Arithmetic Progressions Test

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Arithmetic Progressions Test - 35

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Question 1
1 / -0

$$10^{th}$$ term of AP: $$2, 7, 12, .....$$ is

A
$$35$$

B
$$47$$

C
$$55$$

D
None of the above

Solution

First term $$(a_1)= 2$$
difference $$(d)= 5$$

The $$10$$th term of A.P is
$$a_{10} = 2 +(10-1)\times(5)$$
$$a_{10} = 2 +(9)\times(5)$$
$$a_{10} = 2 + 45$$
$$a_{10} = 47$$
Question 2
1 / -0

The sum to $$200$$ terms of the series $$1+4+6+5+11+6+.....$$ is

A
$$29,600$$

B
$$29,800$$

C
$$30,200$$

D
None of these

Solution

Given, series: $$1+4+6+5+11+6....$$ is a combination of two AP's i.e.,
$$(1 + 6 + 11 + .....) + (4 + 5 + 6......)$$
Now, each of the AP will have 100 terms.
$$\Longrightarrow \quad { S=S }_{ 1 }+{ S }_{ 2 }\quad where{ \quad S }_{ 1 }=1+6+11...\quad and\quad { S }_{ 2 }=4+5+6...\quad \\ Now,{ S }_{ 1 }=\dfrac { n }{ 2 } [2a+(n-1)d]=\dfrac { 100 }{ 2 } [2\times 1+(100-1)(5)]\\ \Longrightarrow { S }_{ 1 }=50(497)=24850\\ Again,{ S }_{ 2 }=\dfrac { n }{ 2 } [2a+(n-1)d]=\dfrac { 100 }{ 2 } [2\times 4+(100-1)(1)]\\ \Longrightarrow { S }_{ 1 }=50(107)=5350\quad \\ \therefore { \quad S=S }_{ 1 }+{ S }_{ 2 }=24850+5350=30,200$$
Hence, the sum of 200 terms to the given series is 30,200.
Question 3
1 / -0

The sum of $$11$$ terms of an A.P. whose middle term is $$30,$$ is

A
$$320$$

B
$$330$$

C
$$340$$

D
$$350$$

Solution

$$\textbf{Step -1: Formulating for middle term.}$$
$$\text{Given that, an A.P. of 11 terms, whose middle term is 30.}$$
$$\text{Middle term = }6^{th}\text{ term.}$$
$$\therefore a_n = a+ (n-1)d$$
$$\text{Where, }a_n\text{ is the }n^{th}\text{ term.}$$
$$\text{where }a \text{ is the first term and }d\text{ is the common difference.}$$
$$a_6 = a + (6-1)d = 30$$
$$\Rightarrow a + 5d = 30$$ $$...\text{(i)}$$
$$\textbf{Step -2: Finding the sum of 11 terms.}$$
$$S_{n} = \dfrac{n}{2}(2a + (n-1)d)$$
$$\therefore S_{11} = \dfrac{11}{2}(2a + (11-1)d)$$
$$\Rightarrow S_{11} = \dfrac{11}{2}(2a + 10d)$$
$$\Rightarrow S_{11} = \dfrac{11}{2}\times 2(a + 5d)$$
$$\text{From (i), }a + 5d = 30$$
$$\therefore S_{11} = 11\times 30$$
$$\Rightarrow S_{11} = 330$$
$$\textbf{Hence option B is correct.}$$
Question 4
1 / -0

If the sum of the series $$54+51+48+.....$$ is $$513$$, then the number of terms are

A
$$18$$

B
$$20$$

C
$$17$$

D
None of these

Solution

Sum of an AP = $${\dfrac{n}{2}}{(2a+(n-1)\times d)}$$ = 513
where $$a= 54$$ and $$d=-3$$
$${\dfrac{n}{2}}{(2\times54+(n-1)\times(-3))}$$ = 513
$${\dfrac{n}{2}}{(111-3n)}$$ = 513
$$111n-3n^2 = 1026$$
Solving for n,
$$n=18$$
Question 5
1 / -0

