Arithmetic Progressions Test

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Arithmetic Progressions Test - 36

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Question 1
1 / -0

$$30$$ trees are planted in a straight line at intervals of $$5\ m.$$ To water them, the gardener needs to bring water for each tree, separately from a well, which is $$10\ m$$ from the first tree in line with the trees. How far will he have to walk in order to water all the trees beginning with the first tree? Assume that he starts from the first well, and he can carry enough water to water only one tree at a time.

A
$$4785\ m$$

B
$$4795\ m$$

C
$$4800\ m$$

D
None of these

Solution

To water the first tree he has to walk $$= 10$$ metres
To water the second tree he has to walk $$= 25$$ metres
To water the third tree he has to walk $$= 35$$ metres
To water the fourth tree he has to walk $$= 45$$ metres

So, if we observe the above distances except for the first they form an arithmetic progression,
$$25,35,45,55,\dots\ \ \text{upto 29 terms}$$

Let the total distance travelled by the gardener be $$d.$$ So,
$$d= 10 + 25 + 35 +\dots$$ to $$30$$ terms
$$= 10 + (25+35+\dots29 \ \text{terms})$$
$$=10 + \dfrac{29}{2}\left[2\times25 + (29-1)\times10\right]$$
$$=10 + \dfrac{29}{2}\left(50 + 28\times10\right)$$
$$=10 + 4785$$
$$=4795\ m$$

Hence, option $$B$$ is correct.
Question 2
1 / -0

Identify the progression:
$$A : 4, 7, 10, 13, 16, 19, 22, 25, .....$$
$$B : 4, 7, 9, 10, 13, 14, .........$$

A
Only $$A$$

B
Only $$B$$

C
Both $$A$$ and $$B$$

D
None of these

Solution

Only sequence $$A$$ is progression as its terms follow same pattern
Question 3
1 / -0

Let $$S_n$$ denote the sum of the first $$'n'$$ terms of an A.P. $$S_{2n} = 3S_n$$. Then, the ratio $$\dfrac{S_{3n}}{S_n}$$ is equal to :

A
$$4$$

B
$$6$$

C
$$8$$

D
$$10$$

Solution

We know, $$S_n = \dfrac{n}{2}[2a+(n-1)d]$$
[where $$a$$ is the first term and $$d$$ is the common difference]
$$S_{2n}=\dfrac{2n}{2}[2a+(2n-1)d]$$
$$S_{3n}=\dfrac{3n}{2}[2a+(3n-1)d]$$
Given, $$S_{2n}=3S_n$$
$$\Rightarrow n[2a+2nd-d]$$
$$=3 \left [ \dfrac{n}{2} (2a+nd-d) \right ]$$
Therefore, $$\displaystyle d(n+1)= 2a$$

The value of ratio, $$\frac{S_{3n}}{S_n}= \frac{{\frac{3n}{2}}{(2a + (3n-1)d)}}{{\frac{n}{2}}{(2a + (n-1)d)}}$$
= $$\frac{{\frac{3n}{2}}{(nd+d + (3n-1)d)}}{{\frac{n}{2}}{(nd+d + (n-1)d)}}$$
= $$\frac{{\frac{3n}{2}}{(4nd)}}{{\frac{n}{2}}{(2nd)}}$$
= 6
Question 4
1 / -0

If $$\displaystyle \frac{3+5+7+......+ n(terms)}{5+8+11+.....+ 10(terms)}=7$$, then the value of $$n$$ is

A
$$35$$

B
$$36$$

C
$$37$$

D
$$40$$

Solution

$$S_n =$$ Sum of n terms of an A.P.

$$\displaystyle = \frac{n}{2} [2a + (n-1)d]$$

where $$a=$$ first term

$$d=$$ common difference

$$\therefore \displaystyle \frac{3+5+7+ ...... + n terms}{5+8+11+...... + 10 terms}=7$$

$$\Rightarrow \displaystyle \dfrac{\dfrac{n}{2}[2\times 3 + (n-1) \times 2]}{\dfrac{10}{2} [2 \times 5 + (10-1) \times 3]}=7$$

$$\Rightarrow \displaystyle \frac{n(2n+4)}{370}=7$$

$$\Rightarrow 2n^2 + 4n - 2590 = 0$$

$$\Rightarrow n^2+2n-1295=0$$

$$\Rightarrow n^2+37n-35n-1295=0$$

$$\Rightarrow n(n+37)-35(n+37)=0$$

$$\Rightarrow (n-35)(n+37)=0$$

$$\Rightarrow n=35$$
Question 5
1 / -0

The sum up to $$9$$ terms of the series $$\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{6}+ ...$$ is

