Arithmetic Progressions Test

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Arithmetic Progressions Test - 55

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Question 1
1 / -0

The tenth term of an A.P. : $$\sqrt {2}, 3\sqrt {2}, 5\sqrt {2}, 7\sqrt {2}$$, ..... is:

A
$$10\sqrt {2}$$

B
$$12$$

C
$$19\sqrt {2}$$

D
$$11\sqrt {2}$$

Solution

Given A.P. is $$\sqrt {2}, 3\sqrt {2}, 5\sqrt {2}, 7\sqrt {2}....$$

Its first term, $$a = \sqrt {2}$$

Common difference, $$d = 3\sqrt {2} - \sqrt {2} = 2\sqrt {2}$$

To find out: The tenth term of the A.P.

We know that, the $$n^{th}$$ term of an A.P. is $$t_n=a+(n-1)d$$

$$\therefore \ T_{10} = a + (10 - 1)d$$

$$\Rightarrow a + 9d$$

$$\Rightarrow \sqrt {2} + 9(2\sqrt {2})$$

$$\Rightarrow \sqrt {2} + 18\sqrt {2}$$

$$\therefore \ t_{10}= 19\sqrt {2}$$

Hence, the tenth term of the given A.P. is $$19\sqrt {2}$$.
Question 2
1 / -0

Match the APs given in column A with suitable common differences given in column B.

Column A Column B

A. $$-2, 2, 6, 10,...$$ 1. $$\dfrac{2}{3}$$

B. $$a = 18, n = 10, a_n = 0$$ 2. $$-5$$

C. $$a = 0, a_{10} = 6$$ 3. $$4$$

D. $$a_2 = 13, a_4 =3$$ 4. $$-4$$

5. $$-2$$

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

7. $$5$$

A
A-3, B-5, C-1, D-2

B
A-4, B-1, C-3, D-2

C
A-2, B-7, C- 5, D-6

D
A-7, B-4, C-2, D- 2

Solution

1. Given sequence $$A:-2,\ 2,\ 6,\ 10,\ ...$$
Here, first term $$a=-2$$ and common difference $$d=6-2=4$$
$$\therefore\ A\rightarrow 3$$

2. Given, $$B:a=18,n=10,a_n=0$$
Now, $$a_n=a+(n-1)d=0$$
$$\Rightarrow 9d=-18$$
$$\Rightarrow d=-2$$
$$\therefore\ B\rightarrow 5$$

3. Given, $$C:a=0,a_{10}=6$$
$$\Rightarrow a+9d=6$$
$$\Rightarrow d=\dfrac69$$
$$\Rightarrow d=\dfrac23$$
$$\therefore C\rightarrow1$$

4. Given, $$D:a_2=13,a_{4}=3$$
$$\Rightarrow a+d=13$$ ...(i)
$$\Rightarrow a+3d=3$$ ...(ii)
On subtracting (i) from (ii), we get
$$2d=-10$$
$$\Rightarrow d=-5$$
$$\therefore D\rightarrow 2$$
Question 3
1 / -0

Find the sum of the first $$15$$ terms of the sequences having $$n$$th term as
$${ x }_{ n }=6-n$$

A
$$30$$

B
$$-30$$

C
$$20$$

D
$$-20$$

Solution

Formula,

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

Given,

$$a_n=6-n$$

$$a_1=6-1=5$$

$$a_{2}=6-2=4$$

$$d= 4-5 = -1 $$

$$S_{15} $$= $$\frac{15}{2}[2(5)+(15-1) \times (-1)]$$

$$\therefore S_{15}=-30$$
Question 4
1 / -0

Identify which of the following list of numbers is an arithmetic progression?

A
$$1,1,2,3,5,...$$

B
$$2,3,5,7,11,...$$

C
$$10,100,1000,...$$

D
$$12,18,24,30,...$$

Solution

The sequence given in option '$$a$$' to '$$b$$' do not have a fixed difference between its consecutive numbers. Only the sequence $$12,18,24, 30,...$$ has a difference of $$6$$ between its two consecutive numbers. Thus the correct answer is option '$$(D)$$'.
Question 5
1 / -0

If $$7th$$ and $$13th$$ terms of an $$A.P$$. Be $$34$$ and $$64$$, respectively, then its $$18th$$ terms is:

A
$$87$$

B
$$88$$

C
$$89$$

D
$$90$$

Solution

The general term of an $$AP$$ is given by $$a_n = a+(n-1)d$$, where $$ a $$ & $$ d $$ are the first and common difference of an $$AP$$ respectively.
$${a}_{7}=34$$
$${a}_{13}=64$$
$$a+6d=34$$
$$a+12d=64$$
$$-6d=-30$$
$$d=\cfrac{30}{6}=5$$
$$\therefore a= 4.$$
$${a}_{18}=a+17d=4+17(5)=89$$
Question 6
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If two terms of an arithmetic progression are known, then the two terms can be represented using which of the formula below?

A
$$t_{n1} = a - (n_1-1)d\ $$ and $$\ t_{n2} = a - (n_2-1)d$$.

