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

a proper option (a), (b), (c) or (d) from given options and write in the box given that so that the statement becomes correct : (All the problems refer to A.P)
${ T } _{ 3 }=8,{ T } _{ 7 }=24,$ then ${ T }$

  1. -4

  2. 28

  3. 32

  4. 36
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

T_n = a + (n-1)d. T_3 = a + 2d = 8 and T_7 = a + 6d = 24. Subtracting gives 4d = 16, so d = 4. Then a + 8 = 8, so a = 0. T_n = 0 + (n-1)4 = 4n - 4. If the question asks for T_1, it is 0; if it asks for T_0 (not standard), it is -4.

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

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

  1. $q + 9p$
  2. $p - 9q$
  3. $p + 9q$
  4. $2p + 9q$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

First term of $AP = P$


Common difference of $AP= q$


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

                             $= p + 9q$

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

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

  1. A.P.

  2. H.P

  3. G.P

  4. A.G.P
Reveal answer Fill a bubble to check yourself
A Correct answer
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
 

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

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

  1. $n$ is odd
  2. $n$ is of the form $3k+1$
  3. $n$ is even
  4. $n$ is of the form $3k +2$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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?

  1. $5, 7$ and $9$
  2. $9, 7$, and $5$
  3. $3, 7$, and $11$

  4. Both (1) and (2)

  5. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  1. $2$
  2. $1$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $2$.
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

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

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $9\times { 10 }^{ 2 }$
  2. $9\times { 10 }^{ 3 }$
  3. $10\times { 9 }^{ 2 }$
  4. $10\times { 9 }^{ 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
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] $
           $=12\left( 75 \right) =900=9\times { 10 }^{ 2 }$