1. Find:

(i) 10th term of the A.P. 1, 4, 7, 10, .......
(ii) 18th term of A.P. √2, 3√2, 5√2, .......
(iii) nth term of the A.P. 13, 8, 3, -2, ......
(iv) 10th term of the A.P. -40, -15, 10, 35, .....
(v) 8th term of the A.P. 11, 104, 91, 78, .......
(vi) 11th term of the A.P. 10.0, 10.5, 11.0, 11.5, .....
(vii) 9th term of the A.P. 3/4, 5/4, 7/4, 9/4 , .......

Solution

(i) Given A.P. is 
1, 4, 7, 10, ......
First term (a) = 1 
Common difference (d) = second term - first term 
= 4 - 1 = 3 
nth term in an A.P = a (n -1)d
10th term in an A.P = 1 + (10 - 1)3
= 1 + 9×3
= 1 + 27 = 28

(ii) Given, A.P is √2, 3√2, 5√2, .......
First term (a) = √2
common difference(d) = second term - first term 
3√2 - √2 = 2√2
n th term in an A.P = a + (n -1)d
18 th term of A.P = √2 + (18 - 1)2√2
= √2 + 17 × 2√2
= √2(1 + 34)
= 35√2

(iii) Given, A.P is 13, 8, 3, -2, .......
First term (a) = 13 
Common difference (d) = Second term - first term 
8 - 13 = - 5
nth term of an A.P = a + (n -1)d
= 13 + (n -1) - 5
= 13 - 5n + 5
= 18 - 5n

(iv) Given A.P is 
-40, -15, 10, 35, .......
First term (a) = -40
Common difference (d) = Second term - first term 
= -15 - (-40)
= 40 - 15 = 25
nth term of an A.P = a + (n -1)d
10th term of A.P = -40 + (10 - 1)25
= -40 + 9×25
= -40 + 225 = 185

(v) Given sequence is 
117, 104, 91, 78, .........
First learn can = 117
Common difference (d) = Second term - first term 
= 104 - 117 = - 13 
nth term = a + (n -1)d
8th term = a + (8 - 1)d
= 117 + 7(-13)
= 117 - 91 = 26

(vi) Given A.P is 10.0, 10.5, 11.0, 11.5, ..... 
First term (a) = 10.0
Common difference (d) = Second term - First term 
= 10.5 - 10.0 = 0.5
nth term = a + (n - 1)d 
11th term = 10.0 + (11 - 1) 0.5
= 10.0 + 10 ×0.5
= 10.0 + 5= 15.0

(vii) Given A.P. is 3/4, 5/4, 7/4, 9/4 , ........
First term (a) = 3/4 
Common difference (d) = Second term - First Term 
= 5/4 - 3/4 = 2/4 
nth term = a + (n - 1)d 
9th term = a + (9 -1)d 
= 3/4 + 8× 2/4
= 3/4 + 16/4 
= 19/4

2. (i) Which term of the AP 3, 8, 13, ..... is 248?
(ii) Which term of the AP 84, 80, 76, ...... is 0 ?
(iii) Which term of the AP 4, 9, 14, ...... is 254 ? 
(iv) Which term of the AP 21, 42, 63, 84, ...... is 420? 
(v) Which term of the AP 121, 117, 113, ...... is its first negative term ? 

Solution

(i) Given A.P is 3, 8, 13, ..... 
First term (a) = 3 
common difference (d) = Second term - first term 
= 8 - 3 = 5 
n th term (a_n) = a + (n - 1)d 
Given nth term a_n = 248 
248 = 3 + (n - 1)5
⇒ 248 = -2 + 5n
⇒ 5n = 250 
⇒ n = 250/5 = 50 
50th term is 248

(ii) Given A.P is 84, 80, 76, .......
First term (a) = 84 
Common difference (d) = a₂ - a 
= 80 - 84 = -4
nth term (a_n) = a + (n - 1)d 
Given nth term is 0 
0 = 84 + (n - 1)× -4
⇒ 0 = 84 -4(n - 1)
⇒ 4(n-1) = 84 
⇒ n - 1= 84/4 = 21 
⇒ n = 21 + 1= 22
⇒ 22 nd term is 0.

