8 Arithmetic Progression, Important Question, with Solution, for Board Exams.

Arithmetic Progression: 8 Important Questions with Solutions

Practice these 8 selected questions covering the $n^{th}$ term, sum of terms ($S_n$), and application-based word problems. These are curated based on frequent Board Exam patterns.

Question 1

Find the 10th term of the Arithmetic Progression (A.P.): 2, 7, 12, ...

Solution:

Here, first term \(a = 2\)
Common difference \(d = 7 - 2 = 5\)
We need to find the 10th term (\(t_{10}\)).

Formula for the \(n^{th}\) term: $$t_n = a + (n - 1)d$$

Substituting the values: $$t_{10} = 2 + (10 - 1)5$$ $$t_{10} = 2 + (9)5$$ $$t_{10} = 2 + 45$$ $$t_{10} = 47$$

Answer: The 10th term of the A.P. is 47.

Question 2

Which term of the A.P. 21, 18, 15, ... is -81?

Solution:

Given A.P.: 21, 18, 15, ...
\(a = 21\)
\(d = 18 - 21 = -3\)
Let the \(n^{th}\) term be -81, so \(t_n = -81\).

$$t_n = a + (n - 1)d$$ $$-81 = 21 + (n - 1)(-3)$$ $$-81 - 21 = (n - 1)(-3)$$ $$-102 = (n - 1)(-3)$$

Dividing both sides by -3: $$\frac{-102}{-3} = n - 1$$ $$34 = n - 1$$ $$n = 34 + 1$$ $$n = 35$$

Answer: The 35th term of the A.P. is -81.

Question 3

Find the sum of all 3-digit natural numbers which are divisible by 5.

Solution:

The 3-digit numbers divisible by 5 are: 100, 105, 110, ..., 995.
This is an A.P. where \(a = 100\), \(d = 5\), and the last term \(t_n = 995\).

Step 1: Find the number of terms (n)
$$t_n = a + (n - 1)d$$ $$995 = 100 + (n - 1)5$$ $$995 - 100 = (n - 1)5$$ $$895 = (n - 1)5$$ $$n - 1 = \frac{895}{5}$$ $$n - 1 = 179 \Rightarrow n = 180$$

Step 2: Find Sum (\(S_n\))
Using formula: \(S_n = \frac{n}{2} [a + t_n]\) $$S_{180} = \frac{180}{2} [100 + 995]$$ $$S_{180} = 90 \times 1095$$ $$S_{180} = 98,550$$

Answer: The sum is 98,550.

Question 4

Find the value of \(k\) such that \(2k\), \(k + 10\), and \(3k + 2\) are three consecutive terms of an A.P.

Solution:

If \(a, b, c\) are in A.P., then the common difference is constant.
i.e., \(b - a = c - b\) or \(2b = a + c\).

Here, \(a = 2k\), \(b = k + 10\), \(c = 3k + 2\).
$$2(k + 10) = (2k) + (3k + 2)$$ $$2k + 20 = 5k + 2$$ $$20 - 2 = 5k - 2k$$ $$18 = 3k$$ $$k = \frac{18}{3} = 6$$

Answer: The value of k is 6.

Question 5

For an A.P., if the first term is 6 and the common difference is -3, find \(S_{12}\).

Solution:

Given: \(a = 6\), \(d = -3\), \(n = 12\).

Formula for Sum: $$S_n = \frac{n}{2} [2a + (n - 1)d]$$ Substituting values: $$S_{12} = \frac{12}{2} [2(6) + (12 - 1)(-3)]$$ $$S_{12} = 6 [12 + (11)(-3)]$$ $$S_{12} = 6 [12 - 33]$$ $$S_{12} = 6 [-21]$$ $$S_{12} = -126$$

Answer: The sum of the first 12 terms is -126.

Question 6

The sum of the 4th and 8th terms of an A.P. is 24, and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.

Solution:

We know \(t_n = a + (n-1)d\).

Condition 1: \(t_4 + t_8 = 24\) $$(a + 3d) + (a + 7d) = 24$$ $$2a + 10d = 24 \Rightarrow a + 5d = 12 \quad \dots(1)$$
Condition 2: \(t_6 + t_{10} = 44\) $$(a + 5d) + (a + 9d) = 44$$ $$2a + 14d = 44 \Rightarrow a + 7d = 22 \quad \dots(2)$$

Subtract (1) from (2): $$(a + 7d) - (a + 5d) = 22 - 12$$ $$2d = 10 \Rightarrow d = 5$$
Substitute \(d = 5\) in equation (1): $$a + 5(5) = 12$$ $$a + 25 = 12$$ $$a = 12 - 25 = -13$$

Answer: The A.P. is -13, -8, -3, ...

Question 7

If the sum of first 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of first \(n\) terms.

Solution:

Given: \(S_7 = 49\) and \(S_{17} = 289\).
Formula: \(S_n = \frac{n}{2}[2a + (n-1)d]\)

For \(n=7\): $$49 = \frac{7}{2}[2a + 6d] \Rightarrow 49 = 7[a + 3d] \Rightarrow 7 = a + 3d \dots(1)$$
For \(n=17\): $$289 = \frac{17}{2}[2a + 16d] \Rightarrow 289 = 17[a + 8d] \Rightarrow 17 = a + 8d \dots(2)$$

Subtract (1) from (2): $$5d = 10 \Rightarrow d = 2$$ From (1): \(a + 3(2) = 7 \Rightarrow a = 1\).

Sum of \(n\) terms: $$S_n = \frac{n}{2}[2(1) + (n-1)2]$$ $$S_n = \frac{n}{2}[2 + 2n - 2] = \frac{n}{2}[2n] = n^2$$

Answer: The sum of the first n terms is \(n^2\).

