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

10th Grade > Mathematics

ARITHMETIC PROGRESSIONS MCQs

Total Questions : 107 | Page 8 of 11 pages

Question 71. The sum of first ten terms of an A.P. is four times the sum of its first five terms. What is the ratio of first term and common difference?

Discuss Question

Answer: Option B. -> 12
:
B
Let

S_{10}

be the sum of first 10 terms and

S_{5}

be thesum of first 5 terms.
The sum upto

n

terms of an AP of first term

a

and common difference

d

is given by

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

Given,

S_{10} = 4 S_{5} .

\Rightarrow \frac{10}{2} [2 a + (10 - 1) d] = 4 \times \frac{5}{2} [2 a + (5 - 1) d]

\Rightarrow \frac{10}{2} [2 a + 9 d] = 4 \times \frac{5}{2} [2 a + 4 d]

\Rightarrow 2 a + 9 d = 4 a + 8 d

\Rightarrow 2 a = d

\Rightarrow \frac{a}{d} = \frac{1}{2}

Question 72. 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 73. If the

11^{th}

and

13^{th}

terms of an AP are 35 and 41 respectively, then its common difference is ___.

Discuss Question

Answer: Option D. -> 3
:
D
The

n^{th}

term of an A.P, with first term

a

and common difference

d

is given by

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

\Rightarrow a_{11} = a + 10 d = 35... (1)

a_{13} = a + 12 d = 41... (2)

Solving equations (1) and (2), we get

2 d = 6.

\Rightarrow d = 3

Question 74. 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, thenumbersare2, 4, 6, and8.

Question 75. An AP consists of 50 terms of which the third term is 12 and the last term is 106. Find the

29^{t h}

term.
___

Discuss Question

:
It is also given that

a_{3} = 12

and

a_{50} = 106

Using formula

a_{n} = a + (n - 1) d

to find

n^{t h}

term of AP,we get

a_{50} = a + (50 - 1) d \Rightarrow 106 = a + 49 d \dots (1)

a_{3} = a + (3 - 1) d \Rightarrow 12 = a + 2 d \dots (2)

Subtracting (1) and (2), we get,

\Rightarrow 47 d = 94

\Rightarrow d = 2

Substituting

d

in (2), we get,

a + 2 (2) = 12

a = 12 - 4 = 8

The

29^{t h}

term is,

a_{29} = a + (29 - 1) d = 8 + 28 (2) = 8 + 56 = 64

Question 76. Find the sum to 100 terms of the series:
1 + 4 + 7 + 5 + 13 + 6 ...

7825
8825
9825
10825

Discuss Question

Answer: Option B. -> 8825
:
B
The series can be clubbed into two AP's, (1 + 7 + 13......) and (4 + 5 + 6.......). So, we can observe that the series can be taken as two AP's by simple rearrangementof numbers. Now, we can find the Sum of 50 terms for each AP formed.
Formula for sum of first n terms of AP is:

S_{n} = \frac{n}{2} (2 a + (n - 1) d)

So, for first AP

S_{1} = \frac{50}{2} (2 + 49 \times 6) = 7400

and similarly,

S_{2} = \frac{50}{2} (8 + 49) = 1425

So, Sum of first 100 terms

= 7400 + 1425 = 8825

Question 77. The

17^{t h}

term of an AP exceeds its

10^{t h}

term by 7. Find the common difference.
___

Discuss Question

:
It is given that

17^{t h}

term exceeds its

10^{t h}

term by 7
It means

a_{17}

a_{10}

+ 7......(1)
Here,

a_{17}

is the

17^{t h}

term anda₁₀is the

10^{t h}

term of an AP.
Using formula

a_{n} = a + (n - 1) d

to find

n^{t h}

term of arithmetic progression,we get

a_{17}

a + (16) d

.....(2)

a_{10}

a + (9) d

.....(3)
Putting(2)and(3)in equation(1), we get

a + 16 d = a + 9 d + 7

\Rightarrow 7 d = 7

\Rightarrow d = \frac{7}{7} = 1

Question 78.

The sequence of numbers 5, - 2, -9, -16, ... is

an AP with common difference -7.
an AP with common difference 7.
an AP with common difference 3.
not an AP.

Discuss Question

Answer: Option A. -> an AP with common difference -7.
:
A
For a sequence to be an AP, the same common difference must exist between any two consecutive terms of AP.
Here,