Arithmetic Progressions Test

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Arithmetic Progressions Test - 48

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Question 1
1 / -0

Find the sum of the first $$15$$ positive odd numbers.

A
$$225$$

B
$$200$$

C
$$210$$

D
$$217$$

Solution

First $$15$$ positive odd numbers will be,
$$1,3,5,7,9,\dots$$

We can observe that these numbers are in A.P.
The sum of first $$n$$ terms of an A.P whose first term is $$a$$ and common difference is $$d$$ can be written as, $$S_n = \dfrac{n}{2} [2a + (n - 1)d].$$

From the series of first positive odd numbers we can conclude that, $$a=1$$ and $$d=2.$$
Substitute the values of $$a,\ d$$ and $$n$$ in the formula of sum of first $$n$$ terms of an A.P,
$$\begin{aligned}{}{S_{15}} &= \frac{{15}}{2}[2 \times 1 + (15 - 1)2]\\& = 7.5[2 + 14\times2]\\& = 7.5[2 + 28]\\ &= 7.5[30]\\& = 225\end{aligned}$$
So, the sum of the first $$15$$ positive odd numbers of the A.P. is $$225.$$
Question 2
1 / -0

Find the sum of first 32 terms of the arithmetic series if $$a_1 = 12$$ and $$a_{32} = 40$$.

A
$$830$$

B
$$831$$

C
$$832$$

D
$$834$$

Solution

Given:
First term $$= 12$$
last term $$= 40$$

The sum of $$n$$ terms of arithmetic sequence can be calculated by the formula $$S_n = \dfrac{n}{2} (a_1 + a_n)$$
Where $$n$$ is the number of terms, $$a_1$$ is the first term and $$a_n$$ is the last term.

Here, $$n = 32$$

$$\therefore\ \ S_{32} = \dfrac{32}{2} (12 + 40)$$

$$= 16 (12 + 40)$$

$$= 16 (52)$$

$$= 832$$

Hence, the sum of the first $$32$$ terms is $$832$$
Question 3
1 / -0

Directions For Questions

In an AP,

...view full instructions

Given $$d=5, S_9=75$$, find a

A
$$\dfrac{-35}{3}$$

B
$$\dfrac{-28}{3}$$

C
$$\dfrac{-54}{3}$$

D
$$\dfrac{-34}{3}$$

Solution

We know that 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.
$$\therefore S_9=\dfrac {9}{2}(2a+d(9-1))$$
$$\Rightarrow 75=4.5(2a+8d)$$
$$\Rightarrow 2a+8d=\dfrac {75}{4.5}$$
$$\Rightarrow 2a=\dfrac {150}{9}-40=\dfrac {150-360}{9}=\dfrac {-210}{9}=\dfrac {-70}{3}$$
$$\Rightarrow a=\dfrac {-35}{3}$$
Question 4
1 / -0

Find the sum of first 20 even terms of the arithmetic sequence.

A
$$420$$

B
$$401$$

C
$$402$$

D
$$403$$

Solution

The even series will be $$2,4,6,8,10......40$$

The sum of n terms of an arithmetic sequence can be calculated by $$S_n = \dfrac{n}{2} [2a+ (n - 1) d]$$
Where,
$$n = $$ number of terms $$\Rightarrow$$ 20
$$a = $$ first even term $$\Rightarrow$$ 2
$$d = $$ common difference of A.P. $$\Rightarrow$$ 2

substitute the given data in the formula,

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

$$ = 10[4 + 38]$$

$$= 10 [42]$$

$$= 420$$

$$\therefore$$ the sum of first $$20$$ even terms of the A.P. is $$420$$.
Question 5
1 / -0

Find the sum of first 10 terms of the arithmetic series if $$a_1 = 25$$ and $$a_{10} = 100$$.

A
$$300$$

B
$$320$$

C
$$625$$

D
$$312$$

Solution

The sum of n terms of arithmetic sequence can be calculated by the formula $$S_n = \dfrac{n}{2} (a_1 + a_n)$$
Where is the number of terms, $$a_1$$ is the firth term and $$a_n$$ is the last term.
$$n = 50$$
First term $$= 25$$
Tenth term $$= 100$$
Therefore, $$S_{10} = \dfrac{10}{2} (25 + 100)$$
$$= 5 (25 + 100)$$
$$= 5 (125)$$
$$= 625$$
Hence, the sum of first 10 terms is $$625.$$
Question 6
1 / -0

In an Arithmetic progression, the first term $$a = 10$$, and last term, $$l = 100$$. The number of terms is $$1000$$. Find their sum.

A
$$52,000$$

B
$$53,000$$

C
$$54,000$$

D
$$55,000$$

Solution

Given, $$a=10, \text{last term }=100$$ and $$\text{number of terms}=1000$$
Sum of first $$n$$ terms in an A.P. is
$$S_n = \dfrac{n}{2}$$ [First term $$+$$ Last term]
$$= \dfrac{1000}{2} [10 + 100]$$
$$= 500 [110]$$
$$S_{1000} = 55,000$$
Question 7
1 / -0

Find the sum of even numbers between $$11$$ and $$51$$.