(iii) Given A.P. 4, 9, 14, .......
First term (a) = 4
Common difference (d) = a₂ - a 
= 9 - 4 = 5
nth term (a_n) = a + (n -1)d 
Given nth term is 254
4 + (n - 1)5 = 254 
⇒ (n - 1)5 = 250 
⇒ n - 1= 250/5 = 50 
= 50 + 1 = 51 
∴ 51 th term is 254.

(iv) Given A.P 21, 42, 63, 84, ......
a = 21, d = a₂ - a 
= 42 - 21 = 21
nth term (a_n) = a + (n - 1)d
Given nth term = 420
21 + (n - 1)21 = 420 
⇒ (n - 1)21 = 399
⇒ n - 1 = 399/21 = 19 
⇒ n = 19 + 1 = 20 
∴ 20th term is 420.

(v) Given A.P is 121, 117, 113,........
First term (a) = 121 
Common difference (d) = 117 - 121 = -4 
nth term (a) = a + (n - 1)d 
Given nth term is negative i.e., a_n < 0 
121 + (n - 1)-4 < 0 
⇒ 121+ 4 - 4n < 0 
⇒ 125 < 4n
⇒ n > 125/4 
⇒ n > 31.25
The integer which comes after 31.25 is 32, 
∴ 32 nd term is first negative term.

3. (i) Is 68 a term of the AP 7, 10, 13, ...... ?
(ii) Is 302 a term of the AP 3, 8, 13,...... ? 
(iii) Is -150 a term of the AP 11, 8, 5, 2, ...... ? 

Solution 

In the given problem, we are given an A.P and the value of one of its term 
We need to find whether it is a term of the AP or not so here we will use the formula
a_n = a + (n - 1)d

(i) Here, AP is 7, 10, 13, ......
a_n = 68, a = 7, and d = 10 - 7 = 3 
Using the above mentioned formula, we get 
68 = 7 + (n - 1)3 
⇒ 68 - 7 = 3n - 3 
⇒ 31 + 3 = 3n 
⇒ 64 = 3n 
⇒ n = 64/3 
Since, the value of n is a fraction. 
Thus, 68 is not the team of the given A.P.

(ii) Here, AP is 3, 8, 13,....
a_n = 302, a = 3 
d = 8 - 3= 5 using the above mentioned formula, we get 
302 = 3 + (n -1)5 
⇒ 302 - 3 = 5n - 5 
⇒ 299 = 5n - 5 
⇒ 5n = 304
⇒ n = 304/5
Since, the value of 'n' is a fraction. Thus, 302 is not the term of the given A.P.

(iii) Here, A.P. is 11, 8, 5, 2, ......
a_n = -150, a = 1 and d = 8 - 11= -3
Thus, using the above mentioned formula, we get 
-150 = 11 + (n - 1)(-3)
⇒ -150 - 11 = -3n + 3 
⇒ -161 - 3 = -3n 
⇒ n = -164/-3
Since, the value of n is a fraction. Thus, -150 is not the term of the given A.P.

4. How many term are there in the AP ? 
(i) 7, 10, 13,..... 43
(ii)-1, -5/6, -2/3, -1/2,....., 10/3
(iii) 7, 13, 19,...... 05
(iv) 18, 15(1/2), 13, ...... , -47 

Solution

(i) 7, 10, 13,..... 43
From given A.P 
a = 7, d = 10 - 7 = 3, a_n = a + (n - 1)d. 
Let, a_n = 43 (last term)
7 + (n - 1)3 = 43 
⇒ (n-1) = 39/3=13
⇒ n = 13 
∴ 13 terms are there in given A.P.

(ii) -1, -5/6, -2/3, -1/2,........, 10/3
From given A.P 
a = -1, d = -5/6 + 1 = 1/6 
a_n = a + (n - 1)d
Let, a_n = 10/3 (last term)
-1 + (n - 1)1/6 = 10/3
⇒ (n - 1) × 1/6 = 13/3 
⇒ n - 1 = (13 × 6)/3 = 26
⇒ n = 27 
∴ 27 terms are there in given A.P.

