Practice Questions on Arithmetic Progression ( AP ) Advanced (original) (raw)

Last Updated : 20 Jan, 2025

An arithmetic progression (AP), also known as an arithmetic sequence, is a sequence of numbers in which the difference between consecutive terms is constant. This constant difference is called the "common difference."

Important Formulas on Arithmetic Progression

Various formulas on arithmetic progression are:

Nth term of an AP	a n = a 1 + (n-1)d**
Sum of first n terms of an AP	S n = n/2(a 1 + a n ) S n = n/2[2a 1 + (n-1)d]**
Nth Term from Sum of an AP	a n = S n - S n-1**
Number of Terms in an AP	**n = (a n - a 1 )/d + 1

where,

**a nis the nth term,
**a 1 is the first term,
**d is the common difference,
**S n is the sum of the first n terms,
**S n-1is the sum of the first n - 1 terms,
**n is the number of terms.

Solved Practice Questions on Arithmetic Progression (AP) - Advanced

**Question 1: Write the first three terms in each of the following sequences defined by

**A n **= 5n + 2(n - 1)
**A n **= 2n + 4(n - 2)

**Solution :

**For A n **= 5n + 2(n - 1)

Put n = 1, we get
a1 = 5 · 1 + 2(1 - 1) = 5+ 0 = 5

Put n = 2, we get
a2 = 5 · 2 + 2(2 - 1) = 10 + 2 = 12

Put n = 3, we get
a3 = 5 · 3 + 2 (3 - 1) = 15 + 4 = 19

So first three terms are 5, 12, 19.
**For A n = 2n + 4(n - 2)

Put n = 1, we get
a1 = 2 · 1 + 4(1 - 2) = 2 - 4 = -2

Put n = 2, we get
a2 = 2 · 2 + 4(2 - 2) = 4+ 0 =4

Put n = 3,we get
a3 = 2 · 3 + 4(3 - 2) = 6 + 4 = 10

So the first three terms are -2, 4, 10.

**Question 2: Find the 20 th **Term of the given expression A n **= (n - 1)(2 - n)(3 + n).

**Solution:

For An = (n - 1)(2 - n)(3 + n)

Put n = 20 in given expression,
a20 = (20 - 1)(2 - 20)(20 + 3)
⇒ a20 = 19 × (-18) × (23)
⇒ a20 = -7886.

**Question 3: Find the middle term of the A.P. 6, 13, 20, …, 216.

**Solution:

In the given AP,

First term a1 = 6
Last term an = 216
Common difference = 7

Now, to find the number of terms, n = (an - a1)/d + 1

n = (216 - 6)/7 + 1
n = 210/7 + 1
n = 30 + 1
n = 31

So, middle term is (n + 1) / 2

= (31 + 1)/2
= 16

Now to calculate the middle term

a16 = a1 + 15 × d
a16 = 6 + 15 × 7
a16 = 6 + 105 = 111

So the middle term of the given AP is 111

**Question 4: Find the sum of all natural numbers lying between 100 and 1000 (inclusive of both 100 and 1000) which are multiples of 5.

**Solution:

Solve: first term to be 100 and last terms is 1000 and common difference is 5.

So our formula is Sn = (n/2)[2a + (n - 1) × d] .

Using an = a1 + (n - 1)d
⇒ 1000 = 100 + (n - 1)5
⇒ 900 = (n - 1)5
⇒ 180 = n - 1
⇒ n = 181

Thus, there are 181 such number. Now for sum of all the 181 terms of sequence can be calculated as follows:

S181 = (181/2)[2 · 100 + (181 - 1) × 5].
⇒ S181 = (181/2)[200 + 180 × 5]
⇒ S181 = (181/2) × 1100
⇒ S181 = 181 × 550
= 99,550

**Question 5: If a, b, c are in GP and log a − log 2b, log 2b − log 3c and log 3c − log a are in AP, then a, b, c are the lengths of the sides of a triangle which is? [NIMCET - 2019]

**A) Acute Angle
**B) Obtuse Angled
**C) Right Angled
**D) Equilateral

**Solution:

**If a, b, c are in GP:

b = √ac

**Given that:
**log⁡ a − log⁡ 2b, log ⁡2b − log⁡3c, and log⁡ 3c − log⁡ a are in AP:

\log\frac{a}{2b} = \log\frac{a}{2b}+\log\frac{3c}{a}

or \log{(\frac{2b}{3c})^2} = \log{(\frac{a}{2b},\frac{3c}{a})}

\frac{2b}{3c} = \frac{3c}{2b}

\frac{4b^2}{9c^2} = \frac{3c}{2b} or 8b3 = 27c3

2b = 3c . . . . . . . (i)

**a, b, c are in GP:

b2 = ac . . . . . . . (ii)

\frac{9c^2}{4} = ac form (i) and (ii)

a = (9/4)c , b = (3/2)c, c= c

sides are 9c/4, 6c/4, 4c/4

Side C is the greatest side so

CosA = \frac{b^2+c^2-a^2}{2bc}=\frac{36+16-81}{48} < 0

Since A is an Obtuse angle and if any angle of a triangle is obtuse it is called an obtuse angled triangle.

