Exercise 5.2 Solutions
Arithmetic Progressions
1. Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:
The formula to use is: an = a + (n-1)d
(i) a = 7, d = 3, n = 8, an = ?
a8 = 7 + (8-1) × 3 = 7 + 7 × 3 = 7 + 21 = 28.
(ii) a = -18, n = 10, an = 0, d = ?
0 = -18 + (10-1)d => 18 = 9d => d = 2.
(iii) d = -3, n = 18, an = -5, a = ?
-5 = a + (18-1)(-3) => -5 = a – 51 => a = 51 – 5 = 46.
(iv) a = -18.9, d = 2.5, an = 3.6, n = ?
3.6 = -18.9 + (n-1)2.5 => 22.5 = (n-1)2.5 => n-1 = 22.5/2.5 = 9 => n = 10.
(v) a = 3.5, d = 0, n = 105, an = ?
a105 = 3.5 + (105-1) × 0 = 3.5 + 0 = 3.5.
2. Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, … is
a = 10, d = 7 – 10 = -3, n = 30.
a30 = 10 + (30-1)(-3) = 10 + 29(-3) = 10 – 87 = -77.
Answer: (C) -77
(ii) 11th term of the AP: -3, -1/2, 2, … is
a = -3, d = -1/2 – (-3) = 5/2, n = 11.
a11 = -3 + (11-1)(5/2) = -3 + 10(5/2) = -3 + 25 = 22.
Answer: (B) 22
3. In the following APs, find the missing terms in the boxes:
(i) 2, [ ], 26
a = 2, a3 = 26. So, a + 2d = 26 => 2 + 2d = 26 => 2d = 24 => d = 12.
Missing term = a + d = 2 + 12 = 14.
(ii) [ ], 13, [ ], 3
a2 = 13, a4 = 3. So, a+d=13, a+3d=3.
Subtracting the first from the second: 2d = -10 => d = -5.
Substitute d: a + (-5) = 13 => a = 18.
Missing terms: a = 18 and a3 = a+2d = 18+2(-5) = 8.
(iii) 5, [ ], [ ], 9 1/2
a = 5, a4 = 19/2. So, a+3d = 19/2 => 5+3d = 19/2 => 3d = 9/2 => d = 3/2.
Missing terms: a2 = 5+3/2 = 13/2 and a3 = 13/2+3/2 = 8.
(iv) -4, [ ], [ ], [ ], [ ], 6
a = -4, a6 = 6. So, a+5d = 6 => -4+5d=6 => 5d=10 => d = 2.
Missing terms: -2, 0, 2, 4.
(v) [ ], 38, [ ], [ ], [ ], -22
a2 = 38, a6 = -22. So, a+d=38, a+5d=-22.
Subtracting: 4d = -60 => d = -15.
Substitute d: a + (-15) = 38 => a = 53.
Missing terms: 53, 38, 23, 8, -7, -22.
4. Which term of the AP: 3, 8, 13, 18, … is 78?
a=3, d=5, an=78. Find n.
78 = 3 + (n-1)5 => 75 = (n-1)5 => 15 = n-1 => n = 16.
Answer: The 16th term is 78.
5. Find the number of terms in each of the following APs:
(i) 7, 13, 19, …, 205
a=7, d=6, an=205. 205 = 7 + (n-1)6 => 198 = (n-1)6 => 33 = n-1 => n = 34.
Answer: 34 terms.
(ii) 18, 15 1/2, 13, …, -47
a=18, d=15.5-18=-2.5, an=-47. -47 = 18 + (n-1)(-2.5) => -65 = (n-1)(-2.5) => 26 = n-1 => n = 27.
Answer: 27 terms.
6. Check whether –150 is a term of the AP: 11, 8, 5, 2, …
a=11, d=-3. Let an = -150.
-150 = 11 + (n-1)(-3) => -161 = (n-1)(-3) => 161/3 = n-1.
n = 161/3 + 1 = 164/3. Since n is not a positive integer, -150 is not a term of this AP.
Answer: No.
7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
a11 = a+10d = 38. a16 = a+15d = 73.
Subtracting the equations: 5d = 35 => d = 7.
