Get complete, step-by-step solutions for NCERT Class 10 Maths Chapter 5 Exercise 5.2.Class 10 Maths Exercise 5.2, $n$-th Term AP Formula, Arithmetic Progression Solutions, Missing Terms AP, Multiples of a Number AP, Salary Increment Problem, Equal $n$-th Terms AP
Master the use of the $n$-th term formula, $a_n = a + (n-1)d$, to find missing terms $a, d, n$, or $a_n$ (Q.1). Practice finding specific terms of an AP (Q.2, Q.7, Q.8) and determining the term number (Q.4). Solutions cover advanced problems including checking if a number is an AP term (Q.6), finding the difference between terms (Q.10), solving for AP terms when given two equations (Q.18), and solving application problems like salary increments and savings (Q.19, Q.20). Crucial topics include finding the number of terms (Q.5), multiples between two numbers (Q.13, Q.14), and $n$-th terms of two APs being equal (Q.15)
The formula used to find the $n$-th term of an AP is:
$$a_n = a + (n-1)d$$
where $a$ is the first term, $d$ is the common difference, and $n$ is the term number.
1. Fill in the Blanks
We use the formula $a_n = a + (n-1)d$ to find the missing value in each case.
| a | d | n | an | Calculation | Answer | |
| (i) | 7 | 3 | 8 | … | $a_8 = 7 + (8-1)3 = 7 + 21 = 28$ | $\mathbf{28}$ |
| (ii) | -18 | … | 10 | 0 | $0 = -18 + (10-1)d \implies 18 = 9d \implies d = 2$ | $\mathbf{2}$ |
| (iii) | … | -3 | 18 | -5 | $-5 = a + (18-1)(-3) \implies -5 = a – 51 \implies a = 46$ | $\mathbf{46}$ |
| (iv) | -18.9 | 2.5 | … | 3.6 | $3.6 = -18.9 + (n-1)2.5 \implies 22.5 = (n-1)2.5 \implies n-1 = 9 \implies n = 10$ | $\mathbf{10}$ |
| (v) | 3.5 | 0 | 105 | … | $a_{105} = 3.5 + (105-1)0 = 3.5 + 0 = 3.5$ | $\mathbf{3.5}$ |
2. Choose the Correct Choice
(i) 30th term of the AP: $10, 7, 4, \dots$
Here $a=10$ and $d = 7 – 10 = -3$. We need $a_{30}$.
$$a_{30} = a + (30-1)d = 10 + 29(-3) = 10 – 87 = -77$$
Answer: (C) -77
(ii) 11th term of the AP: $-3, -1/2, 2, \dots$
Here $a=-3$ and $d = -1/2 – (-3) = -1/2 + 3 = 5/2$. We need $a_{11}$.
$$a_{11} = a + (11-1)d = -3 + 10\left(\frac{5}{2}\right) = -3 + 5(5) = -3 + 25 = 22$$
Answer: (B) 22
3. Finding Missing Terms
(i) $2, \mathbf{\square}, 26$
This is $a_1, a_2, a_3$.
$a_1 = 2$, $a_3 = 26$.
$a_3 = a_1 + 2d \implies 26 = 2 + 2d \implies 24 = 2d \implies d = 12$.
Missing term: $a_2 = a_1 + d = 2 + 12 = \mathbf{14}$.
(ii) $\mathbf{\square}, 13, \mathbf{\square}, 3$
$a_2 = 13$, $a_4 = 3$.
$a_4 – a_2 = (a+3d) – (a+d) = 2d$.
$3 – 13 = 2d \implies -10 = 2d \implies d = -5$.
$a_1 = a_2 – d = 13 – (-5) = 18$.
$a_3 = a_4 – d = 3 – (-5) = 8$.
Missing terms: $\mathbf{18}, 13, \mathbf{8}, 3$.
(iii) $5, \mathbf{\square}, \mathbf{\square}, 9\frac{1}{2}$
$a_1 = 5$, $a_4 = 9.5$.
$a_4 = a_1 + 3d \implies 9.5 = 5 + 3d \implies 4.5 = 3d \implies d = 1.5$.
