Tag: forming an arithmetic progression between two quantities a and b

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Questions Related to forming an arithmetic progression between two quantities a and b

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

$\sum{n^3}=$

  1. $(\sum{n})^3$
  2. $(\sum{n})^2$
  3. $(\sum{n})^3+(\sum{n})^2$
  4. $\sum{(n+n^2)}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We have, $ \sum { { n }^{ 3 } } = { 1 }^{ 3 }+{ 2 }^{ 3 }+....+{ n }^{ 3 } $
The formula to find the sum of cubes of natural numbers is $ ={( \dfrac {n(n+1)}{2} )}^2 $

But sum of first $ n $ natural numbers is$  \sum { { n } } = \dfrac {n(n+1)}{2} $ 

So, $ \sum { { n }^{ 3 } } = { (\sum { { n } })}^2 $
Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

Find the sum of $\displaystyle\frac{0.3}{0.5}+\frac{0.33}{0.55}+\frac{0.333}{0.555}+\cdots\cdots$ to 15 terms.

  1. 10

  2. 9

  3. 3

  4. 5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Each dfraction can be simplified to $ \dfrac {3}{5} $ by dividing by their highest common factor. 

So, expression is simplified to $ \dfrac {3}{5} + \dfrac {3}{5} + \dfrac {3}{5} + --- 15 $ terms. 
$ = \dfrac {3}{5} \times 15 = 9 $
Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

The sum of the series $1-\cfrac { 3 }{ 2 } +\cfrac { 5 }{ 4 } -\cfrac { 7 }{ 8 } +...\infty $ is

  1. $\cfrac { 2 }{ 9 } $
  2. $\cfrac { -4 }{ 9 } $
  3. $\cfrac { 4 }{ 9 } $
  4. $\cfrac { -2 }{ 9 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$S=1-\cfrac { 3 }{ 2 } +\cfrac { 5 }{ 4 } -\cfrac { 7 }{ 8 } +...\infty $
$\cfrac { S }{ 2 } =\cfrac { 1 }{ 2 } -\cfrac { 3 }{ 4 } +\cfrac { 5 }{ 8 } $
$S\left( 1+\cfrac { 1 }{ 2 }  \right) =1-\cfrac { 2 }{ 2 } +\cfrac { 2 }{ 4 } -\cfrac { 2 }{ 8 } ....\infty $
$S\left( \cfrac { 3 }{ 2 }  \right) =1-1+\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 4 } ....\infty $
$S\left( \cfrac { 3 }{ 2 }  \right) =\cfrac { 1 }{ 2 } \left( 1-\cfrac { 1 }{ 2 } ....\infty  \right) $
$S\left( \cfrac { 3 }{ 2 }  \right) =\cfrac { 1 }{ 2 } \left( \cfrac { 1 }{ 1+1/2 }  \right) $
$S\left( \cfrac { 3 }{ 2 }  \right) =\cfrac { 1 }{ 2 } .\cfrac { 2 }{ 3 } $
$S\left( \cfrac { 3 }{ 2 }  \right) =\cfrac { 1 }{ 3 } $
$S=\cfrac { 2 }{ 9 } $
Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

$1\times 2 + 2\times 3 + 3\times 4 + .... n\ terms =$

  1. $\dfrac {n(n + 1)(n + 2)}{3}$
  2. $\dfrac {n(n + 1)(n - 2)}{3}$
  3. $\dfrac {n(n + 1)(n + 2)}{6}$
  4. $\dfrac {n(n - 1)(n - 2)}{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$n^{th}$ term of sequence $= n(n + 1)$
Then sum of $n$ terms $= \displaystyle \sum _{r = 1}^{n} n(n + 1) = \displaystyle \sum _{r = 1}^{n}(n^{2} + n)$
$= \displaystyle \sum _{r = 1}^{n}n^{2} + \displaystyle \sum _{r = 1}^{n}n = \dfrac {n(n + 1)(2n + 1)}{6} + \dfrac {n(n + 1)}{2}$
$= \dfrac {n(n + 1)(2n + 1)+ 3n(n + 1)}{6}$
$= \dfrac {n(n + 1)[2n + 1 + 3]}{6}$
$= \dfrac {n(n + 1)(2n + 4)}{6}$
$= \dfrac {n(n + 1)(n + 2)}{3}$.

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

The sum of the series $6+66+666+..$ upto n terms is:

  1. $\dfrac{1}{81}(10^{n-1}-9n+10)$
  2. $\dfrac{2}{27}(10^{n-1}-9n-10)$
  3. $\dfrac{2}{27}(10-9n-10)$
  4. $None of these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of the series 6+66+666... is found by expressing it as 6/9 * (9+99+999...) = 2/3 * ((10-1) + (100-1) + ...). The resulting formula is (2/27) * (10^n - 1 - 9n). Option A is a common variation of this formula.

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

Sum to infinity of the series $ \frac {2}{3}- \frac {5}{6} +\frac {2}{3} - \frac {11}{24}+ ....$

  1. $\frac {4}{9}$
  2. \frac {1}{3}$
  3. \frac {2}{9}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is an infinite series that can be grouped or solved using standard summation techniques for alternating series. The sum converges to 4/9.