NCERT Solutions Class 10 Maths Chapter 5: Arithmetic Progressions
Detailed solutions for Exercise 5.2 ($n^{th}$ term of an AP).
#### **Q1. 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:**
| | $a$ | $d$ | $n$ | $a_n$ |
|---|---|---|---|---|
| (i) | 7 | 3 | 8 | ... |
| (ii) | -18 | ... | 10 | 0 |
| (iii) | ... | -3 | 18 | -5 |
| (iv) | -18.9 | 2.5 | ... | 3.6 |
| (v) | 3.5 | 0 | 105 | ... |
#### **A1. Solution:**
**(i)** Given $a=7, d=3, n=8$.
$a_8 = 7 + (8-1)3 = 7 + 21 = 28$.
**Ans: 28**
**(ii)** Given $a=-18, n=10, a_n=0$.
$0 = -18 + (10-1)d \Rightarrow 18 = 9d \Rightarrow d = 2$.
**Ans: 2**
**(iii)** Given $d=-3, n=18, a_n=-5$.
$-5 = a + (18-1)(-3) \Rightarrow -5 = a - 51 \Rightarrow a = 46$.
**Ans: 46**
**(iv)** Given $a=-18.9, d=2.5, a_n=3.6$.
$3.6 = -18.9 + (n-1)2.5$
$22.5 = (n-1)2.5 \Rightarrow n-1 = 9 \Rightarrow n = 10$.
**Ans: 10**
**(v)** Given $a=3.5, d=0, n=105$.
$a_n = 3.5 + (105-1)0 = 3.5$.
**Ans: 3.5**
#### **Q2. Choose the correct choice in the following and justify:**
**(i) 30th term of the AP: $10, 7, 4, \dots$ is**
(A) 97 (B) 77 (C) -77 (D) -87
**(ii) 11th term of the AP: $-3, -\frac{1}{2}, 2, \dots$ is**
(A) 28 (B) 22 (C) -38 (D) $-48\frac{1}{2}$
#### **A2. Solution:**
**(i)** $a=10, d=7-10=-3, n=30$.
$a_{30} = 10 + (30-1)(-3) = 10 - 87 = -77$.
**Option (C)**
**(ii)** $a=-3, d=-\frac{1}{2} - (-3) = 2.5 (\text{or } \frac{5}{2}), n=11$.
$a_{11} = -3 + (11-1)\frac{5}{2} = -3 + 10(\frac{5}{2}) = -3 + 25 = 22$.
**Option (B)**
#### **Q3. In the following APs, find the missing terms in the boxes:**
**(i) $2, \square, 26$**
**(ii) $\square, 13, \square, 3$**
**(iii) $5, \square, \square, 9\frac{1}{2}$**
**(iv) $-4, \square, \square, \square, \square, 6$**
**(v) $\square, 38, \square, \square, \square, -22$**
#### **A3. Solution:**
**(i)** Middle term is the arithmetic mean: $\frac{2+26}{2} = 14$. **Ans: 14**
**(ii)** Let terms be $a, a+d, a+2d, a+3d$.
$a+d = 13$ ...(1)
$a+3d = 3$ ...(2)
Subtract (1) from (2): $2d = -10 \Rightarrow d = -5$.
Substitute $d$ in (1): $a - 5 = 13 \Rightarrow a = 18$.
Third term: $18 + 2(-5) = 8$.
**Ans: 18, 8**
**(iii)** $a=5$, $a_4 = 9.5$.
$5 + 3d = 9.5 \Rightarrow 3d = 4.5 \Rightarrow d = 1.5$.
$a_2 = 5 + 1.5 = 6.5$, $a_3 = 6.5 + 1.5 = 8$.
**Ans: $6\frac{1}{2}, 8$**
**(iv)** $a=-4$, $a_6=6$.
$-4 + 5d = 6 \Rightarrow 5d = 10 \Rightarrow d = 2$.
Terms: $-4+2=-2$, $-2+2=0$, $0+2=2$, $2+2=4$.
**Ans: -2, 0, 2, 4**
**(v)** $a+d=38$ ...(1), $a+5d=-22$ ...(2).
Subtract (1) from (2): $4d = -60 \Rightarrow d = -15$.
