Table of Contents
Exercise 6.2 Solutions – Class X Mathematics
These solutions are based on the Telangana State Class X Mathematics textbook, focusing on arithmetic progressions (AP), finding terms, common differences, and solving related problems. Mathematical expressions are rendered using MathJax.
1. Fill in the blanks in the following table, given that \( a \) is the first term, \( d \) the common difference, and \( a_n \) the \( n \)th term of the AP:
| S.No. | \( a \) | \( d \) | \( n \) | \( a_n \) |
|---|---|---|---|---|
| (i) | 7 | 3 | 8 | … |
| (ii) | … | -3 | 18 | -5 |
| (iii) | -18.9 | 2.5 | … | 3.6 |
| (iv) | 3.5 | 0 | 105 | … |
(i) \( a = 7, d = 3, n = 8, a_n = ? \)
Formula: \( a_n = a + (n – 1)d \).
Substitute: \( a_8 = 7 + (8 – 1) \cdot 3 = 7 + 7 \cdot 3 = 7 + 21 = 28 \).
\( a_n = 28 \)
(ii) \( d = -3, n = 18, a_n = -5, a = ? \)
Formula: \( a_n = a + (n – 1)d \).
Substitute: \( -5 = a + (18 – 1) \cdot (-3) \implies -5 = a + 17 \cdot (-3) \implies -5 = a – 51 \).
Solve: \( a = -5 + 51 = 46 \).
\( a = 46 \)
(iii) \( a = -18.9, d = 2.5, a_n = 3.6, n = ? \)
Formula: \( a_n = a + (n – 1)d \).
Substitute: \( 3.6 = -18.9 + (n – 1) \cdot 2.5 \).
Solve: \( 3.6 + 18.9 = (n – 1) \cdot 2.5 \implies 22.5 = (n – 1) \cdot 2.5 \implies n – 1 = \frac{22.5}{2.5} = 9 \implies n = 10 \).
\( n = 10 \)
(iv) \( a = 3.5, d = 0, n = 105, a_n = ? \)
Formula: \( a_n = a + (n – 1)d \).
Substitute: \( a_{105} = 3.5 + (105 – 1) \cdot 0 = 3.5 \).
\( a_n = 3.5 \)
2. Find the:
(i) 30th term of the AP: 10, 7, 4, …
First term (\( a \)): 10, common difference (\( d \)): \( 7 – 10 = -3 \).
Formula: \( a_n = a + (n – 1)d \).
For \( n = 30 \): \( a_{30} = 10 + (30 – 1) \cdot (-3) = 10 + 29 \cdot (-3) = 10 – 87 = -77 \).
30th term: -77
(ii) 11th term of the AP: \(-3, -\frac{1}{2}, 2, \ldots\)
First term: \(-3\), common difference: \( -\frac{1}{2} – (-3) = \frac{5}{2} \).
Formula: \( a_n = a + (n – 1)d \).
For \( n = 11 \): \( a_{11} = -3 + (11 – 1) \cdot \frac{5}{2} = -3 + 10 \cdot \frac{5}{2} = -3 + 25 = 22 \).
11th term: 22
3. Find the respective terms for the following APs:
(i) \( a_1 = 2, a_3 = 26 \) find \( a_2 \)
\( a_1 = a = 2 \), \( a_3 = a + 2d = 26 \).
Solve: \( 2 + 2d = 26 \implies 2d = 24 \implies d = 12 \).
\( a_2 = a + d = 2 + 12 = 14 \).
\( a_2 = 14 \)
(ii) \( a_2 = 13, a_4 = 3 \) find \( a_1, a_3 \)
\( a_2 = a + d = 13 \), \( a_4 = a + 3d = 3 \).
Subtract: \( (a + 3d) – (a + d) = 3 – 13 \implies 2d = -10 \implies d = -5 \).
Substitute: \( a + (-5) = 13 \implies a = 18 \).
\( a_1 = a = 18 \), \( a_3 = a + 2d = 18 + 2 \cdot (-5) = 8 \).
\( a_1 = 18, a_3 = 8 \)
(iii) \( a_1 = 5, a_4 = 9\frac{1}{2} \) find \( a_2, a_3 \)
\( a_1 = 5 \), \( a_4 = a + 3d = \frac{19}{2} \).
Solve: \( 5 + 3d = \frac{19}{2} \implies 3d = \frac{19}{2} – 5 = \frac{9}{2} \implies d = \frac{3}{2} \).
\( a_2 = 5 + \frac{3}{2} = \frac{13}{2} \), \( a_3 = \frac{13}{2} + \frac{3}{2} = 8 \).