Question 6: If the 2nd, 5th, and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is? [JEE(Main)-2016]
**A) 7/4
**B) 8/5
**C) 4/3
**D) 1

**Solution:

Let a be the first term of an AP and d be the common difference, then:

2nd term: a2 = a + d

5th term: a5 = a + 4d

9th term: a9 = a + 8d

Given: a2, a5, a9 are in G.P., so:

(a + 4d)2 = (a +d)(a + 8d)
d(8d - a) = 0
8d = a

Hence the common ratio of G.P \frac{a+4d}{a+d} = \frac{8d+4d}{8d+d} = \frac{12d}{9d} =\frac{4}{3}

Question 7: For three positive integers p, q, r, x^{pq^2} = y^{qr}=z^{p^2r} and r = pq + 1 such that 3, 3 log y x, 3 log z y, 7 log x z are in A.P. with common difference 1/2. Then r - p - q is equal to? [JEE(Main)-2023 January]

**A) 12
**B) -6
**C) 2
**D) 6

**Solution:

Given:

x^{pq^2} = y^{qr}=z^{p^2r}

r = pq + 1,

3, 3log⁡yx, 3log⁡zy, 7log⁡xz are in **A.P. with a common difference of 1/2.

3 logyx =7/2, 3 logzy = 4, 7 logxz = 9/2

x = y7/6, y = z4/3, z = x9/2

y^{\frac{7}{6}pq^2} = y^{qr}=z^{p^2r}

(7/6)pq2 = qr = (3/4)p2r

7pq = 6r, 4q = 3p2

r = pq + 1
r = (6r/7)+ 1
r = 7
pq = 6

p(\frac{3p^2}{4}) = 6

p =2, q = 3

r - p - q = 7 - 5 = 2

Question 8: If 3, a, b, c are in A.P. and 3, a - 1, b - 1, b + 1 be in G.P. Then, the arithmetic mean of a, b, and c is? [JEE(Main)-2024 February]

**A) -4
**B) -1
**C) 13
**D) 11

**Solution:

We are given two conditions:

3, a, b, c are in A.P

a = 3 + d, b = 3 + 2d, c = 3 + 3d

3, a − 1, b + 1, c + 9 are in G.P

3, 2 + d, 4 + 2 d, 12 + 3 d

GP condition:

Common ratio:

(a - 1)/3 = (b + 1) / (a - 1)
(2 + d)2 = 3(4 + 2d)
d2 − 2d − 8 = 0
d = 4 or d = -2(Rejected for positive G.P. values)

If d = 4 G.P ⇒ 3, 6, 12, 24 .
a = 7
b = 11
c = 15
Arithemetic Mean:

(a + b + c)/3 = 7 + 11 + 15/3 = 11

Practice Questions on Arithmetic Progression (AP)

**Question 1: If the sum of the first n terms of an A.P. is 3n2 + 2n, find the common difference and the first term.

**Question 2: The 5th term of an A.P. is 18, and the sum of its first 7 terms is 105. Find the first term and the common difference.

**Question 3: Three numbers a, b, c are in A.P. If the sum of their squares is 70 and their product is 120, find a, b, and c.

**Question 4: The 3rd, 8th, and 15th terms of an A.P. form a geometric progression (G.P.). If the first term is 2, find the common difference and the value of the 20th term.

**Question 5: In an A.P., the sum of the first 10 terms is equal to the sum of the next 5 terms. If the first term is 4, find the common difference.

**Question 6: The sum of all the terms of an A.P. (up to the n-th term) is equal to the sum of all the terms up to the (n + 4)-th term. Prove that the common difference of the A.P. is zero.

**Question 7: A sequence a1, a2, a3,… is in A.P. The sum of the first n terms is 240. The sum of the first 2n terms is 960. Find the common difference of the sequence.

**Question 8: If the sum of the first n terms of an A.P. is given as Sn = 2n2 + 3n, find the first term and the common difference.

**Answer Key

**First term a = 5, Common difference d = 6.

**First term a = 6, Common difference d = 3.

**a = 2, b = 4,c = 6.

**Common difference d = 2, 20th term a 20 = 40.

**Common difference d = 4/3

**Common difference d = 0.

**Common difference d = 8.

**First term a = 5, Common difference d = 4.