NCERT10th MathematicsExercise 5.2Topic: 5 Arithmetic Progressions
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EXERCISE 5.2
Q1. Fill in the blanks in the following table, given that a is the first term, d is the common difference, and an the nth term of the AP:
|
|
a |
d |
n |
an |
|
i |
7 |
3 |
8 |
----- |
|
ii |
-18 |
----- |
10 |
0 |
|
iii |
----- |
-3 |
18 |
-5 |
|
iv |
-18.9 |
2.5 |
----- |
3.6 |
|
v |
3.5 |
0 |
105 |
----- |
Explanation:
1) The nth term an of an AP with the first term 'a' and common difference 'd' is given
by an = a + (n – 1) d.
2) In general, an is known as the nth term of an AP, and ar is known as the general
term or the rth term of an AP.
3) Sometimes an is also called the last term of an AP and is denoted by 'l'.
Solution:
i) a = 7, d = 3, n = 8, find an.
1) According to the problem,
a = 7, d = 3, n = 8,
an = a + (n – 1) d
an = 7 + 3(8 – 1)
an = 7 + 3(8 – 1)
an = 7 + 3(7)
an = 7 + 21an = 28
2) So, here an = 28.
ii) a = - 18, an = 0, n = 10, find d.
1) According to the problem,
a = - 18, an = 0, n = 10,
an = a + (n – 1) d
0 = - 18 + d(10 – 1)
0 = - 18 + 9d
9d = 18
d = 18/9d = 2.
2) So, here d = 2.
iii) d = - 3, an = - 5, n = 18, find a.
1) According to the problem,
d = - 3, an = - 5, n = 18,
an = a + (n – 1) d
- 5 = a + (- 3)(18 – 1)
- 5 = a + (- 3)(17)
- 5 = a - 51
a = 51 - 5a = 46.
2) So, here a = 46.
iv) a = - 18.9, d = 2.5, an = 3.6, find n.
1) According to the problem,
a = - 18.9, d = 2.5, an = 3.6,
an = a + (n – 1) d
3.6 = - 18.9 + (2.5)(n – 1)
(2.5)(n – 1) = 3.6 + 18.9
(2.5)(n – 1) = 22.5(n – 1) = 22.5/2.5
(n – 1) = 225/25
(n – 1) = 9
n = 9 + 1n = 10.
2) So, here n = 10.
v) a = 3.5, d = 0, n = 105, find an.
1) According to the problem,
a = 3.5, d = 0, n = 105,
an = a + (n – 1) d
an = 3.5 + 0(105 – 1)
an = 3.5 + 0(104)
an = 3.5 + 0
an = 3.5
2) So, here an = 3.5.
Q2. Choose the correct choice in the following and justify :
(i) 30th term of the AP: 10, 7, 4, . . . , is
(A) 97 (B) 77 (C) –77 (D) – 87
(ii) 11th term of the AP: – 3, - 1/2, 2, . . ., is
(A) 28 (B) 22 (C) –38 (D) – 48 and 1/2
Explanation:
1) The nth term an of an AP with the first term 'a' and common difference 'd' is given
by an = a + (n – 1) d.
2) In general, an is known as the nth term of an AP, and ar is known as the general
term or the rth term of an AP.
3) Sometimes an is also called the last term of an AP and is denoted by 'l'.
Solution:
(i) 30th term of the AP: 10, 7, 4, . . . , is
(A) 97 (B) 77 (C) –77 (D) – 87
1) According to the problem,
d = a2 - a1
d = 7 - 102) So here,
d = - 3
a = 10, d = - 3, n = 30,
an = a + (n – 1) d
a30 = 10 + (30 – 1) (- 3)
a30 = 10 - 3(29)
a30 = 10 - 87
a30 = - 77
3) So, the answer is B i.e. a30 = - 77.
(ii) 11th term of the AP: - 3, - 1/2, 2, . . ., is
(A) 28 (B) 22 (C) –38 (D) – 48 and 1/2
1) According to the problem,
d = a2 - a1
d = (- 1/2) - (- 3)
d = - 1/2 + 3
d = (6 - 1)/2d = 5/2
2) So here,
a = - 3, d = 5/2, n = 11,
an = a + (n – 1) d
a11 = - 3 + (11 – 1) (5/2)
a11 = - 3 + (5/2)(10)
a11 = - 3 + 25
a11 = 22
3) So, the answer is B i.e. a11 = 22.
