 # JEE Arithmetic Progression Previous Year Questions With Solutions

Arithmetic Progression is a sequence where the difference of any term and the previous term is always the same. In this section, JEE aspirants get a chance to solve all important past year AP questions. Previous year solved questions on Arithmetic Progression are available here. Students who are preparing for IIT JEE, the previous year solved questions is the right tool for the exam preparation. The subject experts at BYJU’S have prepared the JEE chapter-wise solutions for all the previous year important questions. The Arithmetic progression past year questions and answers PDF can be easily accessed by students 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.

### Important Points to Remember:

 Definition: It is a sequence of numbers such that the difference of any two successive numbers is a constant. For example, the sequence 3,6, 9, 12,… is an arithmetic progression with common difference 3. Notation: Common difference denoted by “d” Nth Term by “an” Sum of nth term by “Sn” nth Term of an AP: an = a + (n − 1) × d Sum of N Terms: S = n/2[2a + (n − 1) × d] Sum of AP when Last Term is known S = n/2 (first term + last term) *Where a = first term, n = number of terms and d = Common difference

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

Question 1: Let a1, a2, a3,…….,a49 be in A.P. such that $\sum_{k=0}^{12} a_{4k+1} = 416 \; \; and \; \; a_9 + a_{43} = 66$. If a12 + a22 + ……..+ a172 = 140 m, then m is equal to

(a) 68                  (b) 34                    (c) 33                 (d) 66

Solution:

We know, nth term of A.P. is an = a + (n – 1) d

a9 + a43 = 66

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

Or a + 25d = 33 ……(1)

Now,

$\sum_{k=0}^{12} a_{4k+1} = 416$

Therefore, 13a + 312d = 416

Or a + 24d = 32 …(2)

Solving (1) and (2), we get

a = 8 and d = 1

So,

$\sum_{k=1}^{17} a_{k}^2 = 8^2 + 9^2 +….+24^2$

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

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

= [24x25x49]/6 – [7x8x15]/6

= 4760 = 140 x 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

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) = (25 − 4d2)(25 − d2) = 504

=> 4d4 − 125d2 + 121 = 0

=> 4d4 − 4d2 − 121d2 + 121 = 0

=> d2 = 1 or d2 = 121/4

=> d = ± 11/2

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 Є R, (21 + x + 21 – x), f(x) and (3x + 3-x) are in A.P., then the minimum value of f(x) is :

(a) 0                                     (b) 4                        (c) 3                          (d) 2

Solution:

Given: (21 + x + 21 – x), f(x) and (3x + 3-x) are in A.P.

Therefore, f(x) ≥ 1 + 2 = 3

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

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

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

or d = 1/200 and a = 1/200

Now,

Sum of first 200 terms:

S200 = 200/2 [2(1/100) + 199(1/200)]

= 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) 41/3                     (c) 8                        (d) 4

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

Therefore, the minimum possible value of b is obtained as b ≥ 2 .

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

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

Solution:

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

And [a12 + a22 + …. + a112]/11 = 90

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

Now,

Let the numbers are, a6 + 5d, a6 + 4d, …… a6, …….., a6 – 5d

11a62 + 110d2 = 990

a6 = 15- 5d

a62 + 10d2 = 90

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) x [30-10(3)]/11 = 0

Question 7: In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression is

(a) [1-√5]/2                       (b) √5/2                    (c) [√5 – 1]/2                       (d) √5

Solution:

Let a geometric progression consisting of positive terms, a, ar, ar2

Then, a = ar + ar2

=> r2 + r – 1 = 0

Solving equation using quadratic formula, we get

r = [√5 – 1]/2

Question 8: A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases 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 after:

(a) 18 months

(b) 19 months

(c) 20 months

(d) 21 months

Solution:

From the question, saving are 200, 200, 200, 240, 280,…….. upto n terms.

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

Using sum of n terms of an AP, we have

400 + (n-2)/2 [2×200 + (n – 2-2)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 9: If 100 times the 100th term of an AP with non-zero common difference 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

Solution:

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

100 x T100 = 50 x T50

=> 100(a + 99d) = 50(a + 49d)

=> 2a + 198d = a + 49d

=> a + 149d = 0

=> T150 = 0

Question 10: If the nth term of an AP be (2n-1), then find the sum of its first n terms.

Solution:

Let an = (2n-1)

a1 = 2×1 – 1 = 1

a2 = 4 – 1 = 3

Now, d = a2 – a1 = 3 – 1 = 2

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

= n2

Question 11: The sum of the infinity of the series 1 + 2/3 + 6/32 + 10/33 + 14/34 + ….. is

(a) 2                 (b) 3                        (c) 4                        (d) 5

Solution:

Let S = 1 + 2/3 + 6/32 + 10/33 + 14/34 +….. …(1)

(1/3)S = 1/3 + 2/32 + 6/33 + 10/34 + 14/35 + ….. ….(2)

Dividing (1) and (2), we get

(2/3)S = 4/3 + 4/32(1 + 1/3 + 1/32 + ….) …(3)

We know, 1 + 1/3 + 1/32 + …. = 1/[1-1/3] = 3/2

(2/3)S = 4/3 + 4/32 x 3/2 = 6/3

=> S = 3

Question 12: The pth , qth and rth terms of an A.P are a, b,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 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

Hence Proved.

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