B
$$t_{n1} = 2a + (n_1-1)d$$ and $$t_{n1} = 2a + (n_2-1)d$$.

C
$$t_{n1} = a + (n_1-1)d$$ and $$t_{n1} = a + (n_2-1)d$$.

D
$$t_{n1} = a + (n_1-2)d$$ and $$t_{n1} = a + (n_2-2)d$$.

Solution

If two terms of an arithmetic progression are known, then the two terms can be represented using $$t_n = a + (n-1)d$$ as the values for tn will be known.
Question 7
1 / -0

The third term of an arithmetic progression is $$18$$, and the seventh term is $$30$$, then the sum of $$17$$ terms is

A
$$612$$

B
$$600$$

C
$$656$$

D
None of these

Solution

Let $$T_n$$ be the $$n^{th}$$ term of the AP.

$$T_{n} = a+(n-1)d$$

Where $$a$$ and $$d$$ are first term and common difference respectively.

According to the given condition

$$T_{3} = a+(3-1)d$$

$$T_{3} = a+2d$$

$$18 = a+2d$$........$$(1)$$

$$T_{7} = a+(7-1)d$$

$$T_{7} = a+6d$$

$$30 = a+6d$$........$$(2)$$

solving $$(1)$$ and $$(2)$$ we get,

$$a = 12$$

$$d = 3$$

Sum of $$n$$ terms of an AP is given as,

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

$$S_{17}= \dfrac {17}{2} [2(12)+(17-1)3]$$

$$S_{17}=612$$

Hence option (A) is correct option
Question 8
1 / -0

Sum of the first $$20$$ odd natural numbers is

A
$$400$$

B
$$40$$

C
$$200$$

D
$$0$$

Solution

An odd number is one that is not divisible by $$2$$ and the first odd natural number is $$1.$$ Since two consecutive odd numbers differ by two, the first $$20$$ odd natural numbers are

$$1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39$$

which is an A.P. with the first term $$a=1,$$ the common difference $$d=2 $$and total number of terms $$n = 20.$$ Applying the general formula for the sum of an A.P.

$$S_{n}= \dfrac{n}{2} [{2a + (n–1)d}] $$

$$\Rightarrow S_{20}= \dfrac{20}{2} [2(1)+(20-1)(2)]$$

$$\Rightarrow S_{20}= 10(2+38) $$

$$\Rightarrow S_{20}= 400$$
Question 9
1 / -0

The sum of first $$p$$ terms of an arithmetic progression is $$q$$, and the sum of first $$q$$ term is $$p$$, then the sum of first $$p + q$$ terms is

A
$$(p + q)$$

B
$$-(p + q)$$

C
$$(q - p)$$

D
$$(p - q)$$
Question 10
1 / -0

The first term of an $$A.P.$$ is $$10$$ and its $$41^{st}$$ term is $$81$$. Find the sum of the series, if $$81$$ is the last term.

A
$$1025$$

B
$$1865.5$$

C
$$858$$

D
$$962$$

Solution

Given, first term, $$a=10$$
last term, $$l=81$$
total number of terms, $$n=41$$

Sum, $$S_n=\dfrac{n}{2}[a+l]$$
$$=\dfrac{41}{2}[10+81]$$
$$=\dfrac{41}{2}\times91$$
$$=1865.5$$

Hence, the sum of the series is $$1865.5$$

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 55

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Arithmetic Progressions Test - 55

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SHARING IS CARING

Question 1

$$10\sqrt {2}$$

$$12$$

$$19\sqrt {2}$$

$$11\sqrt {2}$$

Solution

Question 2

A-3, B-5, C-1, D-2

A-4, B-1, C-3, D-2

A-2, B-7, C- 5, D-6

A-7, B-4, C-2, D- 2

Solution

Question 3

$$30$$

$$-30$$

$$20$$

$$-20$$

Solution

Question 4

$$1,1,2,3,5,...$$

$$2,3,5,7,11,...$$

$$10,100,1000,...$$

$$12,18,24,30,...$$

Solution

Question 5

$$87$$

$$88$$

$$89$$

$$90$$

Solution

Question 6

$$t_{n1} = a - (n_1-1)d\ $$ and $$\ t_{n2} = a - (n_2-1)d$$.

$$t_{n1} = 2a + (n_1-1)d$$ and $$t_{n1} = 2a + (n_2-1)d$$.

$$t_{n1} = a + (n_1-1)d$$ and $$t_{n1} = a + (n_2-1)d$$.

$$t_{n1} = a + (n_1-2)d$$ and $$t_{n1} = a + (n_2-2)d$$.

Solution

Question 7

$$612$$

$$600$$

$$656$$

None of these

Solution

Question 8

$$400$$

$$40$$

$$200$$

$$0$$

Solution

Question 9

$$(p + q)$$

$$-(p + q)$$

$$(q - p)$$

$$(p - q)$$

Question 10

$$1025$$

$$1865.5$$

$$858$$

$$962$$

Solution

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