There are $$60$$ terms in an A.P. of which the first term is $$8$$ and the last term is $$185.$$ The $$31^{st}$$ term is

A
$$56$$

B
$$94$$

C
$$85$$

D
$$98$$

Solution

First term of given AP $$a= 8$$
$$60^{th}$$ term of AP $$t_{60}= 185$$
Formula for $$n^{th}$$ term $$t_{n}=a+(n-1)d$$
Hence, $$185= 8 + (60-1)d$$
$$\Rightarrow 177= 59d$$
$$\Rightarrow d= 3$$
From $$n^{th}$$ term formula, $$31^{st} $$ term $$= 8 + (31-1)\times3$$
Thus , $$t_{31} = 98$$
Question 6
1 / -0

If the sum of the series $$2+5+8+11.....$$ is $$60100$$, then the number of terms are

A
$$100$$

B
$$200$$

C
$$250$$

D
$$300$$

Solution

Given AP: $$2+5+8+11+...$$

Here, first term$$,\ a=2$$

Common difference$$,\ d=3$$

Given: Sum of terms of the AP $$=60100$$

We know that sum of terms of an AP $$={\dfrac{n}{2}}{(2a+(n-1)\times d)}$$

$$\Rightarrow {\dfrac{n}{2}}{(2\times2+(n-1)\times3)}=60100$$

$$\Rightarrow {\dfrac{n}{2}}{(4+3n-3)}=60100$$

$$\Rightarrow {\dfrac{n}{2}}{(3n+1)}=60100$$

$$\Rightarrow 3n^2 + n = 120200$$

$$\Rightarrow 3n^2+n-120200=0$$

$$\Rightarrow 3n^2+601n-600n-120200=0$$

$$\Rightarrow n(3n+601)-200(3n+601)=0$$

$$\Rightarrow (n-200)(3n+601)=0$$

$$\Rightarrow n= 200$$ or $$n=\dfrac{-601}{3}$$

Since $$n$$ cannot be negative, hence $$n=200$$
Question 7
1 / -0

Following two given series are in A.P.
$$2, 4, 6, 8 ...$$$$3, 6, 9, 12 ...$$
First series contains $$30$$ terms, while the second series contains $$20$$ terms. Both of the above given series contains some terms, which are common to both of them.The sum of both the above given A.P. are respectively:

A
$$(930, 630)$$

B
$$(630, 930)$$

C
$$(870, 580)$$

D
$$(580, 870)$$

Solution

The given series are $$2,4,6,8,....$$ and $$3,6,9,12,....$$
For first AP: $$a= 2, d= 2, n=30$$
Sum $$=$$ $${\dfrac{n}{2}}{(2a +(n-1)d)}$$
$$=$$ $${\cfrac{30}{2}}{(4+29\times2)}$$
$$=$$ $${\cfrac{30}{2}}{(4+29\times2)}$$
$$=$$ $$31\times30$$
$$= 930$$

For second AP: $$a= 3, d= 3, n=20$$
Sum $$=$$ $${\cfrac{n}{2}}{(2a +(n-1)d)}$$
$$=$$ $${\cfrac{20}{2}}{(6+19\times3)}$$
$$=$$ $${\dfrac{20}{2}}{(6+19\times3)}$$
$$=$$ $$10\times63$$
$$= 630$$
Question 8
1 / -0

The sum of $$12$$ terms of an A.P., whose first term is $$4,$$ is $$256.$$ What is the last term?

A
$$35$$

B
$$36$$

C
$$37$$

D
None

Solution

Given that, for an A.P., $$n=12,\ a=4$$ and $$S_{12}=256$$
To find out: The last term of the A.P.
We know that, sum of $$n$$ terms of an A.P. is $$S_{n}=\dfrac {n}{2}[2a+(n-1)d]$$
$$\therefore \ 256=\dfrac {12}{2}[2\times 4+(12-1)d]\\$$
$$\Rightarrow 256=6[8+11d]\\$$
$$\Rightarrow 256=48+66d\\$$
$$\Rightarrow 208=66d\\$$
$$\Rightarrow d=\dfrac {104}{33}\\$$
We also know that, the $$n^{th}$$ term of an A.P. is $$t_n=a+(n-1)d$$
$$\therefore \ t_{12}=4+(12-1)\dfrac {104}{33}\\$$
$$\Rightarrow 4+11\times \dfrac {104}{33}\\$$
$$\Rightarrow \dfrac {12+104}{3}\\$$
$$\therefore \ t_{12}= \dfrac {116}{3}\\$$

