Arithmetic Progressions Test

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Arithmetic Progressions Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

The sum of the third and ninth terms of an A.P is $$8$$. Find the sum of the first $$11$$ terms of the progression.

A
$$44$$

B
$$22$$

C
$$19$$

D
$$18$$

E
None of these

Solution

As we know that,
$$n^{th}$$ term, $$a_n = a+(n-1)d$$

Sum of first $$n$$ terms, $$S_n = \dfrac{n}{2} (2a+(n-1)d)$$, where $$a$$ & $$d$$ are the first term & common difference of an AP.
It is given that sum of third and ninth terms is $$8$$.
Therefore,
$$(a+2d)+(a+8d)=8$$
$$\Rightarrow 2a+10d=8$$ ...(1)

Therefore, sum of first $$11$$ terms:
$$S_{11}=\dfrac{11}2[2a+(11-1)d]$$

$$\Rightarrow S_{11}=\dfrac{11}2[2a+10d]$$

$$\Rightarrow S_{11}=\dfrac{11}2[8]$$ {From (1)}

$$\Rightarrow S_{11}=44$$

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

What is the sum of all positive integers up to $$1000$$, which are divisible by $$5$$ and are not divisible by $$2$$?

A
$$10,050$$

B
$$5050$$

C
$$5000$$

D
$$50,000$$

Solution

The positive integers, which are divisible by $$5$$, are $$5, 10, 15, \dots, 1000.$$ Out of these $$10,20,30,\dots, 1000$$ are divisible by $$2.$$

Thus, we have to find the sum of the positive integers $$5, 15, 25, ...., 995$$
Clearly above sequence is an AP with the first term $$5$$ and the common difference of $$10.$$

If $$n$$ is the number of terms in this sequence then,
$$\begin{aligned}{}5 + 10\left( {n - 1} \right) &= 995\\5 + 10n - 10 &= 995\\10n& = 1000\\n &= 100\end{aligned}$$

Now by using the formula for the sum of the first $$n$$ terms of an AP,
$$\begin{aligned}{}{S_{100}} &= \frac{{100}}{2}\left[ {2 \times 5 + \left( {100 - 1} \right)10} \right]\\ &= 50\left( {10 + 990} \right)\\& = 50 \times 1000\\ &= 50000\end{aligned}$$
Question 3
1 / -0

If the $$5th$$ and $$8th$$ term of an AP are $$6$$ and $$15$$ respectively, find the $$19th$$ term

A
$$48$$

B
$$54$$

C
$$19$$

D
$$57$$

Solution

Given: $$a_5=6$$ and $$a_8=15$$

$$\Rightarrow$$ $$a_n=a+(n-1)d$$
$$\Rightarrow$$ $$6=a+(5-1)d$$
$$\Rightarrow$$ $$6=a+4d$$
$$\Rightarrow$$ $$a+4d=6$$ ----- ( 1 )

Now,
$$\Rightarrow$$ $$a_8=a+(8-1)d$$
$$\Rightarrow$$ $$15=a+7d$$
$$\Rightarrow$$ $$a+7d=15$$ ----- ( 2 )

Subtracting equation ( 1 ) from ( 2 ) we get,
$$\Rightarrow$$ $$3d=9$$
$$\Rightarrow$$ $$d=3$$

Substituting value of $$d$$ in equation ( 1 ) we get,
$$\Rightarrow$$ $$a+(4)(3)=6$$
$$\Rightarrow$$ $$a+12=6$$
$$\Rightarrow$$ $$a=-6$$

Now,
$$\Rightarrow$$ $$a_{19}=a+(19-1)d$$
$$\Rightarrow$$ $$a_{19}=-6+18(3)$$
$$\Rightarrow$$ $$a_{19}=-6+54$$
$$\therefore$$ $$a_{19}=48$$
Question 4
1 / -0

The $$17$$th term of an AP exceeds its $$10$$th term by $$7$$. Find the common difference

A
$$\dfrac{3}{7}$$

B
$$\dfrac{-3}{7}$$

C
$$\dfrac{7}{3}$$

D
$$\dfrac{-7}{3}$$

Solution

$$a_7=a+6d$$
And, $$a_{10}=a+9d$$
Or, $$a+9d-a-6d=7$$
Or, $$3d=7$$
Or, $$d=7/3$$
Question 5
1 / -0

If the sum of an AP is the same for $$p$$ terms as for the $$q$$ terms , Find the sum for $$(p+q)$$ terms

A
$$2$$

B
$$0$$

C
$$4$$

D
None of these

Solution

Since $$S_p = S_q$$
$$\cfrac { p }{ 2 } [2a+(p-1)d]=\cfrac { q }{ 2 } [2a+(q-1)d]\\ 2ap+(p-1)pd=2aq+q(q-1)d\\ 2a(p-q)=[q(q-1)-p(p-1)]d\\ 2a(p-q)=[{ q }^{ 2 }-{ p }^{ 2 }+(p-q)]d\\ 2a(p-q)=[(q-p)(q+p)+(p-q)]d\\ 2a=[-(q+p)+1]d\\ -2a=(q+p-1)d\\ \therefore { S }_{ p+q }=\cfrac { (p+q) }{ 2 } [2a+(p+q-1)d]\\ { S }_{ p+q }=\cfrac { (p+q) }{ 2 } (2a-2a)=0$$
Question 6
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Subba Rao started work in $$1995$$ at an annual salary of Rs. $$5000$$ and received an increment of Rs. $$200$$ each year. In which year did his income reached Rs. $$7000$$?