\( a_2 = \frac{13}{2}, a_3 = 8 \)
(iv) \( a_1 = -4, a_6 = 6 \) find \( a_2, a_3, a_4, a_5 \)
\( a_1 = -4 \), \( a_6 = a + 5d = 6 \).
Solve: \( -4 + 5d = 6 \implies 5d = 10 \implies d = 2 \).
\( a_2 = -4 + 2 = -2 \), \( a_3 = -2 + 2 = 0 \), \( a_4 = 0 + 2 = 2 \), \( a_5 = 2 + 2 = 4 \).
\( a_2 = -2, a_3 = 0, a_4 = 2, a_5 = 4 \)
(v) \( a_2 = 38, a_6 = -22 \) find \( a_1, a_3, a_4, a_5 \)
\( a_2 = a + d = 38 \), \( a_6 = a + 5d = -22 \).
Subtract: \( (a + 5d) – (a + d) = -22 – 38 \implies 4d = -60 \implies d = -15 \).
Substitute: \( a + (-15) = 38 \implies a = 53 \).
\( a_1 = 53 \), \( a_3 = 53 + 2 \cdot (-15) = 23 \), \( a_4 = 23 + (-15) = 8 \), \( a_5 = 8 + (-15) = -7 \).
\( a_1 = 53, a_3 = 23, a_4 = 8, a_5 = -7 \)
4. Which term of the AP: 3, 8, 13, 18, … is 78?
\( a = 3 \), \( d = 8 – 3 = 5 \), \( a_n = 78 \).
Formula: \( a_n = a + (n – 1)d \).
Solve: \( 78 = 3 + (n – 1) \cdot 5 \implies 75 = (n – 1) \cdot 5 \implies n – 1 = 15 \implies n = 16 \).
Check: \( a_{16} = 3 + (16 – 1) \cdot 5 = 3 + 15 \cdot 5 = 78 \).
Term: 16th
5. Find the number of terms in each of the following APs:
(i) 7, 13, 19, …, 205
\( a = 7 \), \( d = 13 – 7 = 6 \), last term \( a_n = 205 \).
Formula: \( a_n = a + (n – 1)d \).
Solve: \( 205 = 7 + (n – 1) \cdot 6 \implies 198 = (n – 1) \cdot 6 \implies n – 1 = 33 \implies n = 34 \).
Number of terms: 34
(ii) \( 18, 15\frac{1}{2}, 13, …, -47 \)
\( a = 18 \), \( d = \frac{31}{2} – 18 = \frac{31}{2} – \frac{36}{2} = -\frac{5}{2} \), last term = \(-47\).
Solve: \( -47 = 18 + (n – 1) \cdot \left(-\frac{5}{2}\right) \implies -65 = (n – 1) \cdot \left(-\frac{5}{2}\right) \implies n – 1 = 26 \implies n = 27 \).
Number of terms: 27
6. Check whether -150 is a term of the AP: 11, 8, 5, 2, …
\( a = 11 \), \( d = 8 – 11 = -3 \), term = \(-150\).
Solve: \( -150 = 11 + (n – 1) \cdot (-3) \implies -161 = (n – 1) \cdot (-3) \implies n – 1 = \frac{161}{3} \).
Since \( \frac{161}{3} \approx 53.67 \), not an integer, -150 is not a term.
Is -150 a term: No
7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
\( a_{11} = a + 10d = 38 \), \( a_{16} = a + 15d = 73 \).
Subtract: \( (a + 15d) – (a + 10d) = 73 – 38 \implies 5d = 35 \implies d = 7 \).
Substitute: \( a + 10 \cdot 7 = 38 \implies a + 70 = 38 \implies a = -32 \).
\( a_{31} = -32 + (31 – 1) \cdot 7 = -32 + 30 \cdot 7 = -32 + 210 = 178 \).
31st term: 178
8. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
\( a_3 = a + 2d = 4 \), \( a_9 = a + 8d = -8 \).
Subtract: \( 6d = -12 \implies d = -2 \).
Substitute: \( a + 2 \cdot (-2) = 4 \implies a – 4 = 4 \implies a = 8 \).
Find \( n \) where \( a_n = 0 \): \( 0 = 8 + (n – 1) \cdot (-2) \implies -8 = (n – 1) \cdot (-2) \implies n – 1 = 4 \implies n = 5 \).
Term: 5th
9. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
\( a_{17} = a + 16d \), \( a_{10} = a + 9d \).
Given: \( a_{17} = a_{10} + 7 \implies (a + 16d) = (a + 9d) + 7 \implies 16d – 9d = 7 \implies 7d = 7 \implies d = 1 \).