Substitute d: a + 10(7) = 38 => a = -32.
a31 = a + 30d = -32 + 30(7) = -32 + 210 = 178.
Answer: The 31st term is 178.
8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
a3 = a+2d = 12. Last term a50 = a+49d = 106.
Subtracting: 47d = 94 => d = 2.
Substitute d: a + 2(2) = 12 => a = 8.
a29 = a + 28d = 8 + 28(2) = 8 + 56 = 64.
Answer: The 29th term is 64.
9. If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of this AP is zero?
a3 = a+2d = 4. a9 = a+8d = -8.
Subtracting: 6d = -12 => d = -2.
Substitute d: a + 2(-2) = 4 => a = 8.
Let an = 0. 0 = 8 + (n-1)(-2) => -8 = (n-1)(-2) => 4 = n-1 => n=5.
Answer: The 5th term is zero.
10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
a17 – a10 = 7.
(a+16d) – (a+9d) = 7 => 7d = 7 => d = 1.
Answer: The common difference is 1.
11. Which term of the AP: 3, 15, 27, 39, … will be 132 more than its 54th term?
a=3, d=12. First, find a54 = 3 + (53)12 = 3 + 636 = 639.
The required term is 639 + 132 = 771.
Let an = 771. 771 = 3 + (n-1)12 => 768 = (n-1)12 => 64 = n-1 => n = 65.
Answer: The 65th term.
12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Let the APs be An and Bn with first terms a1 and b1 and common difference d.
A100 – B100 = (a1 + 99d) – (b1 + 99d) = a1 – b1 = 100.
The difference between their 1000th terms is A1000 – B1000 = (a1 + 999d) – (b1 + 999d) = a1 – b1.
Since a1 – b1 = 100, the difference is still 100.
Answer: 100.
13. How many three-digit numbers are divisible by 7?
First three-digit number divisible by 7 is 105. Last is 994.
This forms an AP: a=105, d=7, an=994.
994 = 105 + (n-1)7 => 889 = (n-1)7 => 127 = n-1 => n = 128.
Answer: 128 numbers.
14. How many multiples of 4 lie between 10 and 250?
First multiple of 4 after 10 is 12. Last multiple before 250 is 248.
AP: a=12, d=4, an=248.
248 = 12 + (n-1)4 => 236 = (n-1)4 => 59 = n-1 => n=60.
Answer: 60 multiples.
15. For what value of n, are the nth terms of two APs: 63, 65, 67, … and 3, 10, 17, … equal?
1st AP: an = 63 + (n-1)2.
2nd AP: bn = 3 + (n-1)7.
Set them equal: 63 + 2n – 2 = 3 + 7n – 7 => 61 + 2n = 7n – 4 => 65 = 5n => n = 13.
Answer: n = 13.
16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
a3 = a + 2d = 16.
a7 – a5 = (a+6d) – (a+4d) = 12 => 2d = 12 => d=6.
Substitute d: a + 2(6) = 16 => a = 4.
Answer: The AP is 4, 10, 16, 22, …
17. Find the 20th term from the last term of the AP: 3, 8, 13, …, 253.
To find the nth term from the end, we can reverse the AP.
New AP: 253, …, 13, 8, 3. Here, a=253, d = -5.
Find the 20th term of this new AP: a20 = 253 + (20-1)(-5) = 253 – 95 = 158.
Answer: The 20th term from the last is 158.
18. The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
(a+3d) + (a+7d) = 24 => 2a+10d = 24 => a+5d=12.
(a+5d) + (a+9d) = 44 => 2a+14d=44 => a+7d=22.
Subtracting the two new equations: 2d = 10 => d=5.
Substitute d: a + 5(5) = 12 => a = -13.
Answer: The first three terms are -13, -8, -3.
19. Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?
AP: 5000, 5200, … a=5000, d=200, an=7000.
7000 = 5000 + (n-1)200 => 2000 = (n-1)200 => 10 = n-1 => n=11.
It took 11 years. The year is 1995 + (11-1) = 2005.
Answer: In the year 2005.
20. Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.
AP: 5, 6.75, … a=5, d=1.75, an=20.75.
20.75 = 5 + (n-1)1.75 => 15.75 = (n-1)1.75 => 9 = n-1 => n=10.
Answer: n = 10.