(iii) 7, 13, 19, ....., 05
From the given A.P 
a = 7, d = 13 - 7 = 6, 
a_n = a + (n - 1)d
Let a_n = 205 (last term)
7 + (n - 1)6 = 205 
⇒ (n - 1)6 = 205- 7 = 198 
⇒ n -1 = 198/6 = 33
⇒ n = 33+ 1 = 34 
∴ 34 terms are there in given A.P.

(iv) 18, 15(1/2), 13, ...... , -47 
From the given A.P 
a = 18, d = 31/2 - 18 = 15.5 - 18 = -2.5
a_n = a + (n - 1)d
Let a_n = -47  (last term)
18 + (n - 1) × -2.5 = - 47
⇒ (n -1)× -2.5 = -47 - 18 
⇒ n - 1= -65/-2.5
⇒ n - 1 = (65 × 10)/25 = 26
⇒ n = 26 + 1 = 27 
∴ 27 terms are there in given A.P.

5. The first term of an AP is 5, the common difference is 3 and the last term is 80, find the number of terms. 

Solution

Given,
First term(a) = 5 
Common difference (d) = 3 
Last term (a_n ) = 80 
To calculate no of terms in given A.P 
a_n = a + (n - 1)d
Let a_n = 80, 
80 = 5 + (n - 1)3 
⇒ (n - 1)3 = 75 
⇒ n - 1 = 75/3 = 25 
⇒ n = 25 + 1 = 26 
∴ There are 26 terms.

6. The 6th and 17 terms of an A.P are 19 and 41 respectively, find the 40th term. 

Solution

Given, a₆ = 19, a₁₇ = 41 
⇒ a₆ = a + (6 - 1)d 
19 = a + 5d  ...(1)
a_n = a + (17 - 1)d 
⇒ 41 = a + 16d ...2) 
Subtract (1) from (2)
a + 16d = 41 ...(1)
a + 5d = 19 ...(2)
11d = 22 
⇒ d = 22/11 = 2 
Substitute d = 2 in (1) 
19 = a + 5(2) 
a = 9 
∴ 40th term a₄₀ = a + (40 - 1)d 
 = 9 + 39×2
= 9 + 78 = 87 
∴ a₄₀ = 87

7. If 9th term of an A.P is zero, prove that its 29th term is double the 19th term. 

Solution

Given,
9th term of an A.P a₉ = 0, 
a_n = a + (n - 1)d 
a + (a - 1)d = 0 
⇒ a + 8d = 0 
⇒ a = -8d 
We have to prove, 
24th term is double the 19th term a₂₉ = 2. a₁₉
a + (29 - 1)d = 2[a + (19 - 1)d]
a + 28d = 2[a + 18d]
Put a = -8d 
-8d + 28d = 2[-8d + 18d]
⇒ 20d = 2×10d
⇒ 20d = 20d 
Hence, proved.

8. If 10 times the 10th term of an A.P is equal to 15 times the 15th term, show that 25th term of the A.P. is zero. 

Solution

Given, 
10 times of 10th term is equal to 15 times of 15th term. 
10a₁₀ = 15.a₁₅  
10[a + (10 - 1)d] = 15[a + (15 - 1)d] [∴ a_n = a + (n - 1)d]
⇒ 10(a + 9d) = 15(a + 14.d) 
⇒ a + 9d = 15/10(a + 14d)
⇒ a - (3/2)a = 42d/2 - 9d 
⇒ -1a/2 = (42d - 18d)/2 
⇒ -a/2 = 24d/2 =12d
⇒ a = -24d 
We have to prove 25 th term of A.P is 0
a₂₅ = 0 
⇒ a + (25 - 1)d = 0 
⇒ a + 24d = 0 
Put a = -24d 
-24d + 24d = 0 
⇒ 0 = 0 
Hence proved.