$a_2 = 5 + 1.5 = 6.5$.
$a_3 = 6.5 + 1.5 = 8.0$.
Missing terms: $5, \mathbf{6.5}, \mathbf{8}, 9.5$ (or $\mathbf{13/2}, \mathbf{8}$).
(iv) $-4, \mathbf{\square}, \mathbf{\square}, \mathbf{\square}, \mathbf{\square}, 6$
$a_1 = -4$, $a_6 = 6$.
$a_6 = a_1 + 5d \implies 6 = -4 + 5d \implies 10 = 5d \implies d = 2$.
$a_2 = -4 + 2 = -2$.
$a_3 = -2 + 2 = 0$.
$a_4 = 0 + 2 = 2$.
$a_5 = 2 + 2 = 4$.
Missing terms: $-4, \mathbf{-2}, \mathbf{0}, \mathbf{2}, \mathbf{4}, 6$.
(v) $\mathbf{\square}, 38, \mathbf{\square}, \mathbf{\square}, \mathbf{\square}, -22$
$a_2 = 38$, $a_6 = -22$.
$a_6 – a_2 = 4d$.
$-22 – 38 = 4d \implies -60 = 4d \implies d = -15$.
$a_1 = a_2 – d = 38 – (-15) = 53$.
$a_3 = 38 + (-15) = 23$.
$a_4 = 23 + (-15) = 8$.
$a_5 = 8 + (-15) = -7$.
Missing terms: $\mathbf{53}, 38, \mathbf{23}, \mathbf{8}, \mathbf{-7}, -22$.
4. Which term of the AP: $3, 8, 13, 18, \dots$, is 78?
Here $a=3, d=8-3=5$. We are looking for $n$ such that $a_n = 78$.
$$a_n = a + (n-1)d$$
$$78 = 3 + (n-1)5$$
$$75 = (n-1)5$$
$$15 = n – 1 \implies n = 16$$
The term is the $\mathbf{16^{th} \text{ term}}$.
5. Find the number of terms
(i) $7, 13, 19, \dots, 205$
Here $a=7, d=6, a_n=205$. We need $n$.
$$205 = 7 + (n-1)6$$
$$198 = (n-1)6$$
$$33 = n – 1 \implies n = 34$$
The number of terms is $\mathbf{34}$.
(ii) $18, 15\frac{1}{2}, 13, \dots, -47$
Here $a=18, d = 15.5 – 18 = -2.5$, $a_n=-47$. We need $n$.
$$-47 = 18 + (n-1)(-2.5)$$
$$-65 = -2.5(n-1)$$
$$n – 1 = \frac{-65}{-2.5} = 26$$
$$n = 27$$
The number of terms is $\mathbf{27}$.
6. Check whether $-150$ is a term of the AP: $11, 8, 5, 2, \dots$
Here $a=11, d = 8 – 11 = -3$. We check if $a_n = -150$ gives a positive integer value for $n$.
$$-150 = 11 + (n-1)(-3)$$
$$-161 = -3(n-1)$$
$$n – 1 = \frac{161}{3}$$
Since 161 is not divisible by 3 (sum of digits $1+6+1=8$), $\frac{161}{3}$ is not an integer.
$$n = \frac{161}{3} + 1 = \frac{164}{3} = 54\frac{2}{3}$$
Since $n$ must be a positive integer, $-150$ is not a term of the given AP.
7. Find the 31st term
Given $a_{11} = 38$ and $a_{16} = 73$.
Using $a_n = a + (n-1)d$:
$$a_{11} = a + 10d = 38 \quad \dots (1)$$
$$a_{16} = a + 15d = 73 \quad \dots (2)$$
Subtract (1) from (2):
$$(a + 15d) – (a + 10d) = 73 – 38$$
$$5d = 35 \implies d = 7$$
Substitute $d=7$ into (1):
$$a + 10(7) = 38 \implies a + 70 = 38 \implies a = -32$$
Now find $a_{31}$:
$$a_{31} = a + 30d = -32 + 30(7) = -32 + 210 = 178$$
The $\mathbf{31^{st} \text{ term is } 178}$.
