How to find the nth term of an AP using formula a_n = a + (n - 1)d?

The general term formula is a_n = a + (n - 1)d, where a is the first term, d is the common difference, and n is the position of the term.

Which term of the AP: 5, 10, 15, 20... is 105?

Using a=5, d=5, and a_n=105: 105 = 5 + (n - 1)5 gives n = 21. It is the 21st term.

Class 10 Maths Chapter 5 Exercise 5.2 NCERT Solutions

Questions 1, 2 & 3 Formula Basics & Missing Terms

1. Fill in the blanks in the table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n^{\text{th}}$ term of the AP:

Part	$a$	$d$	$n$	$a_n$
(i)	$3$	$2$	$10$	$\dots$
(ii)	$-10$	$\dots$	$10$	$8$
(iii)	$\dots$	$-3$	$15$	$-5$
(iv)	$-10$	$2.5$	$\dots$	$0$
(v)	$5$	$0$	$100$	$\dots$

2. Choose the correct choice:
(i) $20^{\text{th}}$ term of the AP: $2, 5, 8, \dots$
(ii) $10^{\text{th}}$ term of the AP: $-2, 0, 2, \dots$

3. Find the missing terms in the boxes:
(i) $4, \ \square, \ 12$
(ii) $\square, \ 10, \ \square, \ 0$
(iii) $2, \ \square, \ \square, \ 11$
(iv) $-3, \ \square, \ \square, \ \square, \ 5$
(v) $\square, \ 20, \ \square, \ \square, \ \square, \ -20$

Solutions to Question 1 (Formula: $a_n = a + (n - 1)d$)

(i) $a = 3, d = 2, n = 10 \implies a_{10} = 3 + (10 - 1)(2) = 3 + 18 = 21$.
(ii) $a = -10, a_{10} = 8, n = 10 \implies 8 = -10 + 9d \implies 9d = 18 \implies d = 2$.
(iii) $d = -3, n = 15, a_{15} = -5 \implies -5 = a + 14(-3) \implies -5 = a - 42 \implies a = 37$.
(iv) $a = -10, d = 2.5, a_n = 0 \implies 0 = -10 + (n - 1)(2.5) \implies 2.5(n - 1) = 10 \implies n - 1 = 4 \implies n = 5$.
(v) $a = 5, d = 0, n = 100 \implies a_{100} = 5 + (99)(0) = 5$.

Solutions to Question 2

(i) $a = 2, d = 5 - 2 = 3$. $a_{20} = 2 + (19)(3) = 2 + 57 = 59$.
(ii) $a = -2, d = 0 - (-2) = 2$. $a_{10} = -2 + (9)(2) = -2 + 18 = 16$.

Solutions to Question 3

(i) Arithmetic mean between $4$ and $12$: $\square = \dfrac{4 + 12}{2} = 8$.
(ii) $a_2 = a + d = 10$ and $a_4 = a + 3d = 0$. Subtracting gives $2d = -10 \implies d = -5$. Thus $a = 15$. Boxes are $\mathbf{15}$ and $\mathbf{5}$.
(iii) $a = 2, a_4 = a + 3d = 11 \implies 2 + 3d = 11 \implies 3d = 9 \implies d = 3$. Missing terms are $2 + 3 = \mathbf{5}$ and $5 + 3 = \mathbf{8}$.
(iv) $a = -3, a_5 = a + 4d = 5 \implies -3 + 4d = 5 \implies 4d = 8 \implies d = 2$. Missing terms are $\mathbf{-1}, \mathbf{1}, \mathbf{3}$.
(v) $a_2 = a + d = 20, a_6 = a + 5d = -20$. Subtracting: $4d = -40 \implies d = -10$. First term $a = 30$. Missing terms: $\mathbf{30}, \mathbf{10}, \mathbf{0}, \mathbf{-10}$.

Questions 4, 5, 6 & 7 Term Determination & Counts

4. Which term of the AP: $5, 10, 15, 20, \dots$ is $105$?

5. Find the number of terms in each of the following APs:
(i) $5, 10, \dots, 100$
(ii) $20, 19.5, \dots, -10$

6. Check whether $50$ is a term of the AP: $2, 6, 10, \dots$

7. Find the $31^{\text{st}}$ term of an AP whose $10^{\text{th}}$ term is $20$ and $20^{\text{th}}$ term is $40$.

Question 4 Solution

Here $a = 5, d = 5, a_n = 105$.
$a_n = a + (n - 1)d \implies 105 = 5 + (n - 1)5$
$100 = 5(n - 1) \implies n - 1 = 20 \implies n = 21$.
$\therefore 105$ is the $21^{\text{st}}$ term.

