Arithmetic Progressions Test

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Arithmetic Progressions Test - 38

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Question 1
1 / -0

$$(p + q)^{th}$$ and $$(p - q)^{th}$$ terms of an A.P. are respectively $$m$$ and $$n.$$ The $$p^{th}$$ term is

A
$$\displaystyle\frac{1}{2}(m + n)$$

B
$$\displaystyle\sqrt{mn}$$

C
$$m + n$$

D
$$mn$$

Solution

Let $$a$$ be the first term and $$d$$ be the common difference of the given $$AP$$ then, by using $$n^{th}$$ term formula

nth term $$t_{n}= a+(n-1)d $$ we have,

$$(p+q)^{th}\space=m\Rightarrow a+(p+q-1)d=m$$ ...(1)

$$(p-q)^{th}\space=n\Rightarrow a+(p-q-1)d=n$$ ...(2)

Adding (1) and (2), we get

$$2a+(2p-2)d=m+n$$

$$\Rightarrow a+(p-1)d=\dfrac{m+n}2$$

$$\Rightarrow p^{th}\space term=\dfrac{m+n}2$$

So, Option A is correct.
Question 2
1 / -0

Sum of first $$5$$ terms of an A.P. is one fourth of the sum of next five terms. If the first term is $$ 2,$$ then the common difference of the A.P. is

A
$$6$$

B
$$-6$$

C
$$3$$

D
None of these

Solution

let the A.P. be $$a,a+d,a+2d.......$$
We have $$T_{1}+T_{2}+T_{3}+T_{4}+T_{5}=\dfrac{1}{4}\left [ T_{6}+T_{7}+T_{8}+T_{9}+T_{10} \right ]$$
$$\because$$ Sum $$=\dfrac{n}{2} ($$first term $$+$$ last term$$)$$
Then $$\dfrac{5}{2}\left ( T_{1}+T_{5} \right )=\dfrac{1}{4}\times\dfrac{5}{2}\left ( T_{6}+T_{10} \right )$$
$$\Rightarrow \dfrac{5}{2}\left ( a+\left ( a+4d \right ) \right )=\dfrac{5}{8}\left [ \left (a+5d \right )\left (a+9d \right ) \right ] $$
$$\Rightarrow 2a+4d=\dfrac{1}{4}\left [ 2a+14d \right ]$$
$$\Rightarrow 4a+8d=a+7d$$
$$\Rightarrow d=-3a $$ ....$$(a=2)$$
$$\Rightarrow d=-3\times 2=-6$$
Question 3
1 / -0

The first term of an A.P. of consecutive integers is $$p$$$$^2$$ $$+ 1. $$The sum $$2p + 1$$ terms of this series can be expressed as

A
$$(p + 1)$$$$^2$$

B
$$(2p + 1) (p^2+p + 1)$$

C
$$(p + 1)$$$$^3$$

D
$$p$$$$^3$$$$ + (p + 1)$$$$^3$$

Solution

The first term of AP is set of consecutive integers, then common difference is $$d=1$$ & $$a = p^2+1$$
Here we use the formula for sum of n terms of an AP as given below:
$$S_n = \dfrac {n}{2}[2a+(n-1)d] $$
$$\therefore$$ The sum of $$(2p+1)$$, $$S_{2p+1} $$ =$$\dfrac{2p+1}{2}\left [ 2(p^{2}+1)+(2p+1-1)\times 1 \right ]$$
$$=\dfrac{2p+1}{2}\left [ 2p^{2}+2+2p \right ]$$
$$=(2p+1)(p^{2}+p+1)$$
Question 4
1 / -0

In an A.P. S$$_3$$ $$= 6$$, S$$_6$$ $$= 3$$, then it's common difference is equal to ?