A
$$2005$$

B
$$2002$$

C
$$2004$$

D
$$2007$$

Solution

From the given problem we can form the AP as given below
$$ 5000, 5200, 5400, 5800, ........, 7000.$$
Here $$a_n=7000$$ $$a=5000$$ and $$d=200$$

$$7000=5000+200(n-1)$$

Or, $$200(n-1)=2000$$

Or, $$n-1=10$$

Or, $$n=11$$

So in $$2005$$ his income reaches $$Rs7000$$
Question 7
1 / -0

Find the sum of first 10 terms of the arithmetic sequence $$1, 3, 5, 7, ...........$$

A
90

B
95

C
96

D
100

Solution

The sum of first n terms of arithmetic series formula is given by the formula,
$$S_n = \dfrac{n}{2} [2a + (n - 1) d]$$
Where n = number of terms $$= 10$$
$$a =$$ first term $$=1$$
$$d =$$ common difference of A.P. $$=3-1=2$$
On substituting the given values in the formula, we get
$$S_{10} = \dfrac{10}{2} [2 \times 1 + (10 - 1)2]$$
$$= 5 [2 + 18]$$
$$= 5[20]$$
$$= 100$$
So, the sum of first 10 terms of the A.P. is $$100$$
Question 8
1 / -0

Directions For Questions

In an AP,

...view full instructions

Given $$a_3=15, S_{10}=125$$, Find d and $$a_{10}$$

A
$$18$$

B
$$4$$

C
$$8$$

D
$$6$$

Solution

As we know that the nth term, $$a_n = a+(n-1)d$$
& Sum of first $$n$$ terms, $$S_n = \dfrac{n}{2} (2a+(n-1)d)$$, where $$a$$ & $$d$$ are the first term & common difference of an AP.
$$S_{10}=125=\dfrac {10}{2}(2a+9d)$$

$$\Rightarrow 2a+9d=25$$ ............... (i)

$$a_3=a+2d=15$$ ............. (ii)

Subtracting equation (ii) from (i) we get
$$a+7d=10$$ (the $$8^{th}$$ term)

Subtracting $$3^{rd}$$ term from $$8^{th}$$ term we get

$$(a+7d)-(a+2d)=10-15=-5$$

$$\Rightarrow 5d=-5$$

$$\Rightarrow d=-1$$

So, $$a=17$$

$$a_{10}=17-9=8$$
Question 9
1 / -0

Find the sum of first 10 terms of the arithmetic series if $$a_1 = 2$$ and $$a_{10} = 22$$.

A
$$155$$

B
$$120$$

C
$$165$$

D
$$130$$

Solution

The sum of $$n$$ terms of arithmetic sequence can be calculated by the formula
$$ S_n =\dfrac{n}{2}(a_1+a_n)$$
$$n=10$$
Given,
First term $$a_1=2$$
Tenth term $$a_{10}=22$$
$$\Rightarrow S_n=\dfrac{10}{2}(22+2)$$
$$\Rightarrow S_n=10(12)=120$$
Hence, option $$B$$ is correct
Question 10
1 / -0

What is the sum of first 20 multiples of 2?

A
$$225$$

B
$$420$$

C
$$210$$

D
$$217$$

Solution

The sum of first n terms of arithmetic series formula can be written as,
$$S_n = \dfrac{n}{2} [2a + (n -1)d]$$
Where n = number of terms $$\Rightarrow$$ 20
a = first odd number $$\Rightarrow$$ 2
d = common difference of A.P. $$\Rightarrow$$ 2
Apply the given data in the formula,
$$S_{20} = \dfrac{20}{2} [2 \times 2 + (20- 1)2]$$
$$= 10[4 + (19)2]$$
$$= 10 [4 + 38]$$
$$= 10 [42]$$
$$= 420$$
So, the sum of first 20 multiples of 2 is 420.

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Arithmetic Progressions Test - 47

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Arithmetic Progressions Test - 47

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SHARING IS CARING

Question 1

$$44$$

$$22$$

$$19$$

$$18$$

None of these

Solution

Question 2

$$10,050$$

$$5050$$

$$5000$$

$$50,000$$

Solution

Question 3

$$48$$

$$54$$

$$19$$

$$57$$

Solution

Question 4

$$\dfrac{3}{7}$$

$$\dfrac{-3}{7}$$

$$\dfrac{7}{3}$$

$$\dfrac{-7}{3}$$

Solution

Question 5

$$2$$

$$0$$

$$4$$

None of these

Solution

Question 6

$$2005$$

$$2002$$

$$2004$$

$$2007$$

Solution

Question 7

90

95

96

100

Solution

Question 8

$$18$$

$$4$$

$$8$$

$$6$$

Solution

Question 9

$$155$$

$$120$$

$$165$$

$$130$$

Solution

Question 10

$$225$$

$$420$$

$$210$$

$$217$$

Solution

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