Common difference: 1
10. 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 first AP be \( a, a + d, \ldots \), second AP be \( b, b + d, \ldots \).
100th terms: \( a_{100} = a + 99d \), \( b_{100} = b + 99d \).
Given: \( (a + 99d) – (b + 99d) = 100 \implies a – b = 100 \).
1000th terms: \( a_{1000} = a + 999d \), \( b_{1000} = b + 999d \).
Difference: \( (a + 999d) – (b + 999d) = a – b = 100 \).
Difference: 100
11. How many three-digit numbers are divisible by 7?
Three-digit numbers: 100 to 999. Find numbers divisible by 7.
First number: 105 (since \( 100 \div 7 \approx 14.28 \), \( 7 \cdot 15 = 105 \)).
Last number: 994 (since \( 999 \div 7 \approx 142.71 \), \( 7 \cdot 142 = 994 \)).
Sequence: 105, 112, …, 994. \( a = 105 \), \( d = 7 \), last term = 994.
Solve: \( 994 = 105 + (n – 1) \cdot 7 \implies 889 = (n – 1) \cdot 7 \implies n – 1 = 127 \implies n = 128 \).
Number of terms: 128
12. How many multiples of 4 lie between 10 and 250?
Multiples of 4: First number after 10 is 12, last number before 250 is 248.
Sequence: 12, 16, …, 248. \( a = 12 \), \( d = 4 \), last term = 248.
Solve: \( 248 = 12 + (n – 1) \cdot 4 \implies 236 = (n – 1) \cdot 4 \implies n – 1 = 59 \implies n = 60 \).
Number of multiples: 60
13. For what value of \( n \), are the \( n \)th terms of two APs: 63, 65, 67, … and 3, 10, 17, … equal?
First AP: \( a = 63 \), \( d = 2 \). \( a_n = 63 + (n – 1) \cdot 2 \).
Second AP: \( a = 3 \), \( d = 7 \). \( a_n = 3 + (n – 1) \cdot 7 \).
Set equal: \( 63 + (n – 1) \cdot 2 = 3 + (n – 1) \cdot 7 \).
Simplify: \( 63 + 2(n – 1) = 3 + 7(n – 1) \implies 60 = 5(n – 1) \implies n – 1 = 12 \implies n = 13 \).
\( n = 13 \)
14. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
\( a_3 = a + 2d = 16 \).
\( a_7 = a + 6d \), \( a_5 = a + 4d \), given \( a_7 = a_5 + 12 \implies (a + 6d) = (a + 4d) + 12 \implies 2d = 12 \implies d = 6 \).
Substitute: \( a + 2 \cdot 6 = 16 \implies a + 12 = 16 \implies a = 4 \).
AP: 4, 10, 16, 22, …
AP: 4, 10, 16, 22, …
15. Find the 20th term from the end of the AP: 3, 8, 13, …, 253.
\( a = 3 \), \( d = 5 \), last term = 253.
Find total terms: \( 253 = 3 + (n – 1) \cdot 5 \implies 250 = (n – 1) \cdot 5 \implies n = 51 \).
20th term from end is \( (51 – 20 + 1) = 32 \)nd term from start.
\( a_{32} = 3 + (32 – 1) \cdot 5 = 3 + 31 \cdot 5 = 3 + 155 = 158 \).
20th term from end: 158
16. 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.
\( a_4 = a + 3d \), \( a_8 = a + 7d \), \( a_4 + a_8 = 2a + 10d = 24 \implies a + 5d = 12 \).
\( a_6 = a + 5d \), \( a_{10} = a + 9d \), \( a_6 + a_{10} = 2a + 14d = 44 \implies a + 7d = 22 \).
Subtract: \( (a + 7d) – (a + 5d) = 22 – 12 \implies 2d = 10 \implies d = 5 \).
Substitute: \( a + 5 \cdot 5 = 12 \implies a + 25 = 12 \implies a = -13 \).
First three terms: \(-13, -13 + 5 = -8, -8 + 5 = -3\).
First three terms: -13, -8, -3
17. Subba Rao started his job in 1995 at a monthly salary of ₹5000 and received an increment of ₹200 each year. In which year did his salary reach ₹7000?
Salary forms an AP: 5000, 5200, 5400, …, \( a = 5000 \), \( d = 200 \).
Find \( n \) where \( a_n = 7000 \): \( 7000 = 5000 + (n – 1) \cdot 200 \implies 2000 = (n – 1) \cdot 200 \implies n – 1 = 10 \implies n = 11 \).
\( n = 11 \) means 11th year: \( 1995 + 10 = 2005 \).
Year: 2005