A
$$3$$

B
$$-1$$

C
$$1$$

D
None of these

Solution

We know that for an A.P., sum of n terms is
$${ S }_{ n }=\dfrac { n }{ 2 } [2a+(n-1)d]$$, where $$a$$ is the first term and $$d$$ is the common difference.
Now, here $${ S }_{ 3 }=6$$ and $${ S }_{ 6 }=3$$
$${ S }_{ 3 }=\dfrac { 3 }{ 2 } [2a+(3-1)d]=6$$
$$=>\dfrac { 3 }{ 2 } [2a+2d]=6$$
$$=>\dfrac { 6 }{ 2 } [a+d]=6$$
$$=>3(a+d)=6$$
$$=>a+d=2$$ -------------(i)
Also,
$${ S }_{ 6 }=\dfrac { 6 }{ 2 } [2a+(6-1)d]=3$$
$$=>3[2a+5d]=3$$
$$=>2a+5d=1$$
$$=>2(2-d)+5d=1$$ (using (i))
$$=>4-2d+5d=1$$
$$=>3d=1-4$$
$$=>d=-1$$
Thus, the common difference is $$-1$$
Question 5
1 / -0

The sum of all $$2$$-digit odd number is

A
$$2475$$

B
$$2530$$

C
$$4905$$

D
$$5049$$

Solution

We have to find the sum $$11+13+15+...+99.$$

Clearly, the terms of this series form an AP with
first term $$a=11$$,

common difference $$d=2$$ and

last term $$l=a_n=99$$

$$\Rightarrow a+(n-1)d=99\Rightarrow 11+(n-1)2=99 \Rightarrow n=45$$

$$\therefore$$ Required sum$$=S_n=\dfrac n2[a+l]$$

$$=\dfrac{45}2[11+99]=45\times55=2475$$

Therefore, option A is correct.
Question 6
1 / -0

The sum of first 24 terms of the sequence whose n$$^{th}$$ term is given by $${a}_{n} = 3 + \displaystyle\frac{2n}{3}$$, is

A
$$278$$

B
$$272$$

C
$$270$$

D
$$268$$

Solution

Given, $${a}_{n} = 3 + \displaystyle\frac{2n}{3}$$

Hence, $${a}_{n+1} = 3 + \displaystyle\frac{2n+2}{3}$$

$$\therefore\ $$ Common difference, $$d={a}_{n+1}-{a}_{n}$$

$$=3 + \displaystyle\frac{2n+2}{3}-3 - \displaystyle\frac{2n}{3}$$

$$=\dfrac{2}{3}$$

First term, $$a = {a}_{1} = 3 + \displaystyle\frac{2}{3}$$
$$ = \displaystyle\frac{11}{3}$$

We know that, sum of $$n$$ terms of an arithmetic progression is
$${S}_{n} = \displaystyle\frac{n}{2}\left[2a + \left(n - 1\right) d\right]$$

$$\Rightarrow {S}_{24} = \displaystyle\frac{24}{2}\left[\displaystyle\frac{22}{3} + 23\left(\displaystyle\frac{2}{3}\right)\right]$$

$$= 12\left[\displaystyle\frac{22 + 46}{3}\right]$$

$$= \displaystyle\frac{12\times 68}{3}$$

$$=4\times 68$$

$$=272$$

Hence, the sum of first $$24$$ terms of the given series is $$272$$.
Question 7
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If $$7^{th}$$ and $$13^{th}$$ terms of an A.P. are 34 and 64, respectively, then its $$18^{th}$$ term is

A
$$87$$

B
$$88$$

C
$$89$$

D
$$90$$

Solution

$$7^{th}$$ term of the A.P. is $$= 34 = a_7$$

$$13^{th} $$ term of the A.P. is $$= 64 = a_{13}$$
Let $$a$$ be the 1st term $$d$$ be the common diff.

$$\therefore a_7 = 34$$

$$\therefore a_7 = 34$$

$$\Rightarrow a+ 6d = 34$$ ____(i)

again

$$a_{13} = 64$$

$$\Rightarrow a+ 12d = 64$$ ___(ii)

(ii) - (i)