A
$$\displaystyle -\frac{5}{6}$$

B
$$\displaystyle -\frac{1}{2}$$

C
$$1$$

D
$$\displaystyle -\frac{3}{2}$$

Solution

Given that series is an A.P with first term, $$a=\dfrac{1}{2} $$ and
Common difference, $$d=\dfrac{1}{3} -\dfrac{1}{2} =\dfrac{1}{6}-\dfrac{1}{3} =\dfrac{-1}{6} $$
So using sum of A.P formula $$S_n=\dfrac{n}{2} ( 2a + ( n - 1 ) d )$$
Here $$n =9 \Rightarrow S_9=\dfrac{9}{2}\left ( 2. \dfrac{1}{2} + ( 9 - 1 ).\dfrac{-1}{6}\right )=\dfrac{-3}{2} $$
Question 6
1 / -0

If $$S=\cfrac { n }{ 2 } [2a+(n-1)d]$$ ; find $$d$$ , when $$a=8, S=380$$ and $$n=10$$.

A
$$\displaystyle 7 \frac{1}{6}$$

B
$$\displaystyle 4 \frac{5}{2}$$

C
$$\displaystyle 2\dfrac{1}{6}$$

D
$$\displaystyle 6 \dfrac{2}{3}$$

Solution

We know, $$S=\dfrac { n }{ 2 } [2a+(n-1)d]$$

Given, $$ a=8, S=380 $$ and $$n=10$$

$$\therefore 380= \dfrac { 10 }{ 2 } (2(8)+9d)$$

$$ 380= 5(16 +9d)$$

$$ 300= 45d$$

$$ \therefore d =\dfrac { 300 }{ 45 }$$ $$= 6\dfrac { 30 }{ 45 } $$ $$= 6\dfrac { 2 }{ 3 } $$
Question 7
1 / -0

A sprinter runs 6 meters in the first second of a certain race and increase her speed by 25 cm/sec. in each succeeding second. (This means that she goes 6m 25 cm. the second second, 6m 50 cm. the third second, and so on.) How far does she go during the eight second?

A
8.75 m

B
7.75 m

C
8.25 m

D
9.25 m

Solution

Since the sprinter increase her speed by $$ 25cm/sec $$, so we have an AP corresponding to givem problem as below
$$ 6, 6.25, 6.50, 6.75, ....... $$
$$\therefore$$ $$ a = 6, d= 0.25, n = 8 $$
Required distance $$= 6 + (8-1) \times 0.25$$
$$=6 + 1.75 = 7.75 m$$
Question 8
1 / -0

The $$p^{th}$$ and $$q^{th}$$ terms an $$A.P$$ are respectively $$a$$ and $$b.$$ Then sum of $$(p+q)$$ terms is

A
$$\dfrac {p+q}{2}\left(a+b+\dfrac {a-b}{p-q}\right)$$

B
$$\dfrac {p+q}{2}\left(a-b-\dfrac {a+b}{p+q}\right)$$

C
$$ (p+q)\left(a+b+\dfrac {p-q}{a-b}\right)$$

D
$$\dfrac {p+q}{2}\left(a+b+\dfrac {p-q}{a-b}\right)$$

Solution

Let the first term of the A.P be '$$c$$' and its common difference be '$$d$$'.

So its $$p^{th}$$ term is $$c+(p-1)d=a$$            $$...(1)$$

and its $$q^{th}$$ term is $$c+(q-1)d=b$$          $$...(2)$$

Now, add equation $$(1)$$ and $$(2)$$; we get

$$\Rightarrow 2c+(p+q-2)d=a+b$$

$$\Rightarrow 2c+(p+q-1)d-d=a+b$$

$$\Rightarrow 2c+(p+q-1)d=a+b+d$$         $$...(3)$$

Now, subtracting $$(1)$$ from $$(2)$$; we get

$$\Rightarrow (p-q)d=a-b$$

$$\Rightarrow d=\dfrac { (a-b) }{ (p-q) } $$      $$...(4)$$

Now, the sum of $$(p+q)$$ terms is

$$S_{(p+q)}=\dfrac { (p+q) }{ 2 } \left[ 2c+(p+q-1)d \right] $$ $$...(5)$$

Substitute equation $$(3)$$ and $$(4)$$ in $$(5)$$; we get

$$S_{(p+q)}=\dfrac { (p+q) }{ 2 } \left[ a+b+d \right] $$

$$S_{(p+q)}=\dfrac { (p+q) }{ 2 } \left[ a+b+\dfrac { (a-b) }{ (p-q) } \right] $$
Question 9
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STATEMENT - $$1$$ : The sum of first $$11$$ terms of the A.P: $$2, 6, 10, 14,\dots$$ is $$242.$$
STATEMENT - $$2$$ : The sum of first $$n$$ terms of the A.P. is given by $$S_n = \dfrac{n}{2} [2a + (n-1)d]$$

