Tag: maths

Questions Related to maths

Select the correct option.
The first term of an $AP$ is $p$ and the common difference is $q$, then its $10^{th}$ term is 

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $q + 9p$

  2. $p - 9q$

  3. $p + 9q$

  4. $2p + 9q$

Show answer
Correct Option: C
Explanation:

First term of $AP = P$


Common difference of $AP= q$


$10^{th}$ term of $AP = p + (10 - 1) q$

                             $= p + 9q$

Select the correct option.
The value of $x$ for which $2x, (x + 10) $ and $(3x + 2)$ aree the three consecutive terms of an AP, is 

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $6$

  2. $-6$

  3. $18$

  4. $-18$

Show answer
Correct Option: C
Explanation:

Consecutive terms of $AP \Rightarrow 2x, (x + 10 ) , (3x + 2)$

According to the properties

$2(x + 10) = 2x + 3x + 2$

$4x + 20 = 5x + 2$

$x = 18$


$\displaystyle \frac{b+c-a}{a}, \frac{c+a-b}{b}, \frac{a+b-c}{c}$ are in A.P., then $\displaystyle \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. A.P.

  2. H.P

  3. G.P

  4. A.G.P

Show answer
Correct Option: A
Explanation:

Given,

$\dfrac{b+c-a}{a},\dfrac{c+a-b}{b},\dfrac{a+b-c}{c}$ one in A.P

Now,

$\because \dfrac{b+c-a}{a},\dfrac{c+a-b}{b},\dfrac{a+b-c}{c}$ are in A.P


$\therefore  \dfrac{b+c-a}{a}+2,\dfrac{c+a-b}{b}+2,\dfrac{a+b-c}{c}+2$, must be  in A.P


$\therefore \dfrac{b+c-a+2a}{a},\dfrac{c+a-b+2b}{b},\dfrac{a+b-c+2c}{c}$ are in A.P


$\therefore \dfrac{a+b+c}{a},\dfrac{a+b+c}{b},\dfrac{a+b+c}{c}$ are in A.P


$\because \dfrac{a+b+c}{a},\dfrac{a+b+c}{b},\dfrac{a+b+c}{c}$ are in A.P


$\therefore \dfrac{1}{(a+b+c)}\times \dfrac{(a+b+c)}{a},\dfrac{1}{(a+b+c)}\times \dfrac{(a+b+c)}{b},\dfrac{1}{(a+b+c)}\times \dfrac{(a+b+c)}{c}$ are in A.P


$\therefore \dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$ are in A.P
 

Let $S _n$ be the sum of all integers k such that $2^n < k < 2^{n-1}$, for n > 1, Then $9$ divides $S _n$ if and only if

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $n$ is odd

  2. $n$ is of the form $3k+1$

  3. $n$ is even

  4. $n$ is of the form $3k +2$

Show answer
Correct Option: C
Explanation:

Number of integers between 2$^n$ and $2^{2n+ 1} - 2^n - 11$
and I term $= 2^n + 1$
last term $= 2^{n + 1} - 1$
$\therefore \displaystyle S _n = \frac{(2^{n + 1} - 2^n - 1)}{2} [2^n + 1 + 2^{n + 1} - 1]$
$\displaystyle \frac{(2^{n + 1} - 2^n - 1)}{2} (2^n)(1 + 2)$
$= (2^n - 1) \displaystyle \frac{(2^n).3}{2}$
$S _n= 9 \lambda; \lambda \varepsilon I$
$\therefore 3 \times 2^{n - 1} \times (2^n - 1) = 9 \lambda$
$2^{n - 1} \times (2^n - 1) = 3 \lambda$
$2^n (2^n - 1) = 6 \lambda$
It is possible when n is even.

The sum of the three numbers in A.P is $21$ and the product of the first and third number of the sequence is $45$. What are the three numbers?

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $5, 7$ and $9$

  2. $9, 7$, and $5$

  3. $3, 7$, and $11$

  4. Both (1) and (2)

  5. None of these

Show answer
Correct Option: D
Explanation:

Let the numbers are be $a - d, a, a + d$
Then $a - d + a + a + d = 21$
$3a = 21$
$a = 7$
and $(a - d)(a + d) = 45$
$a^2 - d^2 = 45$
$d^2 = 4$
$d=\pm 2$
Hence, the numbers are $5, 7$ and $9$ when $d = 2$ and $9, 7$ and $5$ when $d = -2$. In both the cases numbers are the same.

The sum of $10$ numbers is $100$. The first term is $1$. Find its common difference.