A
$$600$$

B
$$610$$

C
$$620$$

D
$$630$$

Solution

The sum of first n terms of arithmetic series formula can be written as,
$$S_n = \dfrac{n}{2} [2a + (n - 1)d]$$ ............ (1)
The list of even numbers between $$11$$ and $$51$$ is $$12, 14, 16, 18,....... 50$$
Where $$n$$ = number of terms
a = first even number $$\Rightarrow$$ 12
d = common difference of A.P. $$\Rightarrow $$ 2
$$T_n =$$ Last term $$\Rightarrow $$ 50
$$T_n = a + (n - 1)d$$
$$50 = 12 + (n - 1) 2$$
$$50 = 12 + 2n - 2$$
$$50 - 12 + 2 = 2n$$
$$ 40 = 2n$$
$$n = 20$$
Apply the given data in the formula (1),
$$S_{20} = \dfrac{20}{2} [2 \times 12 + (20 - 1)2]$$
$$= 10 [24 + (19)2]$$
$$= 10 [24 + 38]$$
$$= 10 [62]$$
$$= 620$$
So, the sum of even numbers between $$11$$ and $$51$$ is $$620$$.
Question 8
1 / -0

Find the sum of fterms of AP whose first term is 26 and last term is 14.

A
200

B
250

C
300

D
400

Solution

The sum of first n terms of arithmetic series formula can be written as,
$$S_{n}=\dfrac{n}{2}$$[First term + Last term] ... eq ( 1 )
n = number of terms = 20
First term = 26
Last term = 14
$$\therefore S_{n}=\dfrac{20}{2} [26+14]=400$$ (Ans $$\rightarrow$$ D)
Question 9
1 / -0

Find the sum of odd numbers between $$0$$ and $$90$$.

A
$$2025$$

B
$$2020$$

C
$$2035$$

D
$$2041$$

Solution

The sum of first n terms of arithmetic series formula can be written as,
$$S_n = \dfrac{n}{2} [2a + (n - 1)d]$$ ............ (1)
The list of odd numbers between $$0$$ and $$90$$ is $$1, 3, 5, 7, 9, .......... 89$$
Where $$n =$$ number of terms
$$a =$$ first odd number $$\Rightarrow$$ $$1$$
$$d =$$ common difference of A.P. $$\Rightarrow 2$$
$$T_n =$$ Last term $$\Rightarrow $$ 89
$$T_n = a + (n - 1)d$$
$$89 = 1 + (n - 1) 2$$
$$89 = 1 + 2n - 2$$
$$89 - 1 + 2 = 2n$$
$$ 90 = 2n$$
$$n = 45$$
Apply the given data in the formula (1),
$$S_{45} = \dfrac{45}{2} [2 \times 1 + (45 - 1)2]$$
$$= 22.5 [2 + (44)2]$$
$$= 22.5 [2 + 88]$$
$$= 22.5 [90]$$
$$= 2025$$
So, the sum of odd numbers between $$0$$ and $$90$$ is $$2,025$$.
Question 10
1 / -0

What is the sum of first $$100$$ multiples of $$12$$?

A
$$60600$$

B
$$61200$$

C
$$62900$$

D
$$60500$$

Solution

Frist $$100$$ multiples of $$12$$ will be,
$$12,24,36,\dots$$

So, we can conclude that the multiples of $$12$$ form an arithmetic progression whose first term is $$12$$ and the common difference is $$12.$$

The sum of first $$n$$ terms of an arithmetic series whose first term is $$a$$ and common difference is $$d$$ can be written as,
$$S_n = \dfrac{n}{2} [2a + (n - 1)d]$$

So, by using above formula,
$$S_{100} = \dfrac{100}{2} [2 \times 12 + (100 - 1)12]$$
$$= 50 [24 + (99)12]$$
$$= 50 [24 + 1188]$$
$$= 50 \times1212$$
$$= 60,600$$

So, the sum of the first $$100$$ multiples of $$12$$ is $$60600.$$

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Arithmetic Progressions Test - 48

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Arithmetic Progressions Test - 48

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SHARING IS CARING

Question 1

$$225$$

$$200$$

$$210$$

$$217$$

Solution

Question 2

$$830$$

$$831$$

$$832$$

$$834$$

Solution

Question 3

$$\dfrac{-35}{3}$$

$$\dfrac{-28}{3}$$

$$\dfrac{-54}{3}$$

$$\dfrac{-34}{3}$$

Solution

Question 4

$$420$$

$$401$$

$$402$$

$$403$$

Solution

Question 5

$$300$$

$$320$$

$$625$$

$$312$$

Solution

Question 6

$$52,000$$

$$53,000$$

$$54,000$$

$$55,000$$

Solution

Question 7

$$600$$

$$610$$

$$620$$

$$630$$

Solution

Question 8

200

250

300

400

Solution

Question 9

$$2025$$

$$2020$$

$$2035$$

$$2041$$

Solution

Question 10

$$60600$$

$$61200$$

$$62900$$

$$60500$$

Solution

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