class 10 maths Arithmetic Progressions Hots Questions

Q 11 :

The first term of an AP is p and the common difference is q, then its 10th term is:

$q + 9 p$
$p - 9 q$
$p + 9 q$
$2 p + 9 q$

1

2

3

4

(3)

$G i v e n, a = p, d = q ∴ 10 t h t e r m, a_{10} = a + (10 - 1) d = p + 9 q$

Q 12 :

The $n^{t h} t e r m o f t h e A P a, 3 a, 5 a, is :$

$n a$
$(2 n - 1) a$
$(2 n + 1) a$
$2 n a$

1

2

3

4

(2)

Given AP is $a, 3 a, 5 a, \dots$ where first term $= a a n d d = 3 a - a = 2 a .$

$∴ I t s n^{t h} t e r m, a_{n} = a + (n - 1) d = a + (n - 1) \times 2 a \Rightarrow a_{n} = a + 2 a n - 2 a = 2 a n - a = a (2 n - 1)$

Q 13 :

Read the given statements carefully:

(i) If a,b,c are in AP then bc, ca, ab is also in AP.

(ii) If each number in an AP is multiplied by a constant number, then the resulting pattern of numbers also forms an AP.

(iii) If each term of an AP is divided by a constant number, then the resulting pattern will form an AP.

(iv) If a constant number is subtracted from each term of an AP, then the resulting pattern of numbers also forms an AP.

Choose the correct option from the following:

(i) and (ii)
(ii), (iii) and (iv)
(i), (ii) and (iii)
All are correct

1

2

3

4

(2)

Given, a,b,c are in AP.

We know that if a,b,c are in AP, then $\frac{a}{k}, \frac{b}{k}, \frac{c}{k}$

$a l s o f o r m a n A P w h e r e k \neq 0 . \frac{a}{a b c}, \frac{b}{a b c}, \frac{c}{a b c} a r e i n A P (H e r e, k = a b c) . \Rightarrow \frac{1}{b c}, \frac{1}{a c}, \frac{1}{a b}$

are in AP. Hence, (i) is incorrect.

(ii) Let an AP have first term a and common difference d .
Then, $a, a + d, a + 2 d, a + 3 d, \dots M u l t i p l y i n g e a c h t e r m o f t h e A P b y k, w e g e t a k, a k + d k, a k + 2 d k, a k + 3 d k, \dots N o w, a_{2} - a_{1} = (a k + d k) - (a k) = d k a_{3} - a_{2} = (a k + 2 d k) - (a k + d k) = d k a_{4} - a_{3} = (a k + 3 d k) - (a k + 2 d k) = d k$

Hence, the difference between consecutive terms is constant ( dk ),
so it is an AP.

Similarly, if a constant k is subtracted from each term of an AP or
each term is divided by a non-zero constant k , the new sequence also forms an AP

Therefore, (ii), (iii), and (iv) are correct.

Q 14 :

4 groups in a class were asked to come up with an arithmetic progression (AP).
Shown below are their responses:

(i) Group M $\to$ [4, 2, 0, –2, …]
(ii) Group N $\to$ [41, 38.5, 36, 33.5, …]
(iii) Group O $\to$ [–19, –21, –23, –25, …]
(iv) Group P $\to$ [–3, –3, –3, –3, …]

Which of these groups correctly came up with an AP?

(i) and (iii)
(ii) and (iii)
(i), (ii) and (iii)
All are correct

1

2

3

4

(4)

For Group M, we have 4, 2, 0, -2, ....
Now, IInd term – Ist term = 2 – 4 = -2 = 0 – 2 = IIIrd term – IInd term.

∴ It is an AP $\Rightarrow$ Group M is correct

For Group N, we have 41, 38.5, 36, 33.5 ....
Now, IInd term – Ist term = 38.5 – 41 = -2.5 = 36 – 38.5 = IIIrd term – IInd term.
∴ It is an AP $\Rightarrow$ Group N is correct.

For Group O, we have -19, -21, -23, -25 ....
= -21 – (-19) = -23 – (-21) = -2
Now, IInd term – Ist term = IIIrd term – IInd term = -2
∴ It is an AP $\Rightarrow$ Group O is correct.

For Group P, we have -3, -3, -3, -3 ....
-3 – (-3) = -3 – (-3) = 0
Now, IInd term – Ist term = IIIrd term – IInd term = 0
∴ It is an AP $\Rightarrow$ Group P is correct

Q 15 :

If the sum of the first n terms of an AP is 3n² + n and its common difference is 6, then its first term is:

2
3
1
4

1

2

3

4

(4)

$W e h a v e, S ₙ = 3 n ² + n \Rightarrow n / 2 (2 a + (n " - " 1) \times 6) = 3 n^{2} + n \Rightarrow a + 3 n - 3 = 3 n + 1 \Rightarrow a = 1 + 3 \Rightarrow a = 4$

Q 16 :

The first term of an AP is 5 and the last term is 45. If the sum of all the terms is 400, the number of terms is:

20
8
10
16

1

2

3

4

(4)

$W e h a v e a = 5, a ₙ = 45 = l a n d S ₙ = 400 ∴ S_{n} = n / 2 (a + l) \Rightarrow 400 = n / 2 (5 + 45) \Rightarrow 400 = n / 2 \times 50 \Rightarrow 25 n = 400 ∴ n = 400 / 25 \Rightarrow n = 16$

Q 17 :

The sum of the 1st and the last term of an AP of 150 terms is equal to the sum of which of the following two terms?