Q3. In the following APs, find the missing terms in the boxes :
(i) 2, 囗, 26(ii) 囗, 13, 囗, 3(iii) 5, 囗, 囗, 9½(iv) - 4, 囗, 囗, 囗, 囗, 6(v) 囗, 38, 囗, 囗, 囗, - 22
Explanation:
1) The nth term an of an AP with the first term 'a' and common difference 'd' is given
by an = a + (n – 1) d.
2) In general, an is known as the nth term of an AP, and ar is known as the general
term or the rth term of an AP.
3) Sometimes an is also called the last term of an AP and is denoted by 'l'.
Solution:
(i) 2, 囗, 26
1) Here,
a1 = 2, a2 = ?, a3 = 26
2) Using the formula an = a + (n – 1) d, we can find the value of d and all other terms.
a1 = a = 2 --------- equation 1
a3 = a + (3 - 1)(d)
a3 = 2 + d(2)
26 = 2 + 2d
2d = 26 - 22d = 24
d = 24/2
d = 12 --------- equation 2
3) From equation 1 and equation 2, we have a = 2, d = 12.
4) Using the formula an = a + (n – 1) d, we can find the value of a2.
an = a + (n – 1) d
a2 = 2 + (2 – 1) (12)
a2 = 2 + 12(1)
a2 = 14
5) So, the missing term a2 is 14.
(ii) 囗, 13, 囗, 3
1) Here,
a1 = ? (say a), a2 = 13, a3 = ?, a4 = 3
2) Using the formula an = a + (n – 1) d, we can find the value of d and all other terms.
a1 = a --------- equation 1
a2 = a + (2 - 1)(d)
a2 = a + d
13 = a + d
a + d = 13 --------- equation 2
3) Using formula an = a + (n – 1) d, and a4 = 3 we have,
an = a + (n – 1) d
a4 = a + (4 - 1)(d)
a4 = a + 3d
3 = a + 3d
a + 3d = 3 --------- equation 3
4) Subtract equation 2 from equation 3, and we get,
a + 3d = 3
a + d = 13
( - ) ( - ) ( - )
---------------------------
2d = - 10
d = - 10/2
d = - 5 --------- equation 4
5) Put d = - 5 from equation 4 in equation 2, we get,
a + d = 13
a + (- 5) = 13
a - 5 = 13
a = 13 + 5
a = 18 --------- equation 5
6) Now we will find a3.
an = a + (n – 1) da3 = a + (3 - 1)(d)a3 = 18 + 2 (- 5)a3 = 18 - 10
a3 = 8
7) So, the missing term a1 = 18, and a3 = 8
(iii) 5, 囗, 囗, 9½
1) Here,
a1 = a = 5, a2 = ?, a3 = ?, a4 = 9½
2) Using the formula an = a + (n – 1) d, we can find the value of d and all other terms.
a1 = a = 5 --------- equation 1
an = a + (n – 1) d
a4 = 5 + (4 - 1)(d)
a4 = 5 + 3d
9½ = 5 + 3d
3d = 9½ - 5
3d = 19/2 - 5
3d = (19 - 10)/2
3d = 9/2
d = 3/2 --------- equation 2
3) Using formula an = a + (n – 1) d, a = 5, d = 3/2 we have,
an = a + (n – 1) d
a2 = a + (2 - 1)(d)
a2 = a + d
a2 = 5 + 3/2
a2 = (10 + 3)/2
a2 = 13/2 --------- equation 3
4) Now we will find a2,
an = a + (n – 1) da3 = a + (3 - 1)(d)
a3 = a + 2d
a3 = 5 + 2(3/2)
a3 = 5 + 3
a3 = 8 --------- equation 4
5) So, the missing term a2 = 13/2, and a3 = 8.
(iv) - 4, 囗, 囗, 囗, 囗, 6
1) Here,
a1 = a = - 4, a2 = ?, a3 = ?, a4 = ?, a5 = ?, a6 = 6.
