20 Additional Arithmetic Progression Questions with Solutions

20 Arithmetic Progression Questions (Set 5)

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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.