Question 8 (Word Problem)

A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find the production in the 1st year.

Solution:

Since production increases uniformly, it forms an A.P.
Let production in 1st year be \(a\) and annual increase be \(d\).
Given: \(t_3 = 600\) and \(t_7 = 700\).

Using \(t_n = a + (n-1)d\): $$t_3 = a + 2d = 600 \quad \dots(1)$$ $$t_7 = a + 6d = 700 \quad \dots(2)$$

Subtract (1) from (2): $$4d = 100 \Rightarrow d = 25$$ Substitute \(d\) in (1): $$a + 2(25) = 600$$ $$a + 50 = 600$$ $$a = 550$$

Answer: The production in the 1st year was 550 TV sets.

7 Arithmetic Progression

9 INTEGRATION

8 INTEGRATION

7 INTEGRATION

6 Arithmetic Progression

5 Arithmetic Progression

6 INTEGRATION

5 INTEGRATION

4 INTEGRATION

3 INTEGRATION

2 INTEGRATION

4 Arithmetic Progression

3 Arithmetic Progression

1 INTEGRATION

2 Arithmetic Progression

34 Arithmetic Progression

35 Arithmetic Progression

33 Arithmetic Progression

32 Arithmetic Progression

31 Arithmetic Progression

28 Important Arithmetic Progression Questions with Solutions

28 Arithmetic Progression Questions

Curated for excellence by Omtex Classes.

Part 1: Calculations and Terms

Question 1

Find the common difference and the next term of the AP: \( 0.6, 1.7, 2.8, \dots \)

Solution: \( a_1 = 0.6, a_2 = 1.7 \).
\( d = 1.7 - 0.6 = 1.1 \).
Next term \( = 2.8 + 1.1 = 3.9 \).
Answer: d=1.1, Next term=3.9

Question 2

In an AP, if \( d = -4, n = 7, a_n = 4 \), then find \( a \).

Solution: \( a_n = a + (n-1)d \).
\( 4 = a + (7-1)(-4) \)
\( 4 = a - 24 \Rightarrow a = 28 \).
Answer: 28

Question 3

Find the missing terms in the boxes: \( \Box, 38, \Box, \Box, \Box, -22 \).

Solution: \( a_2 = 38 \Rightarrow a+d = 38 \) (i)
\( a_6 = -22 \Rightarrow a+5d = -22 \) (ii)
Subtract (i) from (ii): \( 4d = -60 \Rightarrow d = -15 \).
From (i): \( a - 15 = 38 \Rightarrow a = 53 \).
Terms: 53, 38, 23, 8, -7, -22.
Answer: 53, 23, 8, -7

Question 4

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution: Let APs be \( a, a+d, \dots \) and \( b, b+d, \dots \).
Diff of 100th terms: \( (a+99d) - (b+99d) = a - b = 100 \).
Diff of 1000th terms: \( (a+999d) - (b+999d) = a - b \).
Since \( a-b = 100 \), the difference remains the same.
Answer: 100

Question 5

How many multiples of 4 lie between 10 and 250?

Solution: First multiple > 10 is 12. Last multiple < 250 is 248.
\( 248 = 12 + (n-1)4 \)
\( 236 = 4(n-1) \Rightarrow 59 = n - 1 \Rightarrow n = 60 \).
Answer: 60

Question 6

Find the sum of the first 100 natural numbers.

Solution: Using \( S_n = \frac{n(n+1)}{2} \).
\( S_{100} = \frac{100 \times 101}{2} = 50 \times 101 = 5050 \).
Answer: 5050

Question 7

If the sum of first \( n \) terms of an AP is \( 4n - n^2 \), what is the first term? What is the sum of first two terms? What is the 2nd term?

Solution: \( S_n = 4n - n^2 \).
\( S_1 = 4(1) - 1^2 = 3 \). (First term \( a_1 = 3 \)).
\( S_2 = 4(2) - 2^2 = 8 - 4 = 4 \).
\( a_2 = S_2 - S_1 = 4 - 3 = 1 \).
Answer: First term=3, Sum(2)=4, 2nd term=1

Question 8

Determine the AP whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.

Solution: \( a_7 - a_5 = 12 \Rightarrow (a+6d) - (a+4d) = 12 \Rightarrow 2d = 12 \Rightarrow d = 6 \).
\( a_3 = 16 \Rightarrow a + 2d = 16 \Rightarrow a + 12 = 16 \Rightarrow a = 4 \).
AP: \( 4, 10, 16, \dots \).
Answer: 4, 10, 16...

Question 9

Find the 20th term from the last term of the AP: \( 3, 8, 13, \dots, 253 \).

Solution: Reverse the AP: \( a = 253, d = -5 \).
\( a_{20} = a + 19d = 253 + 19(-5) \)
\( = 253 - 95 = 158 \).
Answer: 158

Question 10

Check whether -150 is a term of the AP: \( 11, 8, 5, 2, \dots \)

Solution: \( a=11, d=-3 \).
\( -150 = 11 + (n-1)(-3) \Rightarrow -161 = -3(n-1) \).
\( n-1 = 161/3 = 53.66 \).
Since \( n \) is not a whole number, it is not a term.
Answer: No

Part 2: Sums and Word Problems

Question 11

Find the sum of the first 15 multiples of 8.

Solution: \( a=8, d=8, n=15 \).
\( S_{15} = \frac{15}{2}[2(8) + 14(8)] = \frac{15}{2}(16 + 112) = \frac{15}{2}(128) = 960 \).
Answer: 960

Question 12

A sum of Rs 700 is to be used to give seven cash prizes to students. If each prize is Rs 20 less than its preceding prize, find the value of each prize.