8. AP with 50 terms
Given $n=50$, $a_3 = 12$, and last term $a_{50} = 106$. We need $a_{29}$.
$$a_3 = a + 2d = 12 \quad \dots (1)$$
$$a_{50} = a + 49d = 106 \quad \dots (2)$$
Subtract (1) from (2):
$$(a + 49d) – (a + 2d) = 106 – 12$$
$$47d = 94 \implies d = 2$$
Substitute $d=2$ into (1):
$$a + 2(2) = 12 \implies a + 4 = 12 \implies a = 8$$
Now find $a_{29}$:
$$a_{29} = a + 28d = 8 + 28(2) = 8 + 56 = 64$$
The $\mathbf{29^{th} \text{ term is } 64}$.
9. Finding which term is zero
Given $a_3 = 4$ and $a_9 = -8$. We need $n$ such that $a_n = 0$.
$$a_3 = a + 2d = 4 \quad \dots (1)$$
$$a_9 = a + 8d = -8 \quad \dots (2)$$
Subtract (1) from (2):
$$(a + 8d) – (a + 2d) = -8 – 4$$
$$6d = -12 \implies d = -2$$
Substitute $d=-2$ into (1):
$$a + 2(-2) = 4 \implies a – 4 = 4 \implies a = 8$$
Now find $n$ for $a_n = 0$:
$$0 = 8 + (n-1)(-2)$$
$$-8 = -2(n-1)$$
$$4 = n – 1 \implies n = 5$$
The $\mathbf{5^{th} \text{ term is zero}}$.
10. Find the common difference
Given $a_{17} = a_{10} + 7$.
$$a_{17} = a + 16d$$
$$a_{10} = a + 9d$$
$$a + 16d = (a + 9d) + 7$$
$$16d – 9d = 7$$
$$7d = 7 \implies d = 1$$
The common difference is $\mathbf{1}$.
11. Which term will be 132 more than its 54th term?
AP: $3, 15, 27, 39, \dots$. Here $a=3, d=15-3=12$.
We need $n$ such that $a_n = a_{54} + 132$.
$$a_n = a_{54} + 132$$
$$a + (n-1)d = (a + 53d) + 132$$
$$(n-1)d = 53d + 132$$
Substitute $d=12$:
$$(n-1)12 = 53(12) + 132$$
Divide by 12:
$$n – 1 = 53 + \frac{132}{12}$$
$$n – 1 = 53 + 11$$
$$n – 1 = 64 \implies n = 65$$
The $\mathbf{65^{th} \text{ term}}$ will be 132 more than its 54th term.
12. Difference between 1000th terms of two APs
Let the first AP be $A_1$ with first term $a$ and common difference $d$.
Let the second AP be $A_2$ with first term $a’$ and common difference $d$ (same $d$).
Given $A_1(100) – A_2(100) = 100$.
$$(a + 99d) – (a’ + 99d) = 100$$
$$a – a’ = 100$$
We need $A_1(1000) – A_2(1000)$:
$$A_1(1000) – A_2(1000) = (a + 999d) – (a’ + 999d)$$
$$= a – a’$$
$$= 100$$
The difference between their 1000th terms is also $\mathbf{100}$.
13. How many three-digit numbers are divisible by 7?
The smallest three-digit number is 100. The largest is 999.
- First term ($a$): Smallest three-digit multiple of 7. $100/7 \approx 14.28$. $7 \times 15 = 105$. So, $a=105$.
- Last term ($a_n$): Largest three-digit multiple of 7. $999/7 \approx 142.71$. $7 \times 142 = 994$. So, $a_n=994$.The AP is $105, 112, \dots, 994$, with $d=7$.$$994 = 105 + (n-1)7$$$$889 = 7(n-1)$$$$n – 1 = \frac{889}{7} = 127$$$$n = 128$$There are $\mathbf{128}$ three-digit numbers divisible by 7.
