Tag: maths

Questions Related to maths

Evaluate the sum of the first nine terms of the geometric sequence $5, 10, 20,...$

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

  1. $1555$

  2. $2555$

  3. $3555$

  4. $4555$

Show answer
Correct Option: B
Explanation:

Given sequence is $5,10,20,....$
To find the sum of the first $S _n$ terms of a geometric sequence using the formula
Here $a = 5, r = 2, n = 9$
We know $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$\Rightarrow S _9 = \dfrac{5(1-2^{9})}{1-2}$
$\Rightarrow S _9 = \dfrac{-2555}{-1}$
$\Rightarrow S _9 = 2555$

Calculate the sum of first $20$ terms of the G.P. $-1, 1, -1, 1....$

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

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: A
Explanation:
Given series is $-1,1,-1,1,....$
Here $a = -1, r = -1$
We know the formula, $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$S _{20} = \dfrac{1(1-(-1)^{20})}{1-(-1)}$
$ = \dfrac{1(1-(-1)^{20})}{2}$
$= 0$

The sum of $6^{th}$ term in the geometric series $4, 12, 36...$ is

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

  1. $1456$

  2. $2456$

  3. $3456$

  4. $4456$

Show answer
Correct Option: A
Explanation:

Given sequence is $4,12, 36$
To find the sum of the first $S _n$ terms of a geometric sequence using the formula
Here $a = 4, r = 3, n = 6$
We know $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$\Rightarrow S _6 = \dfrac{4(1-3^{6})}{1-3}$
$\Rightarrow S _6 = \dfrac{-2912}{-2}$
$\Rightarrow S _6 = 1456$

What is the sum of G.P. $1, 3, 9, 27,.....$ up to $7$ numbers?

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

  1. $1093$

  2. $2093$

  3. $3093$

  4. $4093$

Show answer
Correct Option: A
Explanation:

Given, $a = 1, r = 3$
$S _n = \dfrac{a _1(1-r^n)}{1-r}$
$S _7 = \dfrac{1(1-3^7)}{1-3}$
$ = \dfrac{1(1-3^7)}{-2}$
$ = 1093$

What is the sum of the first five terms of the geometric sequence $5, 15, 45, ... $?

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

  1. $105$

  2. $305$

  3. $505$

  4. $605$

Show answer
Correct Option: D
Explanation:

Given series is $5,15,45,...$
Here first term $a = 5$ and common ratio $=r = 3$
$S _n = \dfrac{a _1(1-r^n)}{1-r}$
$S _{5} = \dfrac{5(1-(3)^{5})}{1-(3)}$
$ = \dfrac{5(1-(3)^{5})}{-2}$
$ = 605$

Find $4 + 12 + 36 +..... $ upto $6$ terms.

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

  1. $164$

  2. $264$

  3. $364$

  4. $464$

Show answer
Correct Option: C
Explanation:

Given series is $4+12+36+....$ upto $6$ terms
Here $a = 4, r = 3$
We know $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$S _{6} = \dfrac{1(1-(3)^{6})}{1-(3)}$
$ = \dfrac{1(1-(3)^{6})}{-2}$
$ = 364$

If the first term and common ratio is $5$, find the sum of 8th term of infinite G.P.

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

  1. $188280$

  2. $288280$

  3. $388280$

  4. $488280$

Show answer
Correct Option: D
Explanation:

Given, first term $=a = 5$, common ratio $=r = 5$, $n=8$
We know $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$S _{8} = \dfrac{5(1-(5)^{8})}{1-(5)}$
$ = \dfrac{5(1-(5)^{8})}{-4}$
$ = 488280$

The value of $1 + 2 + 4 + 8....$ of G.P., where $n=6$  is

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

  1. $61$

  2. $62$

  3. $63$

  4. $64$

Show answer
Correct Option: C
Explanation:
Given series is $1+2+4+8+....$
Here $a = 1, r = 2$
Also given $n=6$
We know $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$S _{6} = \dfrac{1(1-(2)^{6})}{1-(2)}$
$ = \dfrac{1(1-(2)^{6})}{-1}$
$ = 63$

Calculate sum of eleventh term of the geometric sequence $3, 6, 12, 24, ... $

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

  1. $3141$

  2. $6141$

  3. $2141$

  4. $5141$

Show answer
Correct Option: B
Explanation:

Given series is $3,6,12,24,....$
Here first term $=a = 3$ and common ratio $=r = 2$
$S _n = \dfrac{a _1(1-r^n)}{1-r}$
$S _{11} = \dfrac{3(1-(2)^{11})}{1-(2)}$
$ = \dfrac{3(1-(2)^{11})}{-1}$
$ = 6141$

The sum of first $n$ terms of an G.P. is

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

  1. $S _n = \cfrac{a _1(1-r^n)}{1-r}$

  2. $S _n = \cfrac{a _1(1+r^n)}{1-r}$

  3. $S _n = \cfrac{a _1(1-r^n)}{1+r}$

  4. $S _n = \cfrac{a _1(1-r^n)}{r-1}$

Show answer
Correct Option: A
Explanation:

A GP can be written as:

$a,ar,ar^2, ar^3..............,ar^{n-1}$
$\text{sum} = a+ar+ar^2+ar^3+.........+ar^{n-1}$
$\text{sum} = a(r^{n-1}+r^{n-2}+r^{n-3}+r^{n-4}+...........+r+1)$
We know that:
$\dfrac{x^n -1}{x-1} = x^{n-1}+x^{n-2}+x^{n-3}+.............+x+1$
Thus $\text{sum} = a\left (\dfrac{r^n -1}{r-1}\right)$