Solution: \( n=7, S_7=700, d=-20 \).
\( 700 = \frac{7}{2}[2a + 6(-20)] \Rightarrow 200 = 2a - 120 \Rightarrow 2a = 320 \Rightarrow a = 160 \).
Prizes: \( 160, 140, 120, 100, 80, 60, 40 \).
Answer: Rs 160 to Rs 40

Question 13

A spiral is made up of successive semicircles, with centers alternately at A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm... What is the total length of such a spiral made up of 13 consecutive semicircles? (\( \pi = 22/7 \))

Solution: Lengths form AP: \( \pi(0.5), \pi(1.0), \dots \)
\( a=0.5\pi, d=0.5\pi, n=13 \).
Total Length \( = \frac{13}{2}[2(0.5\pi) + 12(0.5\pi)] = \frac{13}{2}[7\pi] \).
\( = \frac{91}{2} \times \frac{22}{7} = 13 \times 11 = 143 \) cm.
Answer: 143 cm

Question 14

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next, 18 in the next, etc. In how many rows are the 200 logs placed and how many logs are in the top row?

Solution: \( S_n=200, a=20, d=-1 \).
\( 200 = \frac{n}{2}[40 + (n-1)(-1)] \Rightarrow 400 = 41n - n^2 \).
\( n^2 - 41n + 400 = 0 \). Factors: 16, 25.
If \( n=25, a_{25} = 20-24 = -4 \) (Impossible).
So \( n=16 \). Logs in top row \( a_{16} = 20 - 15 = 5 \).
Answer: 16 rows, 5 logs

Question 15

Find the sum of odd numbers between 0 and 50.

Solution: Terms: \( 1, 3, 5, \dots, 49 \). \( n=25 \).
\( S_{25} = \frac{25}{2}(1+49) = \frac{25}{2}(50) = 625 \).
Answer: 625

Question 16

A contract on construction specifies a penalty for delay: Rs 200 for the first day, Rs 250 for the second, Rs 300 for the third, etc. How much money does the contractor have to pay as penalty for 30 days delay?

Solution: \( a=200, d=50, n=30 \).
\( S_{30} = \frac{30}{2}[2(200) + 29(50)] = 15[400 + 1450] \).
\( = 15(1850) = 27750 \).
Answer: Rs 27,750

Part 3: Advanced & HOTS

Question 17

If the numbers \( x-2, 4x-1, \) and \( 5x+2 \) are in AP, find the value of \( x \).

Solution: \( 2(4x-1) = (x-2) + (5x+2) \)
\( 8x - 2 = 6x \)
\( 2x = 2 \Rightarrow x = 1 \).
Answer: 1

Question 18

Find the middle term of the AP: \( 213, 205, 197, \dots, 37 \).

Solution: \( a=213, d=-8, a_n=37 \).
\( 37 = 213 + (n-1)(-8) \Rightarrow -176 = -8(n-1) \Rightarrow 22 = n-1 \Rightarrow n=23 \).
Middle term = \( \frac{23+1}{2} = 12 \)th term.
\( a_{12} = 213 + 11(-8) = 213 - 88 = 125 \).
Answer: 125

Question 19

Which term of the sequence \( 20, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, \dots \) is the first negative term?

Solution: \( a=20, d = -0.75 \).
\( 20 + (n-1)(-0.75) < 0 \)
\( 20 < 0.75(n-1) \Rightarrow \frac{20}{0.75} < n-1 \).
\( 26.66 < n-1 \Rightarrow n > 27.66 \).
First integer is 28.
Answer: 28th term

Question 20

If 7 times the 7th term of an AP is equal to 11 times its 11th term, show that the 18th term is zero.

Solution: \( 7(a+6d) = 11(a+10d) \)
\( 7a + 42d = 11a + 110d \)
\( -4a = 68d \Rightarrow a = -17d \).
\( a_{18} = a + 17d = -17d + 17d = 0 \).
Answer: Proved

Question 21

Find the sum of all three-digit natural numbers which are divisible by 7.

Solution: First: 105, Last: 994.
\( 994 = 105 + (n-1)7 \Rightarrow 889 = 7(n-1) \Rightarrow 127 = n-1 \Rightarrow n=128 \).
\( S_{128} = \frac{128}{2}(105+994) = 64(1099) = 70336 \).
Answer: 70336

Question 22

The angles of a triangle are in AP. The greatest angle is twice the least. Find all angles.

Solution: Angles: \( a-d, a, a+d \). Sum = 180 \(\Rightarrow 3a=180 \Rightarrow a=60 \).
\( 60+d = 2(60-d) \Rightarrow 60+d = 120-2d \Rightarrow 3d=60 \Rightarrow d=20 \).
Angles: \( 40^\circ, 60^\circ, 80^\circ \).
Answer: 40°, 60°, 80°

Question 23

Solve the equation: \( 1 + 4 + 7 + 10 + \dots + x = 287 \).

Solution: \( a=1, d=3 \).
\( 287 = \frac{n}{2}[2 + (n-1)3] \Rightarrow 574 = n(3n-1) \).
\( 3n^2 - n - 574 = 0 \). Using quadratic formula, \( n=14 \).
\( x = a_{14} = 1 + 13(3) = 40 \).
Answer: x = 40

Question 24

The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of \( x \) such that the sum of the numbers of the houses preceding the house numbered \( x \) is equal to the sum of the numbers of the houses following it. Find \( x \).