2) Using the formula an = a + (n – 1) d, we can find the value of d and all other
terms.
a1 = a = - 4 --------- equation 1
an = a + (n – 1) d
a6 = - 4 + (6 - 1)(d)
a6 = - 4 + 5d
6 = - 4 + 5d
5d = 6 + 4
5d = 10
d = 10/5
d = 2 --------- equation 2
3) Using formula an = a + (n – 1) d, a = - 4, d = 2 we have,
an = a + (n – 1) d
a2 = a + (2 - 1)(d)
a2 = a + d
a2 = - 4 + 2
a2 = - 2 --------- equation 3
4) Now we will find a3,
an = a + (n – 1) da3 = a + (3 - 1)(d)
a3 = a + 2d
a3 = - 4 + 2(2)
a3 = - 4 + 4
a3 = 0 --------- equation 4
5) Now we will find a4,
an = a + (n – 1) da4 = a + (4 - 1)(d)
a4 = a + 3d
a4 = - 4 + 3(2)
a4 = - 4 + 6
a4 = 2 --------- equation 5
6) Now we will find a5,
an = a + (n – 1) da5 = a + (5 - 1)(d)a5 = a + 4d
a5 = - 4 + 4(2)
a5 = - 4 + 8a5 = 4 --------- equation 6
7) So, the missing term a2 = - 2, a3 = 0, a4 = 2, a5 = 4.
(v) 囗, 38, 囗, 囗, 囗, - 22
1) Here,
a1 = a = ?, a2 = 38, a3 = ?, a4 = ?, a5 = ?, a6 = - 22.
2) Using the formula an = a + (n – 1) d, we can find the value of d and all other
terms.
a1 = a --------- equation 1
an = a + (n – 1) d
a2 = a + (2 - 1)(d)
a2 = a + d
38 = a + d
a + d = 38 --------- equation 2
3) Using formula an = a + (n – 1) d, a6 = - 22 we have,
an = a + (n – 1) d
a6 = a + (6 - 1)(d)
a6 = a + 5d
- 22 = a + 5d
a + 5d = - 22 --------- equation 3
4) Subtract equation 2 from equation 3, and we get,
a + 5d = - 22
a + d = 38
( - ) ( - ) ( - )
---------------------------
4d = - 60
d = - 60/4
d = - 15 --------- equation 4
5) Put d = - 15 from equation 4 in equation 2, we get,
a + d = 38
a + (- 15) = 38
a - 15 = 38
a = 38 + 15
a = 53 --------- equation 5
6) Now we will find a3,
an = a + (n – 1) da3 = a + (3 - 1)(d)
a3 = a + 2d
a3 = 53 + 2(- 15)
a3 = 53 - 30
a3 = 23 --------- equation 6
7) Now we will find a4,
an = a + (n – 1) da4 = a + (4 - 1)(d)
a4 = a + 3d
a4 = 53 + 3(- 15)
a4 = 53 - 45
a4 = 8 --------- equation 7
8) Now we will find a5,
an = a + (n – 1) da5 = a + (5 - 1)(d)a5 = a + 4d
a5 = 53 + 4(- 15)
a5 = 53 - 60a5 = - 7 --------- equation 6
7) So, the missing term a1 = 53, a3 = 23, a4 = 8, a5 = - 7.
Q4. Which term of the AP : 3, 8, 13, 18, . . . , is 78?
Solution:
1) Here,
a1 = a = 3, a2 = 8, a3 = 13, a4 = 18.
2) Here.
d = a2 - a1
d = 8 - 3
d = 5 --------- equation 1
3) Here.
d = a3 - a2
d = 13 - 8
d = 5 --------- equation 2
4) From equation 1 and equation 2 as d = a2 - a1 = a3 - a2 = 5.
5) Using the formula an = a + (n – 1) d,
an = a + (n – 1) d
78 = 3 + (n - 1)(d)
78 = 3 + 5(n - 1)
5(n - 1) = 78 - 3
5(n - 1) = 75
(n - 1) = 75/5
(n - 1) = 15
n = 15 + 1
n = 16 --------- equation 3
6) So, the 16th term of the given AP is 78.