9. The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term. 

Solution

Given, 
a_n = 41, a_n = 73,
a_n = a +(n - 1)d 
⇒ a₁₀ = a + (10 - 1)d 
41 = a + 9d  ...(1) 
⇒ a_n = a + (18 - 1)d 
73 = a + 17d  ...(2)
Subtract (1) from (2) 
a + 17d = 73  ...(1) 
⇒ a + 9d = 41  ...(2)
⇒ 0 + 8d = 32 
⇒ d = 32/8 = 4 
Substitute d = 4 in (1) 
a + 9×4 = 41 
⇒ a = 41 - 36
⇒ a = 5 
26th term a_n = a + (26 - 1)d 
= 5 + 25×4
= 5 + 100 
= 105
∴ 26th term a₂₆ = 105.

10. In a certain A.P the 24th term is twice the 10th term. Prove that the 72 nd term is twice the 34th term. 

Solution

Given, 
24th term is twice the 10th term
a₂₄ = 2 a₁₀
Let, first term of a square = a
Common difference = d
nth term a_n = a + (n - 1)d
a + (24 - 1)d = (a + (10 - 1)d
⇒ a + 23d = 2(a + 9d)
⇒ (23 - 18)d = a
⇒ a = 5d
We have to prove,
72nd term is twice the 34th term 
a₁₂ = 2a₃₄ 
⇒ a + (12 - 1)d = 2[a + (34 - 1)d]
⇒ a + 71d = 2a + 66d 
Substitute a = 5d 
5d + 71d = 2(5d) + 66d
⇒ 76d = 10 d + 66d 
⇒ 76d = 76d
Hence proved.

11. If (m + 1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n +1)th term. 

Solution

Given,
(m + 1) th term is twice the (m + 1)th term. 
First term = a 
Common difference = d 
n th term a_n = a + (n - 1)d 
a_m+1 = 2a_n + 1 
⇒ a + (m + 1 - 1)d = 2(a + (n + 1 - 1)d
⇒ a + md = 2(a + nd)
⇒ a = (m - 2n)d 
We have to prove 
(3m + 1)th term is twice the (m + n + 1)th term 
a_{3m + 1} = 2.a_m+n+1 
⇒ a + (3m + 1 - 1)d = (a + (m + n + 1 -1)d
⇒ a + 3m.d - 2a + 2(m + n)d 
Substitute a = (m - 2n)d
(m - 2n)d + 3m d = 2(m - 2n)d + 2(m + n)d
⇒ 4m - 2n = 4m - 4n + 2n
⇒ 4m - 2n = 4m - 2n
Hence, proved.

12. If the n term of the A.P 9, 7, 5, ...... is same as the term of the A.P. 15, 12, 9, ..... find n. 

Solution

Given, 
First sequence is 9,7,5, .....
a = 9, d = 7 - 9 = -2, a_n = a + (n + 1)d 
a_n = 9 + (n - 1)×-2 
Second sequence is 15, 12, 9, .....
a = 15, d = 12 - 15 = -3, a_n = a + (n - 1)d 
a_n = 15 + (n - 1) × -3
Given an a_n are equal 
9 -2(n - 1) = 15 - 3(n - 1)
⇒ 3(n - 1) - 2(n - 1) - 15 - 9
⇒ n - 1 = 6 
⇒ n = 7 
∴ 7th term of two sequence are equal.

13. Find the 12th term from the end of the following arithmetic progressions: 
(i) 3, 5, 7, 9, .....201 
(ii) 3, 8,13, ......, 253
(iii) 1, 4,7, 10, ...... 88

Solution

(i) 3, 5, 7, 9, ....., 201 
First term (a) = 3 
Common difference (d) = 5 - 3 = 2 
12th term from the end is can be considered as (l) last term = first term and common difference = d₁ = -d 
nth term from the end = last term + (n - 1)×-d
12 th term from end = 201 + (12 - 1)× -2 
= 201 - 22 
= 179

(ii) 3, 8, 13, .....,253
First term a = 3 
Common difference d = 8 - 3 = 5 
Last term (l) = 253 
n th term of sequence on = a + (n - 1)d 
To find nth term from the end, we put last term (l) as 'a' and common difference as -d 
nth term from the end = last term + (n - 1) -d 
12th term from the end = 253 + (12 - 1) ×-5
= 253 - 55 = 198 
∴ 12 th term from the end = 198

