Arithmetic Progressions(10th Grade > Mathematics ) Questions and answers for exam Preparation

10th Grade > Mathematics

ARITHMETIC PROGRESSIONS MCQs

Total Questions : 107 | Page 10 of 11 pages

Find the number of all natural numbers that lie between 24 and 101, which are divisible by 5.

Discuss Question

Answer: Option B. -> 16
:
B

Natural numbers that lie between 24 and 101,which are divisible by 5 are 25, 30,... and 100.
This is an A.P with first term $a = 25,$ common difference $d = 5$ and the last term $l = 100.$
But, $l = a + (n - 1) d .$
$\Rightarrow 100 = 25 + (n - 1) \times 5$
$\Rightarrow 75 = (n - 1) \times 5$
$\Rightarrow 15 = (n - 1)$
$\Rightarrow 16 = n$
Thus, there are 16 natural numbers between 24 and 101 that are divisible by 5.

Question 92.

If the $7^{th}$ and $13^{th}$ terms of an AP are 34 and 64, then its $18^{th}$ term is ___.

Discuss Question

Answer: Option C. -> 89
:
C
The

n^{th}

term of an AP of first term

a

and common difference

d

a_{n} = a + (n - 1) d . \Rightarrow a_{7} = a + 6 d = 34 - - - (1) a_{13} = a + 12 d = 64 - - - (2) (2) - (1) \Rightarrow 6 d = 30 \Rightarrow d = 5 Substituting d = 5 in (1), we get a = 4. ∴ a_{18} = a + 17 d = 4 + (17) 5 = 89

Question 93.

If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the $20^{t h}$ term.

Discuss Question

Answer: Option C. -> 89
:

Here, $S_{14}$ = 1050, n = 14, a = 10.
As $S_{n}$ = ( $\frac{n}{2}$ ) [2a + (n-1) d]
So,
$1050 = (\frac{14}{2}$ ) [20 + 13d]
$\Rightarrow 1050 = 140 + 91 d$
$\Rightarrow 910 = 91 d$
$\Rightarrow d = 10$
Therefore, $a_{20} = 10 + (20 - 1) \times 10 = 200$
i.e. $20^{t h}$ term is 200.

Question 94.

The sum of $r$ terms of an AP is $2 r^{2} + 3 r .$ The $n^{th}$ term is ___.

$2 n - 1$
$4 n + 1$
$3 n - 1$
$3 n + 1$

Discuss Question

Answer: Option B. ->

4 n + 1

:
B
Given,

S_{r} = (2 r^{2} + 3 r),

the sum upto

r

terms of an AP.
We know that the

n^{th}

term of AP is

T_{n} = S_{n} - S_{n - 1} .

\Rightarrow S_{n} = 2 n^{2} + 3 n

and

S_{n - 1} = 2 {(n - 1)}^{2} + 3 (n - 1) .

∴ T_{n} = S_{n} - S_{n - 1}

= 2 n^{2} + 3 n - [2 (n^{2} - 2 n + 1) + 3 n - 3]

= 2 n^{2} + 3 n - 2 n^{2} + 4 n - 2 - 3 n + 3 = 4 n + 1

Question 95.

Find the difference between the sum of first 100 natural numbers and the sum of even numbers from 1 to 100.

2200
2400
2500
2700

Discuss Question

Answer: Option C. -> 2500
:
C

Using the formula
$S_{n} = \frac{n}{2} [2 n + (n - 1) d]$
Sum of first 100 natural numbers = $\frac{100}{2} (2 + 99) = 5050$

Sum of even numbers from 1 to 100, there are 50 even numbers between 1 and 100.

So, we have
Sum of even numbers from 1 to 100 = $\frac{50}{2} (2 \times 2 + 49 \times 2)$ = 2550

Hence, the difference is 5050 - 2550 = 2500

Question 96.

Find the sum of the first $24$ terms of an AP whose $n^{th}$ term is given by $t_{n} = 3 + 2 n .$

Discuss Question

Answer: Option D. -> 672
:
D

Given, $t_{n} = 3 + 2 n .$
$∴ t_{1} = 3 + 2 \times 1 = 5$

$t_{2} = 3 + 2 \times 2 = 7$

$t_{3} = 3 + 2 \times 3 = 9$

Note that $t_{2} - t_{1} = t_{3} - t_{2} = 2.$

Thus, the common difference of the AP is 2.

Therefore, the AP is 5, 7, 9, 11 . . .

The sum to $n$ terms of an AP with first term $a$ and common difference $d$ is
$S_{n} = \frac{n}{2} [2 a + (n - 1) d] .$

$∴ S_{24} = \frac{24}{2} (2 (5) + (24 - 1) 2)$

$= 12 (10 + 46) = 672$

Question 97.

Find the sum of the first 15 multiples of 8.

1000
960
500
820

Discuss Question

Answer: Option B. -> 960
:
B

The first 15 multiples of 8 are 8, 16, 24, 32, .... 112, 120

First term, $a = 8$

Common difference, $d = 16 - 8 = 8$

$n = 15$

Applying formula, $S_{n} = \frac{n}{2} (2 a + (n - 1) d)$ to find sum of $n$ terms of AP, we get
$S_{15} = \frac{15}{2} (2 (8) + (15 - 1) (8))$
$= \frac{15}{2} (16 + 8 (14))$
$= \frac{15}{2} (16 + 112)$
$= \frac{15}{2} (128) = 960$

Question 98.

Find the sum of the following AP 1, 3, 5, 7 …199.

10000
12000
13333
20000

Discuss Question

Answer: Option A. -> 10000
:
A

G i v e n, a = 1, d = 2, a_{n} = l = 199 a + (n - 1) d = 199

1 + (n - 1) 2 = 199

\Rightarrow 1 + 2 n - 2 = 199

\Rightarrow 2 n = 200

n = 100

S_{n} = \frac{n}{2} (a + l)

S_{n} = \frac{100}{2} (1 + 199)

S_{n} = 10000

Question 99.

The sum of four consecutive numbers in an A.P. with common difference $d > 0$ is 20. If the sum of their squares is 120, then the middle terms are __ and __.

2,4
4,6
6,8
8,10

Discuss Question

Answer: Option B. -> 4,6
:
B

Let the numbers be $a - 3 d, a - d, a + d$ and $a + 3 d .$ These numbers form an AP with first term $a - 3 d$ and common difference $2 d > 0.$
Also, given that $(a - 3 d) + (a - d) + (a + d) + (a + 3 d) = 20.$
$\Rightarrow 4 a = 20$
$\Rightarrow a = 5$
Also, ${(a - 3 d)}^{2} + {(a - d)}^{2} + {(a + d)}^{2} + {(a + 3 d)}^{2} = 120$
$\Rightarrow 4 a^{2} + 20 d^{2} = 120$
$\Rightarrow 4 (5)^{2} + 20 d^{2} = 120$
$\Rightarrow d^{2} = 1$
$\Rightarrow d = \pm 1$
SIince $d > 0,$ we must have $d = 1.$
Hence, the numbers are 2, 4, 6, and 8.