$$\Rightarrow 6d = 30$$

$$\Rightarrow d = 5$$

$$\therefore a = 34 - 6d$$

$$= 34 - 30$$

$$= 4$$

$$\therefore a = 4$$

Now, $$18^{th}$$ term $$= a + (18 - 1) d$$

$$= a + 17 d$$

$$ = a + 17 \times 5$$

$$= 4 + 17 \times 5$$

$$= 4 + 85 $$

$$= 89$$

$$\therefore a_{18} = 89$$
Question 8
1 / -0

In an A.P., $$S_1 = 6, S_7 = 105,$$ then $$S_n$$: $$S_{n-3}$$ is same as

A
$$(n + 3) : (n - 3)$$

B
$$(n + 3) : n$$

C
$$n : (n - 3)$$

D
None of these

Solution

Let a and d be the first term and common difference of the given AP respectively.
Given, $$S_1=a=6$$ and
$$S_7=105\Rightarrow \frac72[2a+(7-1)d]=105$$
$$\Rightarrow \frac72[2\times6+6d]=105$$
$$\Rightarrow 6+3d=15$$
$$\Rightarrow d=3$$
Now, $$S_n=\dfrac n2[2a+(n-1)d]=\dfrac n2[12+(n-1)3]=\dfrac {3n}2[n+3]$$ ...(1)
$$S_{n-3}=\dfrac {n-3}2[2a+((n-3)-1)d]=\dfrac {n-3}2[12+(n-4)3]=\dfrac {n-3}2[3n]$$ ...(2)
From (1) and (2), we get
$$\dfrac{S_n}{S_{n-3}}=\dfrac{\dfrac {3n}2[n+3]}{\dfrac {n-3}2[3n]}=\dfrac{n+3}{n-3}$$
Therefore, $$S_n:S_{n-3}::(n+3):(n-3)$$
Question 9
1 / -0

The sum of all two digit numbers which leave remainder $$1$$ when divided by $$3$$ is

A
$$1616$$

B
$$1602$$

C
$$1605$$

D
None of these

Solution

$$\textbf{Step 1:Find first term, common difference and last term of an A.P}$$

$$\text{Clearly, the two digits numbers which leave remainder $$1$$ when divided by $$3$$ are $$10,13,16,...,97$$. }$$

$$\text{This is an AP with first term $$a=10$$, common difference $$d=3$$ and last term $$l=97$$.}$$

$$\textbf{Step 2:Find total number of terms of anA.P}$$

$$\text{Let there be $$n$$ terms in this AP, then}$$

$$a_n=97 = a+(n-1)d$$

$$\therefore 10+(n-1)\times3=97$$

$$10+3n-3=97$$

$$3n=97+3-10$$

$$3n=90$$

$$\therefore n=30$$

$$\textbf{Step 3:Sum of n terms of an A.P}$$

$$\therefore$$ $$\text{Required sum }$$ $$=\displaystyle \frac n2[a+l]=\frac{30}2[10+97]=15\times107=1605$$

$$\textbf{Hence, Option C is correct.}$$
Question 10
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If the $$nth$$ term of the AP $$9, 7, 5..$$ is same as the $$nth$$ term of the AP $$15, 12, 9,...$$, find $$n.$$

A
$$7$$

B
$$8$$

C
$$9$$

D
$$10$$

Solution

$$n th$$ term of A.P $$9,7,5....=a+(n-1)d$$
$$\Rightarrow 9+(n-1)-2$$
$$\Rightarrow 9-2n+2\Rightarrow 11-2n$$
$$n th$$ term of A.P $$15,12,9,....=a+(n-1)d$$
$$\Rightarrow 15+(n-1)-3$$
$$\Rightarrow 15-3n+3\Rightarrow 18-3n$$
$$\therefore 11-2n=18-3n$$
$$\Rightarrow 3n-2n=18-11$$
$$\Rightarrow n=7$$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 38

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Arithmetic Progressions Test - 38

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SHARING IS CARING

Question 1

$$\displaystyle\frac{1}{2}(m + n)$$

$$\displaystyle\sqrt{mn}$$

$$m + n$$

$$mn$$

Solution

Question 2

$$6$$

$$-6$$

$$3$$

None of these

Solution

Question 3

$$(p + 1)$$$$^2$$

$$(2p + 1) (p^2+p + 1)$$

$$(p + 1)$$$$^3$$

$$p$$$$^3$$$$ + (p + 1)$$$$^3$$

Solution

Question 4

$$3$$

$$-1$$

$$1$$

None of these

Solution

Question 5

$$2475$$

$$2530$$

$$4905$$

$$5049$$

Solution

Question 6

$$278$$

$$272$$

$$270$$

$$268$$

Solution

Question 7

$$87$$

$$88$$

$$89$$

$$90$$

Solution

Question 8

$$(n + 3) : (n - 3)$$

$$(n + 3) : n$$

$$n : (n - 3)$$

None of these

Solution

Question 9

$$1616$$

$$1602$$

$$1605$$

None of these

Solution

Question 10

$$7$$

$$8$$

$$9$$

$$10$$

Solution

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