{n})}{1+xy}$

  4. $\displaystyle x^{2}\frac{(1+x^{2n})}{1+x^{2}}+xy\frac{(1-x^{n}y^{n})}{1-xy}$

Show answer
Correct Option: A
Explanation:

$\displaystyle x(x+y)+x^{2}(x^{2}+y^{2})+x^{3}(x^{3}+y^{3})+$.......to n terms.

$=\displaystyle \left(x^2+x^4+x^6+...n terms\right)+\left(xy+x^2y^2+... n terms\right)$

$=\displaystyle x^{2}\frac{(1-x^{2n})}{1-x^{2}}+xy\frac{(1-x^{n}y^{n})}{1-xy}$
Hence, option A

Find the value of the sum $\displaystyle \sum _{r=1}^{n}\,$ $\displaystyle \sum _{s=1}^{n}\, \delta _{rs}\, 2^r\, 3^s$ where $ \delta _{rs}$ is zero if $r \neq s$ & $\delta _{rs}$ is one if $r=s$

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

  1. $ \dfrac {6(6^n-1)}{5}$

  2. $ \dfrac {6(6^n+1)}{5}$

  3. $ \dfrac {5(6^n+1)}{6}$

  4. $ \dfrac {n(6^n-1)}{6}$

Show answer
Correct Option: A
Explanation:

Given  $\delta _{rs}\, =\, 0   if \quad r\neq s\quad \delta  _{rs}=1\quad if \quad r=s$

$\therefore \displaystyle \sum _{r=1}^{n}\, \sum _{s=1}^{n}\, \delta _{rs}\, 2^r\, 3^s$

$=\displaystyle

\sum _{r=1}^{n} 2^r3^r\, =\, \displaystyle \sum _{r=1}^{n} 6^r\, \,

=6+6^2\, +\, 6^3+..6^n\,=\, \displaystyle \frac {6(6^n-1)}{5}$

The A.M. of the series $1, 2, 4, 8, 16, ......, 2$$^n$ is

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

  1. $\displaystyle \frac{2^n - 1}{n}$

  2. $\displaystyle \frac{2^{n+1} - 1}{n + 1}$

  3. $\displaystyle \frac{2^n - 1}{n+1}$

  4. $\displaystyle \frac{2^{n+1} - 1}{n}$

Show answer
Correct Option: B
Explanation:

$\displaystyle A.M.=\frac { 1+2+4+8+...+{ 2 }^{ n } }{ n+1 } =\frac { { 2 }^{ n+1 }-1 }{ \left( n+1 \right) \left( 2-1 \right)  } =\frac { { 2 }^{ n+1 }-1 }{ n+1 } $

Ans: B

$\displaystyle 1+\frac{1}{4\times 3}+\frac{1}{4\times 3^{2}}+\frac{1}{4\times 3^{3}}$  is equal to 

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

  1. $1.120$

  2. $1.250$

  3. $1.140$

  4. $1.160$

Show answer
Correct Option: A
Explanation:

Given expression = $\displaystyle 1+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}=\frac{108+9+3+1}{108}=\frac{121}{108}=1.120\left ( approx \right )$

$6^{1/2}\, .\, 6^{1/4}\, .\, 6^{1/8}\, ..... \infty\, =\, ?$ 

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

  1. 6

  2. $\infty$

  3. 216

  4. 36

Show answer
Correct Option: A
Explanation:

Given $6^{\frac{1}{2}}.6^{\frac{1}{4}}.6^{\frac{1}{8}}....\infty$
Here power of 6 are in G.P
Sum of $\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8} ...\infty$

$S _{\infty} = \dfrac{a}{1-r}$
Here $a = \dfrac{1}{2}, r = \dfrac{\dfrac{1}{4}}{\dfrac{1}{2}} = \dfrac{1}{2}$
$S _{\infty} = \dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}$
$S _{\infty} = \dfrac{\dfrac{1}{2}}{\dfrac{1}{2}} = 1$
$\therefore S _{\infty} = 6^1 = 6$

In a geometric progression with common ratio 'q', the sum of the first 109 terms exceeds the sum of the first 100 terms by 12. If the sum of the first nine terms of the progression is $\displaystyle \frac {\lambda}{q^{100}}$ then the value of $ \lambda $ equals to

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

  1. $10$

  2. $14$

  3. $12$

  4. $22$

Show answer
Correct Option: C
Explanation:
$r=q$ (common ratio)
${ S } _{ n }=\cfrac { a({ r }^{ n }-1) }{ (r-1) } \\ { S } _{ 109 }={ S } _{ 100 }+12\\ \cfrac { a({ q }^{ 109 }-1) }{ (q-1) } =\cfrac { a({ q }^{ 100 }-1) }{ (q-1) } +12\quad \quad (1)\\ \cfrac { a({ q }^{ 9 }-1) }{ (q-1) } =\cfrac { \lambda  }{ { q }^{ 100 } } \\ \lambda =\cfrac { a({ q }^{ 109 }-{ q }^{ 100 }) }{ (q-1) } \quad \quad \quad (2)$
From $(1)$ and $(2)$
$\lambda =12$