JEE Arithmetic Progression Previous Year Questions With Solutions

Arithmetic Progression is a sequence where the difference between any term and the previous term is always the same, i.e., the difference between any two consecutive terms is the constant. In this section, JEE aspirants get a chance to solve all important past year A.P. questions. Previous year’s solved questions on Arithmetic Progression are available here. For students preparing for IIT JEE, the previous year’s solved questions is the right tool for exam preparation. The subject experts at BYJU’S have prepared the JEE chapter-wise solutions for all the previous year’s important questions. Students can easily access the Arithmetic progression past year questions and answers PDF to start adequate preparation for their upcoming exams. Download the chapter-wise JEE Solutions for Maths now and practice all the questions that can be framed in the examination.

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Important Points to Remember:

JEE Main Past Year Questions With Solutions on A.P.

Question 1: Let a₁, a₂, a₃,…….,a₄₉ be in A.P. such that

(a) 68

(b) 34

(c) 33

(d) 66

Answer: (b)

Solution:

We know that the nth term of A.P. is a_n = a + (n – 1) d

a₉ + a₄₃ = 66

Therefore, a + 8d + a + 42d = 66

Or a + 25d = 33 ……(1)

Now,

Therefore, 13a + 312d = 416

Or a + 24d = 32 …(2)

Solving (1) and (2), we get

a = 8 and d = 1

So,

= (1² + 2² + …+24²) – (1² + 2² + …+7²)

Using the sum of squares of n natural numbers formula, we have

= [24 × 25 × 49]/6 – [7 × 8 × 15]/6

= 4760

= 140 × 34

Thus, the answer is 34.

Question 2: Five numbers are in A.P., whose sum is 25 and product is 2520. If one of these five numbers is −1/2, then the greatest number amongst them is

(a) 16

(b) 27

(c) 7

(d) 21/2

Answer: (a)

Solution:

Let 5 numbers be a − 2d, a − d, a, a + d, a + 2d

so, Sum of numbers = 5a = 25 or a = 5

Product of Numbers = (a − 2d)(a − d)a(a + d)(a + 2d) = 2520

⇒ (25 − 4d²)(25 − d²) = 504

⇒ 4d⁴ − 125d² + 121 = 0

⇒ 4d⁴ − 4d² − 121d² + 121 = 0

⇒ d² = 1 or d² = 121/4

⇒ d = ± 1 or d = ± 11/2

For d = ± 1, none of the terms is equal to -1/2. (Hence, rejected)

For d = 11/2, a + 2d is the greatest term, a + 2d = 5 + 11 = 16

Question 3: Let f: R → R be such that for all x ∈ R, (2^{1 + x} + 2^{1 – x}), f(x) and (3^x + 3^-x) are in A.P., then the minimum value of f(x) is :

(a) 0

(b) 4

(c) 3

(d) 2

Answer: (c)

Solution:

Given: (2^{1 + x} + 2^{1 – x}), f(x) and (3^x + 3^-x) are in A.P.

Therefore,

Adding (1) and (2), we get;

f(x) ≥ 1 + 2 = 3

i.e., f(x) ≥ 3

Thus, the minimum value of f(x) is 3.

Question 4: If the 10th term of an A.P. is 1/20 and its 20th term is 1/10, then the sum of its first 200 terms is:

(a) 201/4

(b) 100

(c) 50

(d) 201/2

Answer: (d)

Solution:

10th term of an A.P. is 1/20, and its 20th term is 1/10.

So, a₁₀ = 1/20 and a₂₀ = 1/10

Now, a₂₀ – a₁₀ = 10 d

(1/20) – (1/10) = 10d

⇒ d = 1/200 and a = 1/200

Now,

Sum of first 200 terms:

S₂₀₀ = 200/2 [2(1/200) + 199(1/200)]

= (200/2) × (1/200) × [2 + 199]

= 201/2

Question 5: If three positive numbers a, b and c are in A.P. such that abc = 8, then the minimum possible value of b is :

(a) 2

(b) 4^1/3                    

(c) 8

(d) 4

Answer: (a)

Solution:

For a set of positive numbers, the AM ≥ GM.

Given: abc = 8

Since the numbers are in AP, their arithmetic mean is b.