Q5. Find the number of terms in each of the following APs :
(i) 7, 13, 19, . . . , 205 (ii) 18, 15½, 13, . . . , – 47
Explanation:
1) The nth term an of an AP with the first term 'a' and common difference 'd' is given
by an = a + (n – 1) d.
2) In general, an is known as the nth term of an AP, and ar is known as the general
term or the rth term of an AP.
3) Sometimes an is also called the last term of an AP and is denoted by 'l'.
Solution:
(i) 7, 13, 19, . . . , 205
1) Here,
a1 = a = 7, a2 = 13, a3 = 19, an = 205.
2) Here.
d = a2 - a1
d = 13 - 7
d = 6 --------- equation 1
3) Here.
d = a3 - a2
d = 19 - 13
d = 6 --------- equation 2
4) From equation 1 and equation 2 as d = a2 - a1 = a3 - a2 = 6.
5) Using the formula an = a + (n – 1) d,
an = a + (n – 1) d
205 = 7 + (n - 1)(6)
205 = 7 + 6(n - 1)
6(n - 1) = 205 - 7
6(n - 1) = 198
(n - 1) = 198/6
(n - 1) = 33
n = 33 + 1
n = 34
6) The number of terms in the given AP is 34.
(ii) 18, 15½, 13, . . . , – 47
1) Here,
a1 = a = 18, a2 = 15½, a3 = 13, an = - 47.
2) Here.
d = a2 - a1
d = 15½ - 18
d = 31/2 - 18
d = (31 - 36)/2d = - 5/2 --------- equation 1
3) Here.
d = a3 - a2
d = 13 - 15½
d = 13 - 31/2
d = (26 - 31)/2d = - 5/2 --------- equation 2
4) From equation 1 and equation 2 as d = a2 - a1 = a3 - a2 = - 5/2.
5) Using the formula an = a + (n – 1) d,
an = a + (n – 1) d
- 47 = 18 + (n - 1)(- 5/2)
- 47 = 18 - 5(n - 1)/2
5(n - 1)/2 = 18 + 47
5(n - 1)/2 = 65
5(n - 1) = 2(65)
(n - 1) = 2(65)/5
(n - 1) = 2(13)
(n - 1) = 26
n = 26 + 1n = 27
6) The number of terms in the given AP is 27.
Q6. Check whether – 150 is a term of the AP: 11, 8, 5, 2 . . .
1) Here,
a1 = a = 11, a2 = 8, a3 = 5, a4 = 2.
2) Here.
d = a2 - a1
d = 8 - 11
d = - 3
3) So d = a2 - a1 = - 3.
4) Using the formula an = a + (n – 1) d,
an = a + (n – 1) d
- 150 = 11 + (n - 1)(- 3)
- 150 = 11 - 3(n - 1)
3(n - 1) = 11 + 150
3(n - 1) = 161
(n - 1) = 161/3
n = (161/3) + 1
n = (161 + 3)/3
n = (164)/3
5) As, n = 164/3 is not an integer, (- 150) is not a term of this AP.
Q7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Explanation:
1) The nth term an of an AP with the first term 'a' and common difference 'd' is given
by an = a + (n – 1) d.
2) In general, an is known as the nth term of an AP, and ar is known as the general
term or the rth term of an AP.
3) Sometimes an is also called the last term of an AP and is denoted by 'l'.
Solution:
1) Here,
a11 = 38, a16 = 73, find a31.
2) Let the first term be a and the common difference be d.
an = a + (n – 1) d
a11 = a + (11 - 1)(d)
a11 = a + 10d
38 = a + 10d
a + 10d = 38 --------- equation 1
3) Using formula an = a + (n – 1) d, a16 = 73 we have,
an = a + (n – 1) d
a16 = a + (16 - 1)(d)
a16 = a + 15d
73 = a + 15d
a + 15d = 73 --------- equation 2
4) Subtract equation 1 from equation 2, and we get,
a + 15d = 73
a + 10d = 38
( - ) ( - ) ( - )
---------------------------
5d = 35
d = 35/5
d = 7 --------- equation 3
5) Put d = 7 from equation 3 in equation 1, we get,
a + 10d = 38
a + 10(7) = 38
a + 70 = 38
a = 38 - 70
a = - 32 --------- equation 4
6) Now we will find a31,
an = a + (n – 1) da31 = a + (31 - 1)(d)
a31 = a + 30d
a31 = - 32 + 30(7)
a31 = - 32 + 210
a31 = 178 --------- equation 5
7) So, here a31 = 178.