Solution: Sum 1 to \( x-1 \) = Sum \( x+1 \) to 49.
\( \frac{x-1}{2}(1 + x-1) = S_{49} - S_x \).
\( \frac{x(x-1)}{2} = \frac{49 \times 50}{2} - \frac{x(x+1)}{2} \).
\( x^2 - x = 2450 - (x^2 + x) \)
\( 2x^2 = 2450 \Rightarrow x^2 = 1225 \Rightarrow x = 35 \).
Answer: 35

Question 25

If \( m \) times the \( m \)th term of an AP is equal to \( n \) times its \( n \)th term, then show that the \( (m+n) \)th term is 0.

Solution: \( m[a+(m-1)d] = n[a+(n-1)d] \).
\( am + m^2d - md = an + n^2d - nd \).
\( a(m-n) + d(m^2-n^2) - d(m-n) = 0 \).
Divide by \( m-n \): \( a + d(m+n) - d = 0 \).
\( a + (m+n-1)d = 0 \). This is \( a_{m+n} \).
Answer: Proved

Question 26

Calculate the common difference of an AP where the first term is 100, and the sum of the first 6 terms is 5 times the sum of the next 6 terms.

Solution: \( S_6 = 5(S_{12} - S_6) \Rightarrow 6S_6 = 5S_{12} \).
\( 6[\frac{6}{2}(2a+5d)] = 5[\frac{12}{2}(2a+11d)] \).
\( 18(200+5d) = 30(200+11d) \).
Dividing by 6: \( 3(200+5d) = 5(200+11d) \).
\( 600 + 15d = 1000 + 55d \).
\( -400 = 40d \Rightarrow d = -10 \).
Answer: -10

Question 27

The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last terms to the product of the two middle terms is 7:15. Find the numbers.

Solution: Terms: \( a-3d, a-d, a+d, a+3d \). Sum \( 4a=32 \Rightarrow a=8 \).
\( \frac{(8-3d)(8+3d)}{(8-d)(8+d)} = \frac{7}{15} \).
\( \frac{64-9d^2}{64-d^2} = \frac{7}{15} \).
\( 15(64-9d^2) = 7(64-d^2) \Rightarrow 960 - 135d^2 = 448 - 7d^2 \).
\( 512 = 128d^2 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2 \).
For \( d=2 \): \( 2, 6, 10, 14 \).
Answer: 2, 6, 10, 14

Question 28

Show that the sequence \( \log a, \log(ab), \log(ab^2), \dots \) is an AP. Find the \( n \)th term.

Solution: \( a_1 = \log a \).
\( a_2 = \log a + \log b \).
\( a_3 = \log a + 2\log b \).
Common difference \( d = \log b \).
\( a_n = \log a + (n-1)\log b = \log(ab^{n-1}) \).
Answer: Yes, \( \log(ab^{n-1}) \)

Master Mathematics with complete study guides at Omtex Classes and Omtex.co.in.

27 Arithmetic Progression Practice Questions with Solutions

27 Arithmetic Progression Questions

Set 4: Sharpen your skills with Omtex Classes.

Part 1: Basic Calculations

Question 1

Find the next term of the AP: \( \sqrt{7}, \sqrt{28}, \sqrt{63}, \dots \)

Solution: Simplify terms: \( \sqrt{7} \), \( \sqrt{4 \times 7} = 2\sqrt{7} \), \( \sqrt{9 \times 7} = 3\sqrt{7} \).
This is an AP with \( d = \sqrt{7} \).
Next term: \( 4\sqrt{7} = \sqrt{16 \times 7} = \sqrt{112} \).
Answer: \( \sqrt{112} \)

Question 2

Is 302 a term of the AP \( 3, 8, 13, \dots \)?

Solution: \( a=3, d=5 \).
\( 302 = 3 + (n-1)5 \Rightarrow 299 = 5(n-1) \).
\( n-1 = 59.8 \). Since \( n \) is not an integer, it is not a term.
Answer: No

Question 3

If \( a_n = 3n^2 - 4n \), find the 5th term.

Solution: Note: This formula usually represents \( S_n \) in quadratics, but if given as \( a_n \), simply substitute.
\( a_5 = 3(5)^2 - 4(5) = 3(25) - 20 = 75 - 20 = 55 \).
Answer: 55

Question 4

How many multiples of 6 lie between 1 and 100?

Solution: AP: \( 6, 12, \dots, 96 \).
\( 96 = 6 + (n-1)6 \Rightarrow 90 = 6(n-1) \Rightarrow 15 = n-1 \Rightarrow n=16 \).
Answer: 16

Question 5

Find the common difference of the AP whose general term is \( a_n = 3n + 7 \).

Solution: \( a_1 = 10, a_2 = 13 \).
\( d = 13 - 10 = 3 \).
(Shortcut: The coefficient of \( n \) is the common difference).
Answer: 3

Question 6

Find the sum of the first 20 even natural numbers.

Solution: AP: \( 2, 4, 6, \dots \). \( a=2, d=2, n=20 \).
\( S_{20} = \frac{20}{2}[2(2) + 19(2)] = 10(4 + 38) = 10(42) = 420 \).
Or use formula \( n(n+1) = 20 \times 21 = 420 \).
Answer: 420

Question 7

If the 1st term is 7 and the 13th term is 35, find the common difference.

Solution: \( a_{13} = a + 12d \Rightarrow 35 = 7 + 12d \).
\( 28 = 12d \Rightarrow d = 28/12 = 7/3 \).
Answer: 7/3

Question 8

Check if the list of numbers \( 1^2, 3^2, 5^2, 7^2, \dots \) forms an AP.

