Arithmetic Progressions MCQ Test — Class 10 CBSE

Solution:

For an AP, the common difference 'd' must be constant for all consecutive terms.

Option A: 4-2=2, 8-4=4. Not an AP.

Option B: 4-1=3, 9-4=5. Not an AP.

Option C: 6-3=3, 9-6=3, 12-9=3. The common difference is constant (d=3), so it is an AP.

Option D: 1/3 - 1/2 = -1/6, 1/4 - 1/3 = -1/12. Not an AP.

Solution:

Statement A is the correct formula for the n-th term of an AP.

Statement B is correct; for example, 5, 5, 5, ... is an AP with d=0.

Statement C is incorrect. The terms of an AP always increase if 'd' is positive. If 'd' is negative, the terms decrease. If 'd' is zero, the terms remain constant.

Statement D is the definition of an AP, hence it is correct.

Solution:

The given AP is 5, 8, 11, 14, ... . Here, the first term a = 5.

The common difference d = 8 - 5 = 3.

To find the 10th term using the formula a_n = a + (n-1)d, we substitute n=10, a=5, d=3.

a_10 = 5 + (10-1) × 3 = 5 + 9 × 3 = 5 + 27 = 32. The student's calculation is correct, so there is no mistake in this specific problem.

Solution:

The formula for the n-th term of an AP is a_n = a + (n-1)d.

So, a_18 = a + (18-1)d = a + 17d.

And a_13 = a + (13-1)d = a + 12d.

Then, a_18 - a_13 = (a + 17d) - (a + 12d) = a + 17d - a - 12d = 5d.

Given d = 5, the value is 5 × 5 = 25.

Solution:

For the terms of an AP to be decreasing, the common difference 'd' must be negative.

Option A: d=3 (positive), terms increase (e.g., 7, 10, 13...).

Option B: d=5 (positive), terms increase (e.g., -2, 3, 8...).

Option C: d=-2 (negative), terms decrease (e.g., 10, 8, 6...).

Option D: d=0, terms remain constant (e.g., 0, 0, 0...).

Solution:

Since the terms are in AP, the common difference between consecutive terms must be equal.

So, (x + 3) - (2x + 1) = 5 - (x + 3).

Simplify both sides: x + 3 - 2x - 1 = 5 - x - 3.

-x + 2 = 2 - x. This is an identity. Let me recheck. Ah, the common difference should be consistent. So, (a_2 - a_1) = (a_3 - a_2).

(x + 3) - (2x + 1) = 5 - (x + 3)

x + 3 - 2x - 1 = 5 - x - 3

-x + 2 = 2 - x. This means the equality holds for ANY x if these are consecutive terms? No, the error is in my solution step interpretation. Let's use 2b = a+c for three terms a,b,c in AP.

Using the property that if a, b, c are in AP, then 2b = a + c.

Here, a = 2x + 1, b = x + 3, c = 5.

2(x + 3) = (2x + 1) + 5.

2x + 6 = 2x + 6. This implies the statement is true for any x. This is a problematic question if I cannot find a specific x. Let's re-evaluate the question premise.

The common difference property: a_2 - a_1 = a_3 - a_2.

(x + 3) - (2x + 1) = 5 - (x + 3)

x + 3 - 2x - 1 = 5 - x - 3

-x + 2 = 2 - x. This equation is an identity. It means any value of x would satisfy this IF the sequence is an AP. This is not how these questions are typically framed. Let me correct the question or the solution. Ah, I miscalculated the 2b = a+c, let me re-do it.

Correct approach: The difference between the second and first term must be equal to the difference between the third and second term.

Let a_1 = 2x + 1, a_2 = x + 3, a_3 = 5.

a_2 - a_1 = a_3 - a_2

(x + 3) - (2x + 1) = 5 - (x + 3)

x + 3 - 2x - 1 = 5 - x - 3

-x + 2 = 2 - x. Oh, this is indeed an identity. This means the question needs to be changed. Let's try 2x+1, x+3, 3x-2.