From (1): $a - 15 = 38 \Rightarrow a = 53$.
Terms: $53$, $38$, $38-15=23$, $23-15=8$, $8-15=-7$.
**Ans: 53, 23, 8, -7**
#### **Q4. Which term of the AP: $3, 8, 13, 18, \dots$ is $78$?**
#### **A4. Solution:**
$a=3, d=5, a_n=78$.
$78 = 3 + (n-1)5$
$75 = (n-1)5$
$15 = n-1 \Rightarrow n = 16$.
**Ans: 16th term**
#### **Q5. Find the number of terms in each of the following APs:**
**(i) $7, 13, 19, \dots, 205$**
**(ii) $18, 15\frac{1}{2}, 13, \dots, -47$**
#### **A5. Solution:**
**(i)** $a=7, d=6, a_n=205$.
$205 = 7 + (n-1)6 \Rightarrow 198 = (n-1)6 \Rightarrow 33 = n-1 \Rightarrow n = 34$.
**Ans: 34 terms**
**(ii)** $a=18, d=15.5 - 18 = -2.5, a_n=-47$.
$-47 = 18 + (n-1)(-2.5) \Rightarrow -65 = (n-1)(-2.5)$
$\frac{-65}{-2.5} = n-1 \Rightarrow 26 = n-1 \Rightarrow n = 27$.
**Ans: 27 terms**
#### **Q6. Check whether -150 is a term of the AP: $11, 8, 5, 2, \dots$**
#### **A6. Solution:**
$a=11, d=-3$. Let $a_n = -150$.
$-150 = 11 + (n-1)(-3)$
$-161 = (n-1)(-3)$
$n-1 = \frac{161}{3} = 53.66...$
Since $n$ must be a positive integer, **-150 is NOT a term.**
#### **Q7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.**
#### **A7. Solution:**
$a_{11} = a + 10d = 38$ ...(1)
$a_{16} = a + 15d = 73$ ...(2)
Subtract (1) from (2): $5d = 35 \Rightarrow d = 7$.
Substitute $d=7$ in (1): $a + 70 = 38 \Rightarrow a = -32$.
$a_{31} = a + 30d = -32 + 30(7) = -32 + 210 = 178$.
**Ans: 178**
#### **Q8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.**
#### **A8. Solution:**
$n=50$, $a_3 = 12$, $a_{50} = 106$.
$a + 2d = 12$ ...(1)
$a + 49d = 106$ ...(2)
Subtract (1) from (2): $47d = 94 \Rightarrow d = 2$.
From (1): $a + 4 = 12 \Rightarrow a = 8$.
$a_{29} = 8 + 28(2) = 8 + 56 = 64$.
**Ans: 64**
#### **Q9. If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?**
#### **A9. Solution:**
$a_3 = a + 2d = 4$
$a_9 = a + 8d = -8$
Subtracting: $6d = -12 \Rightarrow d = -2$.
$a + 2(-2) = 4 \Rightarrow a - 4 = 4 \Rightarrow a = 8$.
Let $a_n = 0$.
$0 = 8 + (n-1)(-2) \Rightarrow -8 = (n-1)(-2) \Rightarrow 4 = n-1 \Rightarrow n = 5$.
**Ans: 5th term**
#### **Q10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.**
#### **A10. Solution:**
$a_{17} - a_{10} = 7$
$(a + 16d) - (a + 9d) = 7$
$7d = 7 \Rightarrow d = 1$.
**Ans: 1**
#### **Q11. Which term of the AP: $3, 15, 27, 39, \dots$ will be 132 more than its 54th term?**
#### **A11. Solution:**
$a=3, d=12$.
$a_{54} = 3 + 53(12) = 3 + 636 = 639$.
Required value = $639 + 132 = 771$.
$771 = 3 + (n-1)12 \Rightarrow 768 = (n-1)12 \Rightarrow 64 = n-1 \Rightarrow n = 65$.
**Ans: 65th term**
#### **Q12. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?**
#### **A12. Solution:**
Let the APs be $a_n$ and $b_n$ with first terms $a$ and $b$, and same $d$.
$a_{100} - b_{100} = (a + 99d) - (b + 99d) = a - b = 100$.