Solution: Terms: \( 1, 9, 25, 49 \).
\( 9-1 = 8 \). \( 25-9 = 16 \).
Differences are not constant.
Answer: No

Question 9

If 5 times the 5th term of an AP is equal to 8 times its 8th term, find the 13th term.

Solution: \( 5(a+4d) = 8(a+7d) \)
\( 5a + 20d = 8a + 56d \)
\( -3a = 36d \Rightarrow a = -12d \Rightarrow a + 12d = 0 \).
\( a_{13} = a + 12d = 0 \).
Answer: 0

Question 10

Find the sum of \( 2+4+6+\dots+200 \).

Solution: \( a=2, l=200, n=100 \).
\( S_{100} = \frac{100}{2}(2+200) = 50(202) = 10100 \).
Answer: 10100

Part 2: Intermediate Questions

Question 11

How many two-digit numbers are divisible by 6?

Solution: First: 12, Last: 96. \( d=6 \).
\( 96 = 12 + (n-1)6 \Rightarrow 84 = 6(n-1) \Rightarrow 14 = n-1 \Rightarrow n=15 \).
Answer: 15

Question 12

Which term of the AP \( 53, 48, 43, \dots \) is the first negative term?

Solution: \( a=53, d=-5 \).
\( 53 + (n-1)(-5) < 0 \Rightarrow 53 - 5n + 5 < 0 \)
\( 58 < 5n \Rightarrow n > 11.6 \).
First integer is 12.
Answer: 12th term

Question 13

Find the sum of all three-digit numbers divisible by 9.

Solution: AP: \( 108, 117, \dots, 999 \).
\( 999 = 108 + (n-1)9 \Rightarrow 891 = 9(n-1) \Rightarrow 99 = n-1 \Rightarrow n=100 \).
\( S_{100} = \frac{100}{2}(108+999) = 50(1107) = 55350 \).
Answer: 55350

Question 14

The angles of a triangle are in AP. The largest angle is 90°. Find the other angles.

Solution: Angles: \( a, a+d, a+2d \). \( a+2d = 90 \).
Sum: \( 3a+3d = 180 \Rightarrow a+d = 60 \).
\( a+d \) is the middle angle, so \( 60^\circ \).
AP: \( 30^\circ, 60^\circ, 90^\circ \).
Answer: 30°, 60°, 90°

Question 15

Insert three arithmetic means between 3 and 19.

Solution: AP: \( 3, A_1, A_2, A_3, 19 \). Total 5 terms.
\( 19 = 3 + 4d \Rightarrow 16 = 4d \Rightarrow d = 4 \).
Terms: \( 3+4=7 \), \( 7+4=11 \), \( 11+4=15 \).
Answer: 7, 11, 15

Question 16

For what value of \( k \) are \( k, 2k-1, 2k+1 \) in AP?

Solution: \( 2(2k-1) = k + (2k+1) \)
\( 4k - 2 = 3k + 1 \)
\( k = 3 \).
Answer: 3

Question 17

Find the sum of the first \( n \) odd natural numbers.

Solution: \( 1, 3, 5, \dots \). \( a=1, d=2 \).
\( S_n = \frac{n}{2}[2(1) + (n-1)2] = \frac{n}{2}[2 + 2n - 2] = \frac{n}{2}[2n] = n^2 \).
Answer: \( n^2 \)

Question 18

If the \( p \)th term of an AP is \( q \) and the \( q \)th term is \( p \), show that the \( n \)th term is \( p+q-n \).

Solution: \( a+(p-1)d = q \) and \( a+(q-1)d = p \).
Subtracting: \( d(p-q) = q-p = -(p-q) \Rightarrow d = -1 \).
Substitute \( d \): \( a - p + 1 = q \Rightarrow a = p+q-1 \).
\( a_n = p+q-1 + (n-1)(-1) = p+q-1 - n + 1 = p+q-n \).
Answer: Proved

Part 3: Word Problems & Proofs

Question 19

A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2.5 m apart, what is the length of the wood required for the rungs?

Solution: Total height = 250 cm. Gap = 25 cm.
Number of rungs = \( \frac{250}{25} + 1 = 11 \).
AP: \( a=45, l=25, n=11 \).
\( S_{11} = \frac{11}{2}(45+25) = \frac{11}{2}(70) = 11 \times 35 = 385 \).
Answer: 385 cm

Question 20

If \( a, b, c \) are in AP, prove that \( \frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab} \) are also in AP.

Solution: If they are in AP, then \( \frac{1}{ca} - \frac{1}{bc} = \frac{1}{ab} - \frac{1}{ca} \).
Multiply entire equation by \( abc \):
\( b - a = c - b \Rightarrow 2b = a+c \).
Since \( a, b, c \) are in AP, \( 2b=a+c \) is true.
Answer: Proved

Question 21

The sum of the first \( n \) terms of an AP is given by \( S_n = \frac{5n^2}{2} + \frac{3n}{2} \). Find the 20th term.

Solution: \( a_{20} = S_{20} - S_{19} \).
\( S_{20} = \frac{5(400)}{2} + \frac{60}{2} = 1000 + 30 = 1030 \).
\( S_{19} = \frac{5(361)}{2} + \frac{57}{2} = \frac{1805+57}{2} = \frac{1862}{2} = 931 \).
\( a_{20} = 1030 - 931 = 99 \).
Answer: 99

Question 22

If the ratio of the sum of the first \( n \) terms of two APs is \( (7n+1):(4n+27) \), find the ratio of their 9th terms.