$15 ᵗ ʰ a n d 135 ᵗ ʰ$
$20 ᵗ ʰ a n d 131 ˢ ᵗ$
$50 ᵗ ʰ a n d 51 ˢ ᵗ$
$90 ᵗ ʰ a n d 60 ᵗ ʰ$

1

2

3

4

(2)

For an AP of 150 terms, the sum of the 1?? term $t e r m a_{1} a n d t h e l a s t t e r m a_{150} i s$ equal to the sum of any two terms that are equidistant from the first and last term.

This means $a_{1} + a_{150} = a_{i} + a_{151 - i}$

Given the AP has 150 terms, the middle terms are $a ₇ ₅ a n d a ₇ ₆,$ which will have the same sum as $a ₁ a n d a_{150}$

Since $a_{75} + a_{76} = a_{1} + a_{150},$ we need to find two terms $a_{i} a n d a_{j}$ from the given options that satisfy

$a_{i} + a_{j} = a_{75} + a_{76}$

From the options, the correct pair is (i, j) = (20, 131), which satisfies

$a_{i} + a_{j} = a_{1} + a_{150}$

Q 18 :

The arithmetic mean of five numbers is zero. They are distinct and non-zero numbers. Which of the following statements are true?

(i) At least one of the numbers is positive.
(ii) The product of numbers is zero.
(iii) Maximum number of positive numbers is four.
(iv) At least one of the numbers is negative

Choose the correct option from the following:

(i) and (iii)
(i) and (iv)
(i), (iii) and (iv)
(ii) and (iv)

1

2

3

4

(3)

The arithmetic mean of five numbers is zero implies that sum of the numbers is zero.

(i) At least one of the number is positive, this is true as the sum should be 0.
Sum of positive terms = sum of negative terms
so at least one number must be positive. (e.g. –1, –3, –5, 0, 9)

(ii) The sum being 0 does not imply that the product is also 0.
For example, 1, 1, 1, 1, –4

(iii) All five numbers can not be positive to give a sum zero, so maximum number
of positive numbers is four.

(iv) All five numbers can not be positive to give a sum zero, so at least one number
must be negative.

∴ (i), (iii) and (iv) are correct

Q 19 :

$S ₙ$ be a series which starts with a (positive odd integer) and has a negative common difference. The sum of first 4 terms is 16 and sum of first 6 terms is 12. Which of the following statements are true?

(i) First term of the series is 7.
(ii) Common difference is –2.
(iii) Sum will be positive for first 10 terms.
(iv) Sum will be positive for first 8 terms.

Choose the correct option from the following:

(i) and (iii)
(i) and (ii)
(ii) and (iii)
(iii) and (iv)

1

2

3

4

(2)

Given a is positive odd integer and d is negative

$S ₄ = 16 a n d S ₆ = 12 W e k n o w t h a t S ₙ = n / 2 (2 a + (n - 1) d) S ₄ = 4 / 2 (2 a + (4 - 1) d) \Rightarrow 2 (2 a + 3 d) = 16 \Rightarrow 2 a + 3 d = 8 . . . (i) A l s o, S ₆ = 6 / 2 (2 a + (6 - 1) d) = 12 \Rightarrow 3 (2 a + 5 d) = 12 \Rightarrow 2 a + 5 d = 4 . . . (i i) S o l v i n g (i) a n d (i i) 2 a + 3 d = 8 2 a + 5 d = 4 - - - - - 2 d = 4 \Rightarrow d = - 2 P u t v a l u e o f d i n e q u a t i o n (i) 2 a + 3 (- 2) = 8 \Rightarrow 2 a - 6 = 8 \Rightarrow 2 a = 14 \Rightarrow a = 7 S ₁ ₀ = 10 / 2 (2 (7) + (9) (- 2)) = 5 (14 - 18) = - 20 (n e g a t i v e s u m) S ₈ = 8 / 2 (2 (7) + (7) (- 2)) = 4 (14 - 14) = 0 H e n c e, (i) a n d (i i) a r e c o r r e c t$

Topic Question Set

Q 11 :

The first term of an AP is p and the common difference is q, then its 10th term is:

Q 12 :

The $n^{t h} t e r m o f t h e A P a, 3 a, 5 a, is :$

Q 15 :

If the sum of the first n terms of an AP is 3n² + n and its common difference is 6, then its first term is:

Q 16 :

The first term of an AP is 5 and the last term is 45. If the sum of all the terms is 400, the number of terms is:

Q 17 :

The sum of the 1st and the last term of an AP of 150 terms is equal to the sum of which of the following two terms?

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 12 : The nth term of the AP a,3a,5a,is: .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

Want Us to Email you About Special Offers & Updates?

Q 12 :

The $n^{t h} t e r m o f t h e A P a, 3 a, 5 a, is :$