(iii) 1, 4, 7, 10, ......., 88 
First term a = 1 
Common difference d = 4 - 1 = 3 
Last term (l) = 88 
n th term a_n = a + (n + 1)d 
n th term from the end = last term + (n - 1)-d 
12 th term from the end = 88 + (12 - 1)× -3 
= 88 - 33 
= 55 
∴ 12th term from the end = 55

14. The 4th term of an A.P is three times the first and the 7 term exceeds twice the third term by 1. Find the first term and the common difference. 

Solution

Given, 
4 th term of an AP is three times the times the first term 
a₄ = 3.a 
n th term of a sequence a_n = a + (n - 1)d 
a + (4 - 1)d = 3a 
⇒ a + 3d = 3a 
⇒ 3d = 2a
⇒ a = 3d/2  ...(1) 
Seventh term exceeds twice the third term by 1. 
a₇ + 1 = 2a₃ 
⇒ a + (7 - 1)d + 1 = 2[a+(3 -1)d]
⇒ a + 6d + 1 = 2a + 4d 
⇒ 2a - a = 6d - 4d + 1 
⇒ a = 2d + 1  ...(2) 
By equating (1), (2) 
3d/2 = 2d + 1 
⇒ 3d/2 - 2d = 1 
⇒ -d/2 = 1 
⇒ d = -2
Put d = -2 in a = 3d/2
= (3/2)(-2)
= -3 
∴ First term a = -3, common difference d = -2.

15. Find the second term and nth term of an A.P. whose 6th term is 12 and the 8th term is 22. 

Solution

Given, 
a₆ = 12, a₈ = 22
nth term of an A.P a_n = a + (n - 1)d 
a₆ = a + (6 - 1)d = a + 5d = 12 ...(1) 
a₈ = a+ (8 -1)d = a + 7d = 22  ...(2) 
Subtracting (1) from (2) 
2d = 10
⇒ d = 5
putting value of d in equation (1)
a + 5d = 12
⇒ a + 5×5 = 12
⇒ a = 12 - 25 = -13
Second term a₂ = a + (2 - 1)d
=a + d
= -13 + 5 = -8
n th term a_n = a + (n - 1)d
= -13 + (n - 1)×5
= -13 + 5n - 5
= -18 + 5n
nth term a_n = a + (n - 1)d
= -13 + (n - 1)-5
a_n = -18 + 5n
∴ a₂ = -8, a_n = -18 + 5n

16. How many number of two digits are divisible by 3 ? 

Solution

We observe that 12 is the firs two - digit number divisible by 3 and 99 is the last two digit number divisible by 3. Thus, the sequence is 
12, 15, 18,..., 99
This sequence is in A.P with
First term (a) = 12
Common difference(d) = 15 - 12 = 3 
nth term a_n = 99
nth term of an A.P
(a_n) = a + (n -1)d 
99 = 12 +(n -1)3
⇒ 99 - 12 = (n - 1)3 
⇒ 87/3 = n = 1 
⇒ n = 30 
∴ 30 term are there in the sequence.

17. An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32 nd term. 

Solution

Given, 
No. of terms = n = 60 
First term (a) = 7 
Last term (a) = 7 
Last term (a₁₀) = 125 
a_n = a + (n - 1)d 
⇒ a₆₀ = a + (60 - 1)d 
⇒ 125 = 7 + 59d 
⇒ 118 = 59d 
⇒ d = 118/59 = 2 
52 nd term a₃₂ = a + (32 - 1)d 
= 7 + 31×2 
= 7 + 62 = 69

18. The sum of 4 and 8th terms of an A.P. is 24 and the sum of the 6th and 10 th terms is 34. Find the first term and the common difference of the A.P. 

