Tag: maths

Questions Related to maths

The sum of the first 12 terms is 100. The first term is 20. Find the last term. (use Gauss method)

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

  1. $\dfrac{-20}{6}$

  2. $\dfrac{-10}{6}$

  3. $\dfrac{-15}{5}$

  4. $\dfrac{-30}{2}$

Show answer
Correct Option: A
Explanation:

Given that $S _n = 100, n = 12, a = 20$. fast term = ?
we know that, $S _n = \dfrac{n}{2}$  [First term + Last term]
$100 = \dfrac{12}{2}$   [20 + Last term]
$100 = 120 + 6 (\text{Last term})$
Last term $= \dfrac{-20}{6}$

Select the correct alternative from the given ones that will complete the series.
$0, 7, 26, 63, 124, ?$

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

  1. $251$

  2. $125$

  3. $215$

  4. $512$

Show answer
Correct Option: C
Explanation:

$\underset {(1^{3} - 1)}{0}\rightarrow \underset {(2^{3} - 1)}{7}\rightarrow \underset {(3^{3} - 1)}{26}\rightarrow \underset {(4^{3} - 1)}{63}\rightarrow \underset {(5^{3} - 1)}{124} \rightarrow \underset {(6^{3} - 1)}{215}$.

Find the first term. The sum of the first 100 terms is 1,200. The last term is 150. (use Gauss method)

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

  1. 126

  2. -126

  3. 125

  4. -125

Show answer
Correct Option: B
Explanation:

Using Gauss method,
$S _n = \dfrac{n}{2} $   [First term + Last term]
$1,200 = \dfrac{100}{2} [a _1 + 150]$
$1,200 = 50a _1 + 7,500$
$1,200 - 7,500 = 50 a _1$
$a _1 = - 126$

Given $a _1 = 100, a _n = 50$ and $n = 200$. Find their sum using Gauss method.

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

  1. 15,000

  2. 14,000

  3. 11,000

  4. 12,000

Show answer
Correct Option: A
Explanation:

Given; $a = 100, a _n = 50, n = 200$
Using Gauss formula,
$S _n = \dfrac{n}{2} $   [First term + Last term]
$= \dfrac{200}{2} [100 + 50]$
$S _n = 15,000$

The famous mathematician associated with finding the sum of the first 100 natural numbers is


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

  1. Pythagoras

  2. Newton

  3. Gauss

  4. Euclid

Show answer
Correct Option: C
Explanation:

Guass found the sum of first 100 natural numbers.

Use Gauss method to find which term of the A.P. 1, 3, 5, 7 ......... is 153?

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

  1. 77

  2. 76

  3. 75

  4. 74

Show answer
Correct Option: A
Explanation:

Gauss formula for nth term of the A.P. is $a _n = a + (n - 1) d$
Given; $a = 1, d= 2 , a _n = 154$
$153 = 1 + (n - 1) 2$
$153 - 1 = 2n - 2$
$153 - 1 + 2 = 2n$
$n = \dfrac{154}{2} = 77$

13, 17, 33, 97, 353,....

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

  1. 1377

  2. 653

  3. 712

  4. 1273

Show answer
Correct Option: A
Explanation:

Add to each number 4, $\displaystyle 4^{2}$, $\displaystyle 4^{3}$,$\displaystyle 4^{4}$, $\displaystyle 4^{5}$, i.e. $4, 16, 64, 256, 1024$ etc Next number $= 353 + 1024 = 1377$

The value of ${ 1 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 1 }+{ 2 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 2 }+{ 3 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 3 }+.....{ (20) }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 20 }$ is

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

  1. $210\times { 2 }^{ 17 }$

  2. $420\times { 2 }^{ 17 }$

  3. $420\times { 2 }^{ 87 }$

  4. $210\times { 2 }^{ 87 }$

Show answer
Correct Option: C
Explanation:

$S={ 1 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 1 }+{ 2 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 2 }+{ 3 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 3 }+.....{ (20) }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 20 }=\sum _{ r=1 }^{ 20 }{ { r }^{ 2 } } .{ _{  }^{ 20 }{ C } } _{ r }$
$=\sum _{ r=1 }^{ 20 }{ { r }^{  } } (r.{ _{  }^{ 20 }{ C } } _{ r })\=20\sum _{ r=1 }^{ 20 }{ { r }^{ 19 } } .{ _{  }^{ 19 }{ C } } _{ r-1 }\=20\sum _{ r=1 }^{ 20 }{ (r-1+1) } .{ _{  }^{ 19 }{ C } } _{ r-1 }\=20\sum _{ r=1 }^{ 20 }{ { (r-1) }^{  } } .{ _{  }^{ 19 }{ C } } _{ r-1 }+20\sum _{ r=1 }^{ 20 }{ { r }^{ 2 } } .{ _{  }^{ 19 }{ C } } _{ r-1 }\=20\times 19\sum _{ r=2 }^{ 20 }{ { _{  }^{ 18 }{ C } } _{ r-1 } } +20\times { 2 }^{ 19 }$
$=20\times 19\times { 2 }^{ 18 }+20\times { 2 }^{ 19 }=20\times { 2 }^{ 18 }(19+2)=20\times 21\times { 2 }^{ 18 }=420\times { 2 }^{ 18 }\quad $
(3) option is correct

Let $a-i=i+\dfrac{1}{i}$ for $i=1, 2,..., 20$. Put $p=\dfrac{1}{20}(a _1+n _2+...+n _{20})$ and $q=\dfrac{1}{20}\left(\dfrac{1}{a _1}+\dfrac{1}{a _2}+...+\dfrac{1}{a _{20}}\right)$. Then?

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

  1. $q\in \left(0, \dfrac{22-p}{21}\right)$

  2. $q\in \left(\dfrac{22-p}{21}, \dfrac{2(22-p)}{21}\right)$

  3. $q\in \left(\dfrac{2(22-p)}{21}, \dfrac{22-p}{7}\right)$

  4. $q\in \left(\dfrac{22-p}{7}, \dfrac{4(22-p)}{21}\right)$

Show answer
Correct Option: A

The sum $\displaystyle\sum _{ 0\le i }^{  }{ \sum _{ j\le 10 }^{  }{ \left( _{  }^{ 10 }{ { C } _{ j } } \right) \left( _{  }^{ j }{ { C } _{ i } } \right)  }  } $ is equal to 

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

  1. $2^{10}-1$

  2. $2^{10}$

  3. $3^{10}-1$

  4. $3^{10}$

Show answer
Correct Option: D
Explanation:
$\displaystyle\sum _{ 0\le i }^{  }{ \sum _{ j\le 10 }^{  }{ \left( _{  }^{ 10 }{ { C } _{ j } } \right) \left( _{  }^{ j }{ { C } _{ i } } \right)  }  } $

$= ^{10}C _{0}(^0C _0) + ^{10}C _1(^1C _1+^1C _0) + ……. + ^{10}C _{10}(^{10}C _{10}+^{10}C _{9}+...…+^{10}C _{0})$

$=2^0.^{10}C _{0} + 2^1.^{10}C _{1} + …….. + 2^{10}.^{10}C _{10}$

Now,


$(1+2x)^{10} = ^{10}C _{0}(1)^{10} + ^{10}C _1(1)^9(2x)^1 + ………….+^{10}C _{10}(2x)^{10} $

Put x = 1,

$(3)^{10} = ^{10}C _{0}(1)^{10} + ^{10}C _1(1)^9(2)^1 + ………….+^{10}C _{10}(2)^{10} $

Thus the required sum is $3^{10}$