Hence, the last term of the given A.P. is $$\dfrac {116}{3}$$ and therefore, option D is correct.
Question 9
1 / -0

A ladder has rungs $$25$$ cm apart. The rungs decrease uniformly in the length from $$45$$ cm at the bottom to $$25$$ cm at the top. If the top and the bottom rungs are $$\displaystyle 2 \frac{1}{2}$$m apart, what is the length of the wood required for the rungs

A
$$280$$ cm

B
$$320$$ cm

C
$$250$$ cm

D
$$385$$ cm

Solution

The distance between the top and the bottom rungs $$=2\dfrac {1}{2}$$ m $$=250$$ cm and the distance between two consecutive rungs $$=25$$ cm.

Therefore, the number of gaps between the rungs $$=\dfrac {250}{25}=10$$.

So, the total number of rungs in the ladder $$=11$$

The length of the bottom rung is $$45$$ cm and the lengths of rungs decrease uniformly from bottom to top.

Therefore, the numbers involved in the lengths (in cm) of the rungs from bottom to top form an AP with $$a=45, l=25$$ and $$n=$$ total number of rungs $$=11$$.
Using the formula for sum of an A.P. to find the total length of the woods required $$S_n$$:

$$S_n=\dfrac {n}{2}(a+l)$$

$$=\dfrac {11}{2}(45+25)$$

$$=\dfrac {11}{2}\times 70$$

$$=11\times 35=385$$

Hence, the length of the wood required for rungs $$=385$$ cm.
Question 10
1 / -0

Find the sum of all natural numbers not exceeding $$1000,$$ which are divisible by $$4$$ but not by $$8.$$

A
$$62500$$

B
$$62800$$

C
$$64000$$

D
$$65600$$

Solution

The natural numbers that are divisible by 4 but not 8 below 1000 are:

$$4,\ 12,\ 20,\ 28,\ ... ,\ 996$$

These numbers form an AP

Here $$a=4, d=8,$$

And, $$a_n = 996$$

As we know for an AP, $$a_n = a+(n-1)d$$

$$\Rightarrow 996 = 4+(n-1) \times 8$$

$$\Rightarrow 8n - 8 + 4 = 996$$

$$\Rightarrow n =125$$

For the terms in an AP: $$ S_n = \dfrac {n}{2} \times [2a+(n-1)d] $$

$$\therefore\ \ S_{125} = \dfrac{125}{2} [2 \times 4 + (125 - 1) \times 8]$$

$$=\dfrac{125}{2}\times [8+992]$$

$$=\dfrac{125}{2}\times 1000$$

$$= 62500$$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 35

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Arithmetic Progressions Test - 35

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SHARING IS CARING

Question 1

$$35$$

$$47$$

$$55$$

None of the above

Solution

Question 2

$$29,600$$

$$29,800$$

$$30,200$$

None of these

Solution

Question 3

$$320$$

$$330$$

$$340$$

$$350$$

Solution

Question 4

$$18$$

$$20$$

$$17$$

None of these

Solution

Question 5

$$56$$

$$94$$

$$85$$

$$98$$

Solution

Question 6

$$100$$

$$200$$

$$250$$

$$300$$

Solution

Question 7

$$(930, 630)$$

$$(630, 930)$$

$$(870, 580)$$

$$(580, 870)$$

Solution

Question 8

$$35$$

$$36$$

$$37$$

None

Solution

Question 9

$$280$$ cm

$$320$$ cm

$$250$$ cm

$$385$$ cm

Solution

Question 10

$$62500$$

$$62800$$

$$64000$$

$$65600$$

Solution

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