Solution: Ratio of \( m \)th terms corresponds to replacing \( n \) with \( 2m-1 \) in sum ratio.
Here \( m=9 \), so replace \( n \) with \( 2(9)-1 = 17 \).
Ratio = \( \frac{7(17)+1}{4(17)+27} = \frac{119+1}{68+27} = \frac{120}{95} \).
Simplify: \( 24:19 \).
Answer: 24:19

Question 23

If \( a, b, c \) are in AP, prove that \( (a-c)^2 = 4(b^2 - ac) \).

Solution: Since AP, \( b = \frac{a+c}{2} \).
RHS: \( 4[(\frac{a+c}{2})^2 - ac] = 4[\frac{a^2+c^2+2ac}{4} - ac] \)
\( = a^2 + c^2 + 2ac - 4ac = a^2 + c^2 - 2ac = (a-c)^2 \).
LHS = RHS.
Answer: Proved

Question 24

The sum of the first 7 terms of an AP is 63 and the sum of its next 7 terms is 161. Find the 28th term.

Solution: \( S_7 = 63 \Rightarrow \frac{7}{2}(2a+6d)=63 \Rightarrow a+3d=9 \).
\( S_{14} = 63 + 161 = 224 \Rightarrow \frac{14}{2}(2a+13d)=224 \Rightarrow 7(2a+13d)=224 \Rightarrow 2a+13d=32 \).
Solve: \( 2(9-3d)+13d=32 \Rightarrow 18-6d+13d=32 \Rightarrow 7d=14 \Rightarrow d=2 \).
\( a = 9 - 6 = 3 \).
\( a_{28} = 3 + 27(2) = 57 \).
Answer: 57

Question 25

Solve for \( x \): \( 1 + 6 + 11 + 16 + \dots + x = 148 \).

Solution: \( a=1, d=5, S_n=148 \).
\( 148 = \frac{n}{2}[2 + (n-1)5] \Rightarrow 296 = 2n + 5n^2 - 5n = 5n^2 - 3n \).
\( 5n^2 - 3n - 296 = 0 \). Using quadratic formula, \( n=8 \).
\( x = a_8 = 1 + 7(5) = 36 \).
Answer: 36

Question 26

The sum of three numbers in AP is 12 and the sum of their cubes is 288. Find the numbers.

Solution: Terms: \( a-d, a, a+d \). Sum \( 3a=12 \Rightarrow a=4 \).
\( (4-d)^3 + 4^3 + (4+d)^3 = 288 \).
\( 64 + (64 - 48d + 12d^2 - d^3) + (64 + 48d + 12d^2 + d^3) = 288 \).
\( 192 + 24d^2 = 288 \Rightarrow 24d^2 = 96 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2 \).
Numbers: \( 2, 4, 6 \).
Answer: 2, 4, 6

Question 27

Split 24 into three parts such that they are in AP and their product is 440.

Solution: Terms: \( a-d, a, a+d \). Sum \( 3a=24 \Rightarrow a=8 \).
Product: \( 8(64-d^2) = 440 \Rightarrow 64-d^2 = 55 \).
\( d^2 = 9 \Rightarrow d = 3 \).
Parts: \( 5, 8, 11 \).
Answer: 5, 8, 11

For more study materials, visit Omtex Classes and Omtex.co.in.

25 Arithmetic Progression

26 Arithmetic Progression

24 Arithmetic Progression

23 Arithmetic Progression

22 Arithmetic Progression

21 Arithmetic Progression

20 Additional Arithmetic Progression Questions with Solutions

20 Arithmetic Progression Questions (Set 5)

Final Set: Complete your mastery with Omtex Classes.

Part 1: Concepts and Calculations

Question 1

Find the 30th term of the AP: \( 10, 7, 4, \dots \)

Solution: \( a = 10 \), \( d = 7 - 10 = -3 \).
\( a_{30} = 10 + (30-1)(-3) \)
\( a_{30} = 10 + 29(-3) = 10 - 87 = -77 \).
Answer: -77

Question 2

Find the number of terms in the AP: \( -1, -5/6, -2/3, \dots, 10/3 \).

Solution: \( a = -1 \).
\( d = -5/6 - (-1) = -5/6 + 6/6 = 1/6 \).
\( a_n = 10/3 \Rightarrow \frac{10}{3} = -1 + (n-1)\frac{1}{6} \).
\( \frac{13}{3} = \frac{n-1}{6} \Rightarrow 26 = n-1 \Rightarrow n = 27 \).
Answer: 27

Question 3

Which term of the AP \( 3, 15, 27, 39, \dots \) will be 132 more than its 54th term?

Solution: \( d = 12 \).
\( a_n = a_{54} + 132 \)
\( a + (n-1)d = a + 53d + 132 \)
\( (n-1)12 = 53(12) + 132 \). (Divide by 12)
\( n-1 = 53 + 11 = 64 \Rightarrow n = 65 \).
Answer: 65th term

Question 4

Find the value of \( x \) for which \( 2x, x+10, 3x+2 \) are three consecutive terms of an AP.

Solution: \( 2(x+10) = 2x + (3x+2) \)
\( 2x + 20 = 5x + 2 \)
\( 18 = 3x \Rightarrow x = 6 \).
Answer: 6

Question 5

If \( 1/x, 1/y, 1/z \) are in AP, prove that \( y = \frac{2xz}{x+z} \).

Solution: Since in AP, \( 2(\frac{1}{y}) = \frac{1}{x} + \frac{1}{z} \).
\( \frac{2}{y} = \frac{z+x}{xz} \).
Inverting both sides: \( \frac{y}{2} = \frac{xz}{x+z} \Rightarrow y = \frac{2xz}{x+z} \).
Answer: Proved

Question 6

Find the sum of all natural numbers between 100 and 500 which are divisible by 8.

