Tag: sums

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

 $1+\displaystyle \frac{x}{a _{1}}+\frac{x(x+a _{1})}{a _{1}a _{2}}+\ldots +\displaystyle \frac{x(x+a _{1})(x+a _{2}.).\cdot.\cdots\cdots\cdot(x+a _{n})}{a _{1}a _{2}...a _{n}}=$

  1. $ \dfrac{(x+a _{1})(x+a _{2})...(x+a _{n-1})}{a _{1}a _{2}...a _{n-1}a _{n}} \left\{x^{2}+ a _{n}x+a _{n}\right\}$
  2. $ \dfrac{(x+a _{1})(x+a _{2})...(x+a _{n-1})}{a _{1}a _{2}...a _{n-1}a _{n}} \left\{x^{2}+ a _{n}x\right\}$
  3. $ \dfrac{(x+a _{1})(x+a _{2})...(x+a _{n-1})}{a _{1}a _{2}...a _{n-1}} \left\{x^{2}+ a _{n}x-a _{n}\right\}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$ 1+ \dfrac{x}{a _{1}}+\dfrac{x(x+a _{1})}{a _{1}a _{2}} + ...+ \dfrac{x(x+a _{1})(x+a _{2})...(x+a _{n})}{a _{1}a _{2}...a _{n}} $
$ = \dfrac{a _{1}+x}{a _{1}}+ \dfrac{x(x+a _{1})}{a _{1}a _{2}} + ...+ \dfrac{x(x+a _{1})(x+a _{2})...(x+a _{n})}{a _{1}a _{2}...a _{n}} $
$ = \dfrac{(x+a _{2})(x+a _{1})}{a _{1}a _{2}} + ...+ \dfrac{x(x+a _{1})(x+a _{2})...(x+a _{n})}{a _{1}a _{2}...a _{n}} $
$ =\dfrac{(x+a _{1})(x+a _{2})...(x+a _{n-1})}{a _{1}a _{2}...a _{n-1}} + \dfrac{x(x+a _{1})(x+a _{2})...(x+a _{n})}{a _{1}a _{2}...a _{n}} $
$ = \dfrac{(x+a _{1})(x+a _{2})...(x+a _{n-1})}{a _{1}a _{2}...a _{n-1}} \left\{1 +\dfrac{x(x+a _{n})}{a _{n}}\right\}$
$ = \dfrac{(x+a _{1})(x+a _{2})...(x+a _{n-1})}{a _{1}a _{2}...a _{n-1}a _{n}} \left\{x^{2}+ a _{n}x+a _{n}\right\}$
Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

Calculate the sum of the given series $1+11+111+1111+11111+.....$ upto $9$ terms: 

  1. $123456789$
  2. $987654321$
  3. $999999999$

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

$S=1+11+111+1111+...$ 9 terms

$\Rightarrow 9S=9+99+999+9999+...$ 9 terms
$\Rightarrow 9S=(10-1)+(100-1)+(1000-1)+(10000-1)+...$ 9 terms
$\Rightarrow9S=[10+100+1000+10000+...$ 9 terms $]-9$
$\Rightarrow9S=[\dfrac{10.(10^{9}-1)}{10-1}]-9$
$\Rightarrow9S=[1111111110]-9$
$\Rightarrow9S=1111111101$
$\therefore S=123456789$

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

Find sum of the first $10$ terms of the series:
$(1)(5)+(2)(6)+(3)(7)+(4)(8)+....$ 

  1. $506$
  2. $605$
  3. $572$
  4. $563$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have to find the value of $1 \times 4+2\times 6+3 \times 7+4 \times 8+............$ , upto $10$ terms

$\Rightarrow1\times 5+2\times 6+3\times 7+4\times 8+........=1(1+4)+2(2+4)+3(3+4)+4(4+4)+.......$
$\Rightarrow { (1 }^{ 2 }+{ 2 }^{ 2 }+....+{ 10 }^{ 2 })+4(1+2+...10)$
$\Rightarrow \frac{10 \times 11 \times 21}{6}+4(\frac{10 \times 11}{2})=605$
Therefore the option is $B$

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

Sum of $n$ terms of the series  $5+7+13+31+85+...,$  is

  1. $4n+\dfrac12(3^n-1)$
  2. $8n+\dfrac12(3^n-1)$
  3. $2n+\dfrac12(3^n-1)$

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

The given series is $5+7+13+31+85+.............$

$\Rightarrow 4n+(1+3+9+27+81+..........)$
$\Rightarrow 4n+(1+3+3^{2}+3^{3}+3^{4}+...........)$
$\Rightarrow 4n+\frac{1}{2}(3^{n}-1)$
Therefore option $A$ is correct

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

If the sum of first $n$ natural numbers is $\dfrac15$ times the sum of their squares, then the value of $n$ is

  1. $5$
  2. $6$
  3. $7$
  4. $8$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\Rightarrow$Sum of first $n$ natural numbers is $\dfrac{n(n+1)}{2}$.
$\Rightarrow$Sum of the squares of the first $n$ natural numbers is $\dfrac{n(n+1)(2n+1)}{6}$.

So from the question, we have

$\Rightarrow$$\dfrac{n(n+1)}{2}=\dfrac{1}{5}.\dfrac{n(n+1)(2n+1)}{6}$

$\Rightarrow 2n=14$

$\Rightarrow n=7$