Question 5 Solution

(i) $a = 5, d = 5, a_n = 100$.
$100 = 5 + (n - 1)5 \implies 95 = 5(n - 1) \implies n - 1 = 19 \implies n = 20$.
There are $20$ terms.

(ii) $a = 20, d = 19.5 - 20 = -0.5, a_n = -10$.
$-10 = 20 + (n - 1)(-0.5) \implies -30 = -0.5(n - 1) \implies n - 1 = 60 \implies n = 61$.
There are $61$ terms.

Question 6 Solution

Here $a = 2, d = 4$. Let $a_n = 50$.
$50 = 2 + (n - 1)4 \implies 48 = 4(n - 1) \implies n - 1 = 12 \implies n = 13$.
Since $n = 13$ is a positive integer:
$\therefore$ Yes, $50$ is the $13^{\text{th}}$ term of the given AP.

Question 7 Solution

$a_{10} = a + 9d = 20$ (1)
$a_{20} = a + 19d = 40$ (2)
Subtracting (1) from (2): $10d = 20 \implies d = 2$.
Substituting into (1): $a + 9(2) = 20 \implies a = 20 - 18 = 2$.
Now, $a_{31} = a + 30d = 2 + 30(2) = 2 + 60 = 62$.
$\therefore$ The $31^{\text{st}}$ term is $62$.

Questions 8, 9 & 10 Linear Systems in $a$ and $d$

8. An AP consists of $50$ terms. The $3^{\text{rd}}$ term is $5$ and the last term is $99$. Find the $29^{\text{th}}$ term.

9. If the $3^{\text{rd}}$ term of an AP is $4$ and the $9^{\text{th}}$ term is $-8$, which term of this AP is zero?

10. The $15^{\text{th}}$ term of an AP exceeds its $10^{\text{th}}$ term by $25$. Find the common difference.

Question 8 Solution

Total terms $n = 50 \implies a_{50} = a + 49d = 99$.
$a_3 = a + 2d = 5$.
Subtracting: $47d = 94 \implies d = 2$.
$a + 2(2) = 5 \implies a = 1$.
$a_{29} = a + 28d = 1 + 28(2) = 1 + 56 = 57$.
$\therefore$ The $29^{\text{th}}$ term is $57$.

Question 9 Solution

$a_3 = a + 2d = 4$ (1)
$a_9 = a + 8d = -8$ (2)
Subtracting (1) from (2): $6d = -12 \implies d = -2$.
Substituting into (1): $a + 2(-2) = 4 \implies a = 8$.
Setting $a_n = 0$:
$a + (n - 1)d = 0 \implies 8 + (n - 1)(-2) = 0 \implies 2(n - 1) = 8 \implies n - 1 = 4 \implies n = 5$.
$\therefore$ The $5^{\text{th}}$ term of the AP is zero.

Question 10 Solution

Given $a_{15} - a_{10} = 25$.
$(a + 14d) - (a + 9d) = 25$
$5d = 25 \implies d = 5$.
$\therefore$ The common difference $d = 5$.

Questions 11 to 15 Term Relationships & Multiples

11. Which term of the AP: $3, 15, 27, 39, \dots$ will be $120$ more than its $20^{\text{th}}$ term?

12. Two APs have the same common difference. The difference between their $100^{\text{th}}$ terms is $100$. What is the difference between their $1000^{\text{th}}$ terms?

13. How many three-digit numbers are divisible by $6$?

14. How many multiples of $4$ lie between $12$ and $248$?

15. For what value of $n$, are the $n^{\text{th}}$ terms of two APs: $10, 12, 14, \dots$ and $5, 8, 11, \dots$ equal?

Question 11 Solution

Here $a = 3, d = 12$.
Condition: $a_n = a_{20} + 120$
$a + (n - 1)d = a + 19d + 120 \implies (n - 1)d = 19d + 120$
$(n - 1)(12) = 19(12) + 120 = 228 + 120 = 348$
$n - 1 = \dfrac{348}{12} = 29 \implies n = 30$.
$\therefore$ The $30^{\text{th}}$ term will be $120$ more than the $20^{\text{th}}$ term.

Question 12 Solution

Let the two APs have first terms $a$ and $b$ and same common difference $d$.
$a_n = a + (n - 1)d$ and $b_n = b + (n - 1)d$.
Difference between $n^{\text{th}}$ terms: $a_n - b_n = a - b$ (independent of $n$).
Given $a_{100} - b_{100} = 100 \implies a - b = 100$.
Therefore, $a_{1000} - b_{1000} = a - b = 100$.
$\therefore$ The difference between their $1000^{\text{th}}$ terms is $100$.