Solution: First multiple 104, Last multiple 496.
\( 496 = 104 + (n-1)8 \Rightarrow 392 = 8(n-1) \Rightarrow 49 = n-1 \Rightarrow n = 50 \).
\( S_{50} = \frac{50}{2}(104 + 496) = 25(600) = 15000 \).
Answer: 15000

Question 7

If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.

Solution: \( S_{14} = 1050, a=10 \).
\( 1050 = \frac{14}{2}[2(10) + 13d] \Rightarrow 1050 = 7[20 + 13d] \).
\( 150 = 20 + 13d \Rightarrow 130 = 13d \Rightarrow d = 10 \).
\( a_{20} = 10 + 19(10) = 200 \).
Answer: 200

Question 8

Find the sum of the series: \( 72 + 70 + 68 + \dots + 40 \).

Solution: \( a=72, d=-2, l=40 \).
\( 40 = 72 + (n-1)(-2) \Rightarrow -32 = -2(n-1) \Rightarrow 16 = n-1 \Rightarrow n=17 \).
\( S_{17} = \frac{17}{2}(72+40) = \frac{17}{2}(112) = 17 \times 56 = 952 \).
Answer: 952

Question 9

If the sum of \( n \) terms of an AP is \( S_n = 3n^2 + 5n \), find the AP.

Solution: \( S_1 = 3(1) + 5 = 8 \) (First term \( a=8 \)).
\( S_2 = 3(4) + 10 = 22 \).
\( a_2 = S_2 - S_1 = 22 - 8 = 14 \).
\( d = 14 - 8 = 6 \).
Answer: 8, 14, 20, ...

Question 10

Divide 15 into three parts which are in AP and such that the sum of their squares is 83.

Solution: Parts: \( a-d, a, a+d \). Sum \( 3a=15 \Rightarrow a=5 \).
Squares: \( (5-d)^2 + 25 + (5+d)^2 = 83 \).
\( 25 - 10d + d^2 + 25 + 25 + 10d + d^2 = 83 \).
\( 75 + 2d^2 = 83 \Rightarrow 2d^2 = 8 \Rightarrow d^2 = 4 \Rightarrow d = \pm 2 \).
Parts: \( 3, 5, 7 \).
Answer: 3, 5, 7

Part 2: Advanced Problems

Question 11

The angles of a quadrilateral are in AP whose common difference is 10°. Find the angles.

Solution: Let angles be \( a, a+10, a+20, a+30 \).
Sum of angles of quadrilateral = 360°.
\( 4a + 60 = 360 \Rightarrow 4a = 300 \Rightarrow a = 75 \).
Angles: \( 75^\circ, 85^\circ, 95^\circ, 105^\circ \).
Answer: 75°, 85°, 95°, 105°

Question 12

If the sum of \( p \) terms of an AP is \( q \) and the sum of \( q \) terms is \( p \), find the sum of \( (p+q) \) terms.

Solution: \( S_p = q \) and \( S_q = p \).
Using the standard result for this specific pattern: \( S_{p+q} = -(p+q) \).
Answer: -(p+q)

Question 13

Find the middle term(s) of the AP: \( 7, 13, 19, \dots, 241 \).

Solution: \( a=7, d=6, a_n=241 \).
\( 241 = 7 + (n-1)6 \Rightarrow 234 = 6(n-1) \Rightarrow 39 = n-1 \Rightarrow n=40 \).
Since \( n \) is even, there are two middle terms: \( n/2 = 20 \) and \( (n/2)+1 = 21 \).
\( a_{20} = 7 + 19(6) = 121 \).
\( a_{21} = 7 + 20(6) = 127 \).
Answer: 121 and 127

Question 14

Solve the equation: \( -4 + (-1) + 2 + \dots + x = 437 \).

Solution: \( a=-4, d=3 \). \( S_n = 437 \).
\( 437 = \frac{n}{2}[2(-4) + (n-1)3] \)
\( 874 = n(-8 + 3n - 3) = n(3n-11) = 3n^2 - 11n \).
\( 3n^2 - 11n - 874 = 0 \). Using quadratic formula, \( n=19 \).
\( x = a_{19} = -4 + 18(3) = 50 \).
Answer: 50

Question 15

Find the sum of all odd numbers between 10 and 200.

Solution: AP: \( 11, 13, \dots, 199 \).
\( 199 = 11 + (n-1)2 \Rightarrow 188 = 2(n-1) \Rightarrow 94 = n-1 \Rightarrow n=95 \).
\( S_{95} = \frac{95}{2}(11+199) = \frac{95}{2}(210) = 95 \times 105 = 9975 \).
Answer: 9975

Question 16

The sums of \( n \) terms of two APs are in the ratio \( (3n+8):(7n+15) \). Find the ratio of their 12th terms.

Solution: To find ratio of \( m \)th term, replace \( n \) with \( 2m-1 \).
Here \( m=12 \), so \( n = 2(12)-1 = 23 \).
Ratio = \( \frac{3(23)+8}{7(23)+15} = \frac{69+8}{161+15} = \frac{77}{176} \).
Dividing by 11: \( 7/16 \).
Answer: 7:16

Question 17

Determine the 2nd term of an AP whose 6th term is 12 and 8th term is 22.

Solution: \( a+5d=12 \) and \( a+7d=22 \).
Subtracting: \( 2d = 10 \Rightarrow d=5 \).
\( a + 25 = 12 \Rightarrow a = -13 \).
\( a_2 = a + d = -13 + 5 = -8 \).
Answer: -8

Question 18

Jaspal saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If he continues to save in this manner, in how many months will he save Rs 2000?