Let's assume the question meant 2x+1, x+3, and say, 3x-2 are consecutive terms. Then: (x+3)-(2x+1) = (3x-2)-(x+3)

-x+2 = 2x-5

7 = 3x

x = 7/3. This works. The original question has an issue. Let's change the question: 'If 2x + 1, x + 3, and 5x - 1 are consecutive terms of an AP, what is the value of x?'

Let's retry the solution with the new question.

If 2x + 1, x + 3, and 5x - 1 are consecutive terms of an AP, then the common difference is constant.

(x + 3) - (2x + 1) = (5x - 1) - (x + 3)

x + 3 - 2x - 1 = 5x - 1 - x - 3

-x + 2 = 4x - 4

2 + 4 = 4x + x

6 = 5x

x = 6/5. This is not in the options. I need integer options. Let's make the original problem work with an integer solution for x.

Original problem: 2x + 1, x + 3, and 5 are consecutive terms of an AP.

The property of an AP states that the middle term is the average of the first and third terms (if there are three terms).

So, (x + 3) = [(2x + 1) + 5] / 2

2(x + 3) = 2x + 6

2x + 6 = 2x + 6. This is still an identity. The question as stated is flawed. I need to make sure the question yields a unique value for x.

Let's change the question to: If x+2, 2x, and 2x+3 are consecutive terms of an AP, what is the value of x?

Using the property 2b = a + c for terms a, b, c in AP.

2(2x) = (x + 2) + (2x + 3)

4x = 3x + 5

x = 5. This works and is an integer. Let's make options accordingly. But the previous options were 1,2,3,4. I need to stick to the previous options. So, let me find an AP with given options.

Let's try (x-1), (x+2), (2x+1). (x+2)-(x-1) = 3. (2x+1)-(x+2) = x-1. So, 3=x-1, x=4.

Okay, new question: If x-1, x+2, and 2x+1 are consecutive terms of an AP, what is the value of x?

If x-1, x+2, 2x+1 are in AP, then the common difference is constant.

(x+2) - (x-1) = (2x+1) - (x+2)

x+2-x+1 = 2x+1-x-2

3 = x-1

x = 4. This fits one of the options. I will use this question.

Solution:

The number of saplings planted each day forms an AP: 15, 18, 21, ...

The first term (a) = 15.

The common difference (d) = 18 - 15 = 3.

We need to find the number of saplings on the 7th day, which is a_7.

Using the formula a_n = a + (n-1)d, we have a_7 = 15 + (7-1) × 3 = 15 + 6 × 3 = 15 + 18 = 33.

Solution:

A sequence is an AP if the difference between any term and its preceding term is constant.

Let's find the (n+1)-th term: a_(n+1) = 3(n+1) - 5 = 3n + 3 - 5 = 3n - 2.

Now, find the difference: a_(n+1) - a_n = (3n - 2) - (3n - 5).

a_(n+1) - a_n = 3n - 2 - 3n + 5 = 3.

Since the difference is a constant (3), the sequence is an AP with a common difference d = 3.

Solution:

Let the AP be a, a+d, a+2d, ... .

Option A: If multiplied by k (non-zero), terms become ka, k(a+d), k(a+2d), ... . The new common difference is k(a+d) - ka = kd, which is constant. So, this statement is TRUE.

Option B: If squared, terms become a², (a+d)², (a+2d)². For example, for 1, 2, 3, it becomes 1, 4, 9. (4-1=3, 9-4=5). Not an AP. So, this statement is FALSE.

Option C: If divided by k (non-zero), terms become a/k, (a+d)/k, (a+2d)/k, ... . The new common difference is (a+d)/k - a/k = d/k, which is constant. So, the resulting sequence IS an AP. This statement is FALSE.

Option D: Consider the AP 1, 2, 3. The product is 1 × 2 × 3 = 6, not zero. This statement is FALSE.

Solution:

The general term is a_n = 5n - 2.

To find the first term (a), substitute n=1: a_1 = 5(1) - 2 = 5 - 2 = 3.

To find the common difference (d), we know that for an AP defined by a_n = Pn + Q, the common difference is P. Here, P=5, so d=5. Alternatively, find a_2: a_2 = 5(2) - 2 = 10 - 2 = 8. Then d = a_2 - a_1 = 8 - 3 = 5.

The sum of the first term and the common difference = a + d = 3 + 5 = 8.