Tag: summing geometric series

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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

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

  1. $10,\ 15,\ 20$
  2. $8,\ 15,\ 22$
  3. $5,\ 15,\ 25$
  4. $12,\ 15,\ 18$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $10,15,20$

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

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 

  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. $S\left[1-\left(1-\dfrac{S}{a}\right)\right]^{n}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an infinite G.P., S = a / (1-r), so r = 1 - a/S. The sum of the first n terms is S_n = a(1-r^n) / (1-r). Substituting r = 1 - a/S and 1-r = a/S, we get S_n = a(1 - (1-a/S)^n) / (a/S) = S(1 - (1-a/S)^n).

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

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?

  1. 2184

  2. 1358

  3. 1456

  4. 728
Reveal answer Fill a bubble to check yourself
D Correct answer
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

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

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

  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. $\displaystyle x^{2}\frac{(1+x^{2n})}{1+x^{2}}+xy\frac{(1-x^{n}y^{n})}{1-xy}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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 $r=s$

  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}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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}$

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

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

  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. $\displaystyle \frac{2^{n+1} - 1}{n}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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

  1. 6
  2. $\infty$

  3. 216

  4. 36
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $a = \dfrac{1}{2}, r = \dfrac{\dfrac{1}{4}}{\dfrac{1}{2}} = \dfrac{1}{2}$
$S _{\infty} = \dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}$
$S _{\infty} = \dfrac{\dfrac{1}{2}}{\dfrac{1}{2}} = 1$
$\therefore S _{\infty} = 6^1 = 6$

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

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

  1. $10$
  2. $14$
  3. $12$
  4. $22$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $(2)$
$\lambda =12$