The geometric mean of the three numbers:

(a+b+c)/3 = b

⇒ b ≥ (abc)^1/3

That means, b ≥ 2.

Therefore, the minimum possible value of b is 2.

Question 6: Let a₁, a₂, a₃,…… a₁₁ be real numbers satisfying a₁ = 15, 27 – 2a₂ > 0 and a_k = 2a_k-1 – a_k-2 for k = 3, 4, …..,11

If [a₁² + a₂² + …. + a₁₁²]/11 = 90 then find the value of [a₁ + a₂ + …. +a₁₁]/11.

Solution:

First term a₁ = 15, 27 – 2a₂ > 0 and a_k = 2a_k-1 – a_k-2 for k = 3, 4, …..,11

And [a₁² + a₂² + …. + a₁₁²]/11 = 90

a_k-1 = [a_k + a_k-2]/2

Now,

Let the 11 terms be, a – 5d, a – 4d, …., a, …., a + 4d, a + 5d.

11a² + 110d² = 990….(i)

a² + 10d² = 90

a – 5d = 15 {since the first term = 15}

a = 15 + 5d

Substituting a = 15 + 5d in (i), we get;

Now, (15 + 5d)² + 10d² = 90

⇒ d = -3, -9/7

For d = -3 => a₂ = 12

For d = -9/7 => a₂ = 13.7 (not possible, as a₂ < 27/2)

Thus, [a₁ + a₂ + …. +a₁₁]/11 = (11/2) × [30 – 10(3)]/11 = 0

Question 7: A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months, his savings increased by Rs.40 more than the saving of immediately previous month. After how many months his total savings from the start of service will be Rs. 11040?

(a) 18 months

(b) 19 months

(c) 20 months

(d) 21 months

Answer: (d)

Solution:

From the question, savings are 200, 200, 200, 240, 280,…….. up to n terms.

But 200, 240, 280,……..(n-2) terms are in AP.

Using the sum of n terms of an AP, we have

400 + (n – 2)/2 [2 × 200 + (n – 2 – 1)40] = 11040

⇒ (n – 2)[200 + 20n – 60] = 10640

⇒ 20(n + 7)(n – 2) = 10640

⇒ n² + 5n – 546 = 0

⇒ (n + 26)(n – 21) = 0

-ve value is not possible.

n = 21

Therefore, total time = 21 months

Question 8: If 100 times the 100th term of an AP with a non-zero common difference is equal to the 50 times its 50th term, then the 150th term of this AP is

(a) -150

(b) 150 times its 50th term

(c) 0

(d) 150

Answer: (c)

Solution:

Given 100 times the 100th term of an AP = 50 times its 50th term.

100 × T₁₀₀ = 50 × T₅₀

⇒ 100(a + 99d) = 50(a + 49d)

⇒ 2a + 198d = a + 49d

⇒ a + 149d = 0

⇒ T₁₅₀ = 0

Question 9: If the nth term of an AP is (2n – 1), then find the sum of its first n terms.

Solution:

Let a_n = (2n – 1)

a₁ = 2 × 1 – 1 = 1

a₂ = 2 × 2 – 1 = 4 – 1 = 3

Now, d = a₂ – a₁ = 3 – 1 = 2

Sum of first n terms = (n/2) [2a + (n – 1)d]

= (n/2) [2 + 2n – 2]

= n²

Question 10: The pth, qth and rth terms of an A.P are a, b, and c, respectively. Show that (q – r)a + (r – p)b + (p – q) c = 0.

Solution:

Let a, a + d, a+ 2d, ..are in A.P.

pth term = a + (p – 1)d = a

qth term = a + (q – 1)d = b and

rth term = a + (r – 1)d = c

L.H.S. = (q – r)a + (r – p)b + (p – q) c

= (q – r)(a + (p – 1)d) + (r – p)(a + (q – 1)d) + (p – q)(a + (r – 1)d)

Solving the above equation, we have;

a(q – r) + b(r – p) + c(p – q) = a[(q – r) + (r – p) + (p – q)] + d[(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)]

= a(0) + d(0)

= 0

Therefore, (q – r)a + (r – p)b + (p – q) c = 0.

Also, Read

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