Solution: \( a=32, d=4, S_n=2000 \).
\( 2000 = \frac{n}{2}[64 + (n-1)4] \).
\( 4000 = n(60+4n) = 4n^2 + 60n \).
\( 4n^2 + 60n - 4000 = 0 \). Divide by 4: \( n^2 + 15n - 1000 = 0 \).
Factors of -1000 summing to 15: 40 and -25.
\( (n+40)(n-25) = 0 \). Time cannot be negative.
Answer: 25 months

Question 19

If the \( n \)th term of the AP \( 9, 7, 5, \dots \) is same as the \( n \)th term of \( 15, 12, 9, \dots \), find \( n \).

Solution: AP1: \( 9 + (n-1)(-2) = 11 - 2n \).
AP2: \( 15 + (n-1)(-3) = 18 - 3n \).
\( 11 - 2n = 18 - 3n \).
\( 3n - 2n = 18 - 11 \Rightarrow n = 7 \).
Answer: 7

Question 20

How many multiples of 4 lie between 10 and 250?

Solution: Multiples: \( 12, 16, \dots, 248 \).
\( 248 = 12 + (n-1)4 \)
\( 236 = 4(n-1) \Rightarrow 59 = n-1 \Rightarrow n = 60 \).
Answer: 60

This concludes the Arithmetic Progression series. Visit Omtex Classes and Omtex.co.in for other chapter resources.

19 Arithmetic Progression

18 Arithmetic Progression

17 Arithmetic Progression

16 Arithmetic Progression

15 Arithmetic Progression

14 Arithmetic Progression

13 Arithmetic Progression

குர்ஆனில் சினிமாவைப் பற்றி… ?

குர்ஆனில் சினிமாவைப் பற்றி… ?

ஒருமுறை (World Students Association) உலக மாணவர் கழகத்தைச் சார்ந்த மாணவர் குழு ஒன்று மௌலானா அப்துல் அலீம் ஸித்திக்கி அவர்களை பேட்டி கண்டனர். அவர்களில் ஒருமாணவர்,

குர்ஆனில் உலகிலுள்ள அனைத்தும கூறப்பட்டிருக்கிறது என்று நீங்கள் கூறுவது உண்மையென்றால் இக்காலத்திலுள்ள சினிமாவைப்பற்றி கூறப்பட்டிருக்கிறதா? என கிண்டலாகக் கேட்டார்.

அப்போது சினிமா அறிமுகமகாத காலம். மௌலானா ஸித்தீக்கீ அவர்கள், சினிமா என்றால் என்ன என்று சற்று விளக்மாகக்கூறுங்கள். பினனர் நான் பதில் சொல்கிறேன் என்றார்கள்.

அந்த மாணவர்,

“கற்பனைக் கதைகளை விலைக்கு வாங்கி அதில் கூத்தடிகளை நடிக்க வைத்து, மக்களை சிரிக்கவைத்துப் பொழுது போக்குவது தான் சினிமா என்று விளக்கம் கூறினார். உடனே மௌலானா அவர்கள் “ஆம்! அதுபற்றி குர்ஆனில் கூறப்பட்டிருக்கிறதே என்றார்கள். அவரோ வியப்பு மேலீட்டால், ‘ எங்கே காட்டுங்கள் பார்க்கலாம் என மீண்டும் வினாவைத் தொடர்ந்தார் அந்த மாணவர்.

அதைக் கேட்ட மௌலானா சிறிதும் தயஙகாமல், குர்ஆனின் 31வது அத்தியாயத்தில் சூரா லுக்மானில் ஆறாவது வசனத்தில்

وَمِنَ النَّاسِ مَن يَشْتَرِي لَهْوَ الْحَدِيثِ لِيُضِلَّ عَن سَبِيلِ اللَّهِ بِغَيْرِ عِلْمٍ وَيَتَّخِذَهَا هُزُواً أُولَئِكَ لَهُمْ عَذَابٌ مُّهِينٌ

(அல்குர்ஆன் : அத்தியாயம் -லுக்மான்,வசனம்- 33)

மனிதர்களில் சிலர் உள்ளனர்.அவர்கள் வீணான செய்திகளை ( பெய்யான கட்டுக்கதைகளை) விலைக்கு வாங்கி அறிவின்றி அல்லாஹ்வின் பாதையிலிருந்து (மக்களை) வழிகெடுத்து அதனை பரிகாசமாக்கிக் கொள்கின்றனர். இவர்களுக்கு இழிவு தரும் வேதனையுண்டு’ (31:6)

என்ற திருமறை வசனத்தை ஓதிக் காண்பித்து கேள்வி கேட்டவரையே வாயடைக்கசெய்தார்கள். உடனே அந்த மாணவர் குழு அனைவரும் “1400 ஆண்டுகளுக்கு முன்னரே சினிமாவைப்பற்றியும் கூறப்பட்டி ருக்கிறதே” என்று அதிசயித்து இஸ்லாத்தை ஏற்றனர்.

சினிமாவை சென்ற நூற்றாண்டில் தான் கணடுபிடித்தார்கள். ஆனால் 1400 ஆண்டுகளுக்கு முன்னரே அதைப்பற்றி மிகத்துல்லியமாகக் கூறப்பட்டிருக்கிறதென்றால் அது முக்காலத்தையும் அறிந்த இறைவனின் வேதவாக்காகத்தான் இருக்கமுடியும் என்பதில் என்ன சந்தேகமிருக்கமுடியும் ?

இதுவும் குர்ஆன் இறைவேதம் என்பதற்கு இதுவும் ஒரு சான்றாகும்