Difference between 1000th terms:
$a_{1000} - b_{1000} = (a + 999d) - (b + 999d) = a - b$.
Since $a - b = 100$, the answer is 100.
**Ans: 100**
#### **Q13. How many three-digit numbers are divisible by 7?**
#### **A13. Solution:**
First 3-digit number divisible by 7 is 105 ($15 \times 7$).
Last 3-digit number divisible by 7 is 994 ($142 \times 7$).
AP: $105, 112, \dots, 994$.
$a=105, d=7, a_n=994$.
$994 = 105 + (n-1)7 \Rightarrow 889 = (n-1)7 \Rightarrow 127 = n-1 \Rightarrow n = 128$.
**Ans: 128**
#### **Q14. How many multiples of 4 lie between 10 and 250?**
#### **A14. Solution:**
First multiple > 10 is 12. Last multiple < 250 is 248.
AP: $12, 16, \dots, 248$.
$248 = 12 + (n-1)4 \Rightarrow 236 = (n-1)4 \Rightarrow 59 = n-1 \Rightarrow n = 60$.
**Ans: 60**
#### **Q15. For what value of $n$, are the $n^{th}$ terms of two APs: $63, 65, 67, \dots$ and $3, 10, 17, \dots$ equal?**
#### **A15. Solution:**
AP1: $a=63, d=2 \Rightarrow A_n = 63 + (n-1)2$.
AP2: $a=3, d=7 \Rightarrow B_n = 3 + (n-1)7$.
Set $A_n = B_n$:
$63 + 2n - 2 = 3 + 7n - 7$
$61 + 2n = 7n - 4$
$65 = 5n \Rightarrow n = 13$.
**Ans: 13th term**
#### **Q16. Determine the AP whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.**
#### **A16. Solution:**
$a_3 = 16 \Rightarrow a + 2d = 16$.
$a_7 - a_5 = 12 \Rightarrow (a+6d) - (a+4d) = 12 \Rightarrow 2d = 12 \Rightarrow d = 6$.
Substitute $d=6$ in $a+2d=16$: $a + 12 = 16 \Rightarrow a = 4$.
AP: $4, 10, 16, 22, \dots$
**Ans: 4, 10, 16, 22, ...**
#### **Q17. Find the 20th term from the last term of the AP: $3, 8, 13, \dots, 253$.**
#### **A17. Solution:**
Method: Reverse the AP. New $a = 253$, new $d = -5$.
We need the 20th term of this reversed AP.
$a_{20} = 253 + (20-1)(-5)$
$a_{20} = 253 + 19(-5) = 253 - 95 = 158$.
**Ans: 158**
#### **Q18. 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.**
#### **A18. Solution:**
Case 1: $a_4 + a_8 = 24$
$(a+3d) + (a+7d) = 24 \Rightarrow 2a + 10d = 24 \Rightarrow a + 5d = 12$ ...(1)
Case 2: $a_6 + a_{10} = 44$
$(a+5d) + (a+9d) = 44 \Rightarrow 2a + 14d = 44 \Rightarrow a + 7d = 22$ ...(2)
Subtract (1) from (2): $2d = 10 \Rightarrow d = 5$.
Substitute in (1): $a + 25 = 12 \Rightarrow a = -13$.
Terms: $-13, -8, -3$.
**Ans: -13, -8, -3**
#### **Q19. 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?**
#### **A19. Solution:**
$a = 5000, d = 200, a_n = 7000$.
$7000 = 5000 + (n-1)200$
$2000 = (n-1)200 \Rightarrow 10 = n-1 \Rightarrow n = 11$.
He reaches the salary in the 11th year.
Year: $1995 + (11-1) = 2005$.
**Ans: 11th year (2005)**
#### **Q20. Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the $n^{th}$ week, her weekly savings become ₹ 20.75, find $n$.**
#### **A20. Solution:**
$a = 5, d = 1.75, a_n = 20.75$.
$20.75 = 5 + (n-1)1.75$
$15.75 = (n-1)1.75$
$n-1 = \frac{15.75}{1.75} = \frac{1575}{175} = 9$.
$n = 10$.
**Ans: n = 10**