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Questions Related to maths

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

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

  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 )$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 )$

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

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

  1. $22$
  2. $23$
  3. $44$
  4. $45$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\sin^2 x^o=\cos^2 (90^o-x^o)$]
$=(\sin^2 2^o+\cos^2 2^o)+(\sin^2 4^o+\cos^2 4^o)+......+(\sin^2 44^o+\cos^2 44^o)+1$
$=1+1+.....+1(22\text{th})+1$
$=22+1$
$=23$.

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

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

  1. 16, 10 and 2

  2. 17, 12 and 3

  3. 17, 12.5 and 3.5

  4. 1, 2 and 3
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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

  1. 7, 5

  2. 6, -5

  3. 3, -5

  4. 2, 0
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. 4,400

  2. 4,100

  3. 4,200

  4. 4,450
Reveal answer Fill a bubble to check yourself
D Correct answer
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,
$a _n = a + (n - 1) d$
$a _{100} = - 5 + (100 - 1)1$
$= - 5 + 99$
$a _{100} =94$
$S _n = \dfrac{100}{2} [-5 + 94]$
$= 50 [89]$
$S _{100} = 4,450$

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

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

  1. 1,983

  2. 1,984

  3. 1,985

  4. 1,986
Reveal answer Fill a bubble to check yourself
B Correct answer
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,
$s _n = \dfrac{n}{2} [2a + (n - 1)d]$
$S _{31} = \dfrac{31}{2} [2 \times 4 + (31 - 1)4]$
$= 15.5 [8 + 30 \times 4]$
$=15.5 [128]$
$S _{31} = 1,984$