Tag: arithmetic sequences

Questions Related to arithmetic sequences

The value of the sum $\dfrac{1}{3^2+1}+\dfrac{1}{4^2+2}+\dfrac{1}{5^2+3}+\dfrac{1}{6^2+4}$.....$\infty$ is equal to 

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

  1. $\dfrac{13}{36}$

  2. $\dfrac{12}{36}$

  3. $\dfrac{15}{36}$

  4. $\dfrac{18}{36}$

Show answer
Correct Option: A

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

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

  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

Show answer
Correct Option: A
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\}$

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

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

  1. $123456789$

  2. $987654321$

  3. $999999999$

  4. None of these

Show answer
Correct Option: A
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 $]-9$
$\Rightarrow9S=[\dfrac{10.(10^{9}-1)}{10-1}]-9$
$\Rightarrow9S=[1111111110]-9$
$\Rightarrow9S=1111111101$
$\therefore S=123456789$

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

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

  1. $506$

  2. $605$

  3. $572$

  4. $563$

Show answer
Correct Option: B
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 $B$

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

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

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

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

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

  4. None of these

Show answer
Correct Option: A
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

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

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

  1. $5$

  2. $6$

  3. $7$

  4. $8$

Show answer
Correct Option: C
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}$

$\Rightarrow 2n=14$

$\Rightarrow n=7$

If $x, | x+1| ,|x-1| $ are the three terms of an AP. Its sum up to $20$ terms is 

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

  1. $90 \ or \ 175$

  2. $180 \ or \ 350$

  3. $360 \ or \  700$

  4. $720 \ or \ 1400$

Show answer
Correct Option: B