Question 13 Solution

Three-digit numbers range from $100$ to $999$.
Smallest three-digit number divisible by $6$ is $102$.
Largest three-digit number divisible by $6$ is $996$ (since $999 = 6 \times 166 + 3$).
The sequence is $102, 108, 114, \dots, 996$, with $a = 102, d = 6, a_n = 996$.
$996 = 102 + (n - 1)6 \implies 894 = 6(n - 1) \implies n - 1 = 149 \implies n = 150$.
$\therefore$ There are $150$ three-digit numbers divisible by $6$.

Question 14 Solution

Multiples of $4$ between $12$ and $248$ (inclusive): $12, 16, 20, \dots, 248$.
Here $a = 12, d = 4, a_n = 248$.
$248 = 12 + (n - 1)4 \implies 236 = 4(n - 1) \implies n - 1 = 59 \implies n = 60$.
$\therefore$ There are $60$ multiples of $4$.

Question 15 Solution

First AP: $10, 12, 14, \dots \implies a_1 = 10, d_1 = 2 \implies a_n = 10 + 2(n - 1) = 2n + 8$.
Second AP: $5, 8, 11, \dots \implies a_2 = 5, d_2 = 3 \implies b_n = 5 + 3(n - 1) = 3n + 2$.
Equating $a_n = b_n$:
$2n + 8 = 3n + 2 \implies 3n - 2n = 8 - 2 \implies n = 6$.
$\therefore$ Their $n^{\text{th}}$ terms are equal when $n = 6$.

Questions 16 to 20 AP Construction & Practical Word Problems

16. Determine the AP whose $3^{\text{rd}}$ term is $5$ and the $7^{\text{th}}$ term exceeds the $5^{\text{th}}$ term by $10$.

17. Find the $10^{\text{th}}$ term from the last term of the AP: $2, 4, 6, \dots, 100$.

18. The sum of the $3^{\text{rd}}$ and $7^{\text{th}}$ terms of an AP is $20$ and the sum of the $5^{\text{th}}$ and $9^{\text{th}}$ terms is $32$. Find the AP.

19. A person started work at an annual salary of ₹$10,000$ with an increment of ₹$500$ each year. In which year did his income reach ₹$20,000$?

20. Priya saved ₹$10$ in the first week and increased her savings by ₹$5$ weekly. If her savings become ₹$110$ in the $n^{\text{th}}$ week, find $n$.

Question 16 Solution

$a_7 - a_5 = 10 \implies (a + 6d) - (a + 4d) = 10 \implies 2d = 10 \implies d = 5$.
Given $a_3 = a + 2d = 5 \implies a + 2(5) = 5 \implies a + 10 = 5 \implies a = -5$.
$\therefore$ The required AP is $-5, 0, 5, 10, \dots$

Question 17 Solution

Given AP: $2, 4, 6, \dots, 100$ with $d = 2$.
Reversing the AP from the end: $100, 98, 96, \dots, 2$.
Here first term $a' = 100$, common difference $d' = -2$.
$10^{\text{th}}$ term from end $= a' + 9d' = 100 + 9(-2) = 100 - 18 = 82$.
$\therefore$ The $10^{\text{th}}$ term from the last term is $82$.

Question 18 Solution

$a_3 + a_7 = (a + 2d) + (a + 6d) = 2a + 8d = 20 \implies a + 4d = 10$ (1)
$a_5 + a_9 = (a + 4d) + (a + 8d) = 2a + 12d = 32 \implies a + 6d = 16$ (2)
Subtracting (1) from (2): $2d = 6 \implies d = 3$.
Substituting into (1): $a + 4(3) = 10 \implies a = 10 - 12 = -2$.
$\therefore$ The required AP is $-2, 1, 4, 7, \dots$

Question 19 Solution

Initial salary $a = ₹10,000$, annual increment $d = ₹500$, target salary $a_n = ₹20,000$.
$a_n = a + (n - 1)d$
$20000 = 10000 + (n - 1)500$
$10000 = 500(n - 1) \implies n - 1 = 20 \implies n = 21$.
$\therefore$ In the $21^{\text{st}}$ year, his income reached ₹$20,000$.

Question 20 Solution

Initial weekly savings $a = ₹10$, weekly increment $d = ₹5$, target savings $a_n = ₹110$.
$110 = 10 + (n - 1)5$
$100 = 5(n - 1) \implies n - 1 = 20 \implies n = 21$.
$\therefore n = 21$ (in the $21^{\text{st}}$ week).

Arithmetic Progressions - Exercise 5.2

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