14. How many multiples of 4 lie between 10 and 250?
- First term ($a$): Multiple of 4 after 10. $4 \times 3 = 12$. So, $a=12$.
- Last term ($a_n$): Multiple of 4 before 250. $250/4 = 62.5$. $4 \times 62 = 248$. So, $a_n=248$.The AP is $12, 16, \dots, 248$, with $d=4$.$$248 = 12 + (n-1)4$$$$236 = 4(n-1)$$$$n – 1 = \frac{236}{4} = 59$$$$n = 60$$There are $\mathbf{60}$ multiples of 4 between 10 and 250.
15. Value of $n$ for equal $n^{th}$ terms
AP 1: $63, 65, 67, \dots$. $a=63, d=2$.
$$a_n = 63 + (n-1)2$$
AP 2: $3, 10, 17, \dots$. $a’=3, d’=7$.
$$a’_n = 3 + (n-1)7$$
Set $a_n = a’_n$:
$$63 + (n-1)2 = 3 + (n-1)7$$
$$63 – 3 = 7(n-1) – 2(n-1)$$
$$60 = (n-1)(7 – 2)$$
$$60 = 5(n-1)$$
$$12 = n – 1 \implies n = 13$$
The $n^{th}$ terms are equal for $\mathbf{n = 13}$.
16. Determine the AP
Given $a_3 = 16$.
Given $a_7 = a_5 + 12$.
$$a_7 – a_5 = 12$$
$$(a + 6d) – (a + 4d) = 12$$
$$2d = 12 \implies d = 6$$
Using $a_3 = 16$:
$$a + 2d = 16$$
$$a + 2(6) = 16 \implies a + 12 = 16 \implies a = 4$$
The first three terms are $a, a+d, a+2d$.
The AP is $\mathbf{4, 10, 16, \dots}$.
17. Find the 20th term from the last
AP: $3, 8, 13, \dots, 253$. Here $a=3, d=5, l=253$.
The formula for the $k^{th}$ term from the last is $a’_k = l – (k-1)d$. We need $k=20$.
$$a’_{20} = 253 – (20-1)5$$
$$a’_{20} = 253 – 19(5)$$
$$a’_{20} = 253 – 95 = 158$$
The $\mathbf{20^{th} \text{ term from the last is } 158}$.
18. Find the first three terms
Given $a_4 + a_8 = 24$ and $a_6 + a_{10} = 44$.
- First equation:$$(a + 3d) + (a + 7d) = 24$$$$2a + 10d = 24 \implies a + 5d = 12 \quad \dots (1)$$
- Second equation:$$(a + 5d) + (a + 9d) = 44$$$$2a + 14d = 44 \implies a + 7d = 22 \quad \dots (2)$$Subtract (1) from (2):$$(a + 7d) – (a + 5d) = 22 – 12$$$$2d = 10 \implies d = 5$$Substitute $d=5$ into (1):$$a + 5(5) = 12 \implies a + 25 = 12 \implies a = -13$$The first three terms are $a, a+d, a+2d$:$$-13, -13 + 5, -13 + 10$$The first three terms are $\mathbf{-13, -8, -3}$.
19. Subba Rao’s Income
AP: $5000, 5200, 5400, \dots$. Here $a=5000, d=200$. We need the year when $a_n = 7000$.
$n$ is the number of years worked.
$$7000 = 5000 + (n-1)200$$
$$2000 = (n-1)200$$
$$10 = n – 1 \implies n = 11$$
Subba Rao reached ₹ 7000 in the 11th year of service.
Year: $1995 + (11 – 1) = 1995 + 10 = 2005$.
His income reached ₹ 7000 in the year $\mathbf{2005}$.
20. Ramkali’s Savings
AP: $5, 5 + 1.75, \dots$. Here $a=5, d=1.75$. We need $n$ when $a_n = 20.75$.
$$20.75 = 5 + (n-1)1.75$$
$$15.75 = 1.75(n-1)$$
$$n – 1 = \frac{15.75}{1.75} = 9$$
$$n = 10$$
Her weekly savings become ₹ 20.75 in the $\mathbf{10^{th} \text{ week}}$.