A
Statement - $$1$$ is True, Statement - $$2$$ is True, Statement - $$2$$ is a correct explanation for Statement - $$1$$

B
Statement - $$1$$ is True, Statement - $$2$$ is True : Statement $$2$$ is NOT a correct explanation for Statement - $$1$$

C
Statement - $$1$$ is True, Statement - $$2$$ is False

D
Statement - $$1$$ is False, Statement - $$2$$ is True

Solution

Checking Statement $$1$$:
Given series: $$2, 6, 10, 14,\dots$$
$$n=11$$
$$a=2$$
$$d=6-2=4$$
$$\therefore S_{11}=\dfrac{11}{2}[2.2+(11-1)4]=242$$
$$\therefore $$ Statement $$1$$ is True
$$\therefore $$ Statement $$2$$ is also True (general formula)
Statement $$2$$ is used to prove Statement $$1$$
Hence, statement $$2$$ is correct explanation for statement $$1.$$
Question 10
1 / -0

The sum of 24 terms of the series $$1, -3, 4, -6, 7, -9, 10, ...$$ is

A
24

B
-24

C
16

D
-16

Solution

The given series S can be divided into two sub-series $$S_{1} $$ and $$ S_{2} $$
where $$ S_1 = 1,4,7,10..$$ and $$ S_2 = -3,-6,-9,... $$
24 terms of S would consist of first 12 terms of $$S_1$$ and first 12 terms of $$S_2$$
$$ \therefore$$ Sum of $${S} $$= Sum of $${S_1} $$+ Sum of $$ {S_2} $$
So using sum of A.P formula $$S_n=\dfrac{n}{2} ( 2a + ( n - 1 ) d )$$
$$ = \frac{12}{2}(2\times1+11\times3)+ \frac{12}{2}(2\times-3+11\times-3) $$
$$ = 210 -234 = -24 $$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 36

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Arithmetic Progressions Test - 36

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SHARING IS CARING

Question 1

$$4785\ m$$

$$4795\ m$$

$$4800\ m$$

None of these

Solution

Question 2

Only $$A$$

Only $$B$$

Both $$A$$ and $$B$$

None of these

Solution

Question 3

$$4$$

$$6$$

$$8$$

$$10$$

Solution

Question 4

$$35$$

$$36$$

$$37$$

$$40$$

Solution

Question 5

$$\displaystyle -\frac{5}{6}$$

$$\displaystyle -\frac{1}{2}$$

$$1$$

$$\displaystyle -\frac{3}{2}$$

Solution

Question 6

$$\displaystyle 7 \frac{1}{6}$$

$$\displaystyle 4 \frac{5}{2}$$

$$\displaystyle 2\dfrac{1}{6}$$

$$\displaystyle 6 \dfrac{2}{3}$$

Solution

Question 7

8.75 m

7.75 m

8.25 m

9.25 m

Solution

Question 8

$$\dfrac {p+q}{2}\left(a+b+\dfrac {a-b}{p-q}\right)$$

$$\dfrac {p+q}{2}\left(a-b-\dfrac {a+b}{p+q}\right)$$

$$ (p+q)\left(a+b+\dfrac {p-q}{a-b}\right)$$

$$\dfrac {p+q}{2}\left(a+b+\dfrac {p-q}{a-b}\right)$$

Solution

Question 9

Statement - $$1$$ is True, Statement - $$2$$ is True, Statement - $$2$$ is a correct explanation for Statement - $$1$$

Statement - $$1$$ is True, Statement - $$2$$ is True : Statement $$2$$ is NOT a correct explanation for Statement - $$1$$

Statement - $$1$$ is True, Statement - $$2$$ is False

Statement - $$1$$ is False, Statement - $$2$$ is True

Solution

Question 10

24

-24

16

-16

Solution

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