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $2$

  2. $1$

  3. $3$

  4. $4$

Show answer
Correct Option: A
Explanation:
The sum of first $n$ terms of arithmetic series formula can be written as,
$S _n = \dfrac{n}{2} [2a + (n - 1)d]$ ............ (1)
$n =$ number of terms $= 10$
$S _n = 100$
First term, $a = 1$
Common difference, $d = ?$
From $(1)$, we have
$100 = \dfrac{10}{2} [2 \times 1 + (10 - 1)d]$
$100 =  5[2 + 9d]$
$100 = 10 + 45d$
$ 100 - 10 = 45d$
$90 = 45d$
$d = \dfrac{90}{45} = 2$
The common difference is $2$.

The sum of first $10$ terms and $20$ terms of an AP are $120$ and $440$ respectively. What is the common difference?

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: B
Explanation:

Let the first term be $a$ and common difference be $d$.
So, sum of first $10$ terms $=\dfrac { 10 }{ 2 } (2a+(10-1)d)$

$\implies 120 =5(2a+9d)$
$\implies 24=2a+9d$ .............. $(i)$
Sum of first 20 terms $=\frac { 20 }{ 2 } (2a+(20-1)d)$
$\implies 440 =10(2a+19d)$
$\implies 44=2a+19d$ ......... $(ii)$
Subtracting equation (i) from (ii) gives
$20=10d$
$\implies d=2$
Common difference =2
Hence, option B is correct.

If ${ a } _{ 1 },{ a } _{ 2 },{ a } _{ 3 },\dots $ are terms of AP such that ${ a } _{ 1 }+{ a } _{ 5 }+{ a } _{ 10 }+{ a } _{ 15 }+{ a } _{ 20 }+{ a } _{ 24 }=225$, then the sum of first $24$ terms is

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $9\times { 10 }^{ 2 }$

  2. $9\times { 10 }^{ 3 }$

  3. $10\times { 9 }^{ 2 }$

  4. $10\times { 9 }^{ 3 }$

Show answer
Correct Option: A
Explanation:

We know that the sum of terms of AP equidistant from the beginning and end is always same and it is always equal to the sum of first and last terms.
$\Rightarrow { a } _{ 1 }+{ a } _{ 24 }={ a } _{ 6 }+{ a } _{ 20 }={ a } _{ 10 }+{ a } _{ 15 }$
$\because { a } _{ 1 }+{ a } _{ 5 }+{ a } _{ 10 }+{ a } _{ 15 }+{ a } _{ 20 }+{ a } _{ 24 }=225$
$\therefore 3\left( { a } _{ 1 }+{ a } _{ 24 } \right) =225\Rightarrow { a } _{ 1 }+{ a } _{ 24 }=75$
$\therefore { S } _{ 24 }=\dfrac { 24 }{ 2 } \left( { a } _{ 1 }+{ a } _{ 24 } \right)$   $\left[ \because { S } _{ n }=\dfrac { n }{ 2 } \left( { a } _{ 1 }+{ a } _{ n } \right)  \right] $
           $=12\left( 75 \right) =900=9\times { 10 }^{ 2 }$

$x _{1}, x _{2}, x _{3}, ....$ are in A.P.
If $x _{1} + x _{7} + x _{10} = -6$ and $x _{3} + x _{8} + x _{12} = -11$, then $x _{3} + x _{8} + x _{22} = ?$

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $-21$

  2. $-15$

  3. $-18$

  4. $-31$

Show answer
Correct Option: A
Explanation:

Let the first term be $a$ and common difference be $d$

$x _{1}+x _{7}+x _{10}=-6$
$\Rightarrow (a)+(a+6d)+(a+9d)=-6$
$\Rightarrow 3a+15d=-6 \rightarrow \text{eqn}(1) $

$x _{3}+x _{8}+x _{12}=-11$
$\Rightarrow (a+2d)+(a+7d)+(a+11d)=-11$
$\Rightarrow 3a+20d=-11 \rightarrow \text{eqn}(2)$
Solving eqn (1) and (2), we get $a=3$ and $d=-1$

Now, $x _{3}+x _{8}+x _{22}=3a+30d = -21$

If the $n^{th}$ term of an AP be $(2n-1)$, then the sum of its first n terms will be.

maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

  1. $n^2-1$

  2. $(2n-1)^2$

  3. $n^2$

  4. $n^2+1$

Show answer
Correct Option: C
Explanation:
$a _n=(2n-1)$
$\Rightarrow$  $a _1=2\times 1-1=1$
$\Rightarrow$  $a _2=2\times 2-1$
            $=4-1$
            $=3$
$\Rightarrow$  $d=a _1-a _1=3-1$
$\therefore$  $d=2$
$\Rightarrow$  $S _1=\dfrac{n}{2}[2a _1+(n-1)d]$
           
            $=\dfrac{n}{2}[2(1)+(n-1)2]$

            $=\dfrac{n}{2}[2+2n-2]$

            $=\dfrac{n}{2}\times 2n$

            $=n^2$