Tag: maths

Questions Related to maths

999, 730, 511, 344, 215, ....

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 123

  2. 126

  3. 125

  4. 130

Show answer
Correct Option: B
Explanation:

The rule is $\displaystyle 10^{3}-1$, $\displaystyle 9^{3}+1$, $\displaystyle 8^{3}-1$, $\displaystyle 7^{3}+1$, $\displaystyle 6^{3}-1$, $\displaystyle 5^{3}+1$ The number is 126

Find the Odd one among : 1, 4, 27, 16, 25, 36

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 4

  2. 27

  3. 16

  4. 25

Show answer
Correct Option: B
Explanation:

The numbers are written as $\displaystyle 1^{2}$, $\displaystyle 2^{2}$, $\displaystyle 3^{2}$, $\displaystyle 4^{2}$, $\displaystyle 5^{2}$, $\displaystyle 6^{2}$ etc It should be 9 in place of 27

Find the sum of the series: 1 + 2 + 3 + .......... + 50 using Gauss method.

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 1,275

  2. 1,200

  3. 1,100

  4. 2,000

Show answer
Correct Option: A
Explanation:

Sum of 'n' series using Gauss method is,
$S _n = \dfrac{n(n + 1)}{2}$
$\therefore$ Given n = 50
So, $\dfrac{50(50 + 1)}{2}$
$S _{50} = 1,275$

Find the sum of the following geometric series:
$ \sqrt{7}, \sqrt{21}, 3\sqrt{7},...$ to n terms

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $ \sqrt{7}\left ( \dfrac{3^{-n/2}-1}{\sqrt{3}-1} \right )$

  2. $ \sqrt{6}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$

  3. $ \sqrt{7}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$

  4. $ \sqrt{5}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$

Show answer
Correct Option: C
Explanation:

Let $ S _n$ denote the sum of n terms of the G.P. $\sqrt{7}, \sqrt{21}, 3\sqrt{7}, ...,$ 


Cleraly, the given series is a $G.P.$ with first term$=a=\sqrt7$ and common ratio$=r=\sqrt 3$

Then,
$ S _n = \sqrt{7} \left { \dfrac{\left ( \sqrt{3} \right )^{n}-1}{\sqrt{3}-1} \right } = \sqrt{7} \left ( \dfrac{3^{n/2}-1}{3^{1/2}-1} \right )$

${\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;.... + \;{\sin ^2}{\text{9}}{{\text{0}}^{\text{o}}}$ is equal to

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $22$

  2. $23$

  3. $44$

  4. $45$

Show answer
Correct Option: B
Explanation:

Now,

${\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;.... + \;{\sin ^2}{\text{9}}{{\text{0}}^{\text{o}}}$
$=({\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;...+\sin^2 44^o)+(\sin^2 46^o+... +\sin^2 98^o)+ \;{\sin ^2}{\text{9}}{{\text{0}}^{\text{o}}}$
$=({\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;...+\sin^2 44^o)+(\cos^2 44^o+... +\cos^2 2^o)+ 1$ [ Since $\sin^2 x^o=\cos^2 (90^o-x^o)$]
$=(\sin^2 2^o+\cos^2 2^o)+(\sin^2 4^o+\cos^2 4^o)+......+(\sin^2 44^o+\cos^2 44^o)+1$
$=1+1+.....+1(22\text{th})+1$
$=22+1$
$=23$.

Find the missing terms :- .........., .........., 8, .........., -1

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 16, 10 and 2

  2. 17, 12 and 3

  3. 17, 12.5 and 3.5

  4. 1, 2 and 3

Show answer
Correct Option: C
Explanation:

Third term, $a _3 = 8$
Fifth term, $a _5 = - 1$
we know that, $a _n = a + (n - 1) d$
$a _3 = a + (3 - 1) d$
$a _3 = a + 2d $     ...... (1)
$a _5 = a + (5 - 1) d$
$-1 = a + 4d $   ....... (2)
comparing equation (1) & (2)
$8    =   a    +   2d$     .... (1)
$-1   =   a    +   4d$     .... (2)
 (+)    (-)    (-)
-------------------
   $a = - 2d$
$d = - 4.5 $
Put $d = -4.5 $ in equation (1)
$8 = a + 2 (- 4.5)$
$8 = a - a$
$a = 17$
$\therefore a _2 = 17 + (2 - 1) d$
$= 17 + 1 (-4.5)$
$a _2 = 12.5$
$a _4 = 17 + (4 - 1) (-4.5)$
$a _4 = 3.5$
The missing terms are 17, 12.5 and 3.5

If the sum of the first n integers is 15. What is n? (use Gauss method)

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 7, 5

  2. 6, -5

  3. 3, -5

  4. 2, 0

Show answer
Correct Option: B
Explanation:

Using Gauss formula,
$S _n = \dfrac{n}{2} (n + 1)$
$15 = \dfrac{n}{2} (n + 1)$
$30 = n (n + 1)$
$30 = n^2 + n$   ....... (1)
By factorization we can write equation (1) as
$n^2 - 6n + 5n - 30 = 0$
$n(n - 6) + 5 (n - 6) = 0$
$n = 6, - 5$
$\therefore$ The n integers will start from -5 to 6.

Find the sum of the first 100 terms -5, -4, -3, -2, -1, 0, 1, 2 ............. using Gauss method

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 4,400

  2. 4,100

  3. 4,200

  4. 4,450

Show answer
Correct Option: D
Explanation:

Given that $a = -5, n = 100, a _n =?, d = 1$
we know that Gauss formula is, $S _n = \dfrac{n}{2}$   [First term + Last term]
To find nth term,
$a _n = a + (n - 1) d$
$a _{100} = - 5 + (100 - 1)1$
$= - 5 + 99$
$a _{100} =94$
$S _n = \dfrac{100}{2} [-5 + 94]$
$= 50 [89]$
$S _{100} = 4,450$

Apply Gauss method to find which term of the A.P. 2, 4, 6, 8 ..... is 108?

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 51

  2. 52

  3. 53

  4. 54

Show answer
Correct Option: D
Explanation:

Gauss formula for nth term of the A.P. is $a _n = a + (n - 1) d$
Given: $a = 2, d= 2, a _n = 108$
$108 = 2 + (n - 1) 2$
$108 - 2 = 2n - 2$
$106 + 2 = 2n$
$N = \dfrac{108}{2} = 54$

Find the sum of first 31 terms of an A.P. whose third term is 12 and fourth term is 16.

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 1,983

  2. 1,984

  3. 1,985

  4. 1,986

Show answer
Correct Option: B
Explanation:

Given that, $a _3 = 12; a _4 = 16$
Common difference, $d = a _4 - a _3 = 16 - 12 = 4$
$a _3 - a _2 = d$
$12 - 4 = a _2 $ $\Rightarrow  8$
$d = a _2 - a _1$ 
$a = 4$
We know the formula,
$s _n = \dfrac{n}{2} [2a + (n - 1)d]$
$S _{31} = \dfrac{31}{2} [2 \times 4 + (31 - 1)4]$
$= 15.5 [8 + 30 \times 4]$
$=15.5 [128]$
$S _{31} = 1,984$