Solution

Given, 
a₄ + a₈ = 24 
a₆ + a₁₀ = 34 
⇒ a + (4 - 1)d + a + (18 - 1)d = 24 
2a + 10d = 24 
a + 5d = 12 ....(1)
⇒ a₆ + a₁₀ = 34 
a + (6 - 1)d + a + (10 - 1)d = 34 
⇒ 2a + 14d = 34 
a + 7d = 17 ....(2) 
Subtract (1) from (2) 
a + 7d = 17 
⇒ a + 5d = 12 
⇒ 2d = 5 
⇒ d = 5/2 
Put d = 5/2 in a + 5d = 12 
a = 12 - 5d 
⇒ a = 12 - 5× 5/2
⇒ a = 12 - 25/2 = -1/2 
∴ a = -1/2, d = b/2

19. The first term of an A.P. is 5 and its 100th term is -292. Find the 50th term of this A.P. 

Solution

Given,
First term of an A.P. = 5 and 100th term = -292

20. Find a₃₀– a₂₀ for the A.P.
(i) -9, -14, -19, -24, …
(ii) a, a + d, a + 2d, a + 3d, …

Solution

21. Write the expression a_n– a_k for the A.P. a, a + d, a + 2d, ……
Hence, find the common difference of the A.P. for which
(i) 11th term is 5 and 13th term is 79.
(ii) a₁₀– a₅ = 200
(iii) 20th term is 10 more than the 18th term.

Solution

In the A.P. a, a + d, a + 2d, ….
a_n = a+(n-1)d and a_k = a+(k-l)d
a_n - a_k = [a + (n-1)d] - [a + (k-1)d]
= a + nd - d - (a + kd - d)
= a + nd - d - a - kd + d
= nd - kd - (n - k)d

22. Find n if the given value of x is the nth term of the given A.P.

(i) 25, 50, 75, 100,....; x = 1000
(ii) -1, -3, -5, -7, ....; x = -151
(iii) 5½, 11, 16½, 22, ....; x = 550
(iv) 1, 21/11, 31/11, 41/11, ....; x = 171/11

Solution

(i) The A.P. is 25, 50, 75, 100, ...; x = 1000
Here, a = 25 and d = 50 - 25 = 25 and a_n = 100
a_n = a+(n-1)d
1000= 25 + (13-1)×25
= 25 + 25n - 25 = 25n

(ii) The A.P. is -1, -3, -5, -7, ...; x = -151
Here, a = -1, d = -3 -(-1) = -3 + 1 = -2 and a_n = -15
a_n = a + (n-1)d
⇒ -151 = -1 + (n-1)(-2)
⇒ -151 = -1 - 2n + 2
⇒ -151 = 1 - 2n
⇒ -2n = -151 - 1 = -152

23. The eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.

Solution

Let a and d be the first term and common difference of an A.P., respectively.

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24. Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12.

Solution

Let a, a + d, a + 2d, a + 3d, …. be the A.P.
a_n = a + (n – 1) d
But a₃ = 16

25. The 7th term of an A.P. is 32 and its 13th term is 62. Find the A.P.

Solution

Let a, a + d, a + 2d, a + 3d, be the A.P.
Here a is the first term and d is the common difference
a_n = a + (n – 1) d
Now, a₇ = a + (7 – 1) d = a + 6d = 32 ….(i)
and a₁₃ = a + (13 – 1) d = a + 12d = 62 ….(ii)
Subtracting (i) from (ii)
6d = 30
⇒ d = 5
a + 6×5 = 32
⇒ a + 30 = 32
⇒ a = 32 – 30 = 2
A.P. will be 2, 7, 12, 17, ……

26. Which term of the A.P. 3, 10, 17, … will be 84 more than its 13th term ?

Solution

The given A.P. is 3, 10, 17,... whose first term (a) = 3 and,
common difference (d) = 10 - 3 = 7
a_n = a + (n-1)d
Let kth term is greater than 13th term by 84.
kth term(a_k) = a + (k-1)d
and a₁₃ = a + (13-1) = a + 12d
But a_k - a₁₃ = 84
a + (k-1)d - (a + 12d) = 84
⇒ a + kd - d - a - 12d = 84
⇒ kd - 13d = 84
⇒ (k-13)d = 84
⇒ 7(k-13) = 84
⇒ 7k - 91 = 84