Q8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
1) Here,
n = 50, a3 = 12, a50 = 106, find a29.
2) Let the first term be "a" and the common difference be "d".
an = a + (n – 1) d
a3 = a + (3 - 1)(d)
a3 = a + 2d
12 = a + 2d
a + 2d = 12 --------- equation 1
3) Using formula an = a + (n – 1) d, a50 = 106 we have,
an = a + (n – 1) d
a50 = a + (50 - 1)(d)
a50 = a + 49d
106 = a + 49d
a + 49d = 106 --------- equation 2
4) Subtract equation 1 from equation 2, and we get,
a + 49d = 106
a + 2d = 12
( - ) ( - ) ( - )
---------------------------
47d = 94
d = 94/47
d = 2 --------- equation 3
5) Put d = 2 from equation 3 in equation 1, we get,
a + 2d = 12
a + 2(2) = 12
a + 4 = 12
a = 12 - 4
a = 8 --------- equation 4
6) Now we will find a29,
an = a + (n – 1) da29 = a + (29 - 1)(d)
a29 = a + 28d
a29 = 8 + 28(2)
a29 = 8 + 56
a29 = 64--------- equation 5
7) So, here a29 = 64.
Q9. If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
Solution:
1) Here,
a3 = 4, a9 = - 8, find which term is 0.
2) Let the first term be "a" and the common difference be "d".
an = a + (n – 1) d
a3 = a + (3 - 1)(d)
a3 = a + 2d
4 = a + 2d
a + 2d = 4 --------- equation 1
3) Using formula an = a + (n – 1) d, a9 = - 8 we have,
an = a + (n – 1) d
a9 = a + (9 - 1)(d)
a9 = a + 8d
- 8 = a + 8d
a + 8d = - 8 --------- equation 2
4) Subtract equation 1 from equation 2, and we get,
a + 8d = - 8
a + 2d = 4
( - ) ( - ) ( - )
---------------------------
6d = - 12
d = - 12/6
d = - 2 --------- equation 3
5) Put d = - 2 from equation 3 in equation 1, we get,
a + 2d = 4
a + 2(- 2) = 4
a - 4 = 4
a = 4 + 4
a = 8 --------- equation 4
6) Now we will find n when an = 0.
an = a + (n – 1) d0 = 8 + (n - 1)(- 2)
0 = 8 - 2(n - 1)
2(n - 1) = 8(n - 1) = 8/2
(n - 1) = 4
n = 5 --------- equation 5
7) So, here the 5th term is 0.
Q10. The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
1) Here,
a17 = a10 + 7 --------- equation 1.
2) Let the first term be "a" and the common difference be "d".
an = a + (n – 1) d
a17 = a + (17 - 1)(d)
a17 = a + 16d --------- equation 2
3) Using formula an = a + (n – 1) d, a9 = - 8 we have,
an = a + (n – 1) d
a10 = a + (10 - 1)(d)
a10 = a + 9d --------- equation 3
4) From equations 1, 2, and 3 we have,
a17 = a10 + 7
a + 16d = a + 9d + 7
a + 16d - a - 9d = 7
16d - 9d = 7
7d = 7
d = 7/7
d = 1 --------- equation 4
5) So, here the common difference is 1.
Q11. Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
Solution:
1) Here,
a1 = a = 3, a2 = 15, a3 = 27, a4 = 39.
2) Here.
d = a2 - a1
d = 15 - 3
d = 12 --------- equation 1
3) Using the formula an = a + (n – 1) d,
an = a + (n – 1) d
a54 = 3 + 12 (54 – 1)
a54 = 3 + 12 (53)
a54 = 3 + 636
a54 = 639 --------- equation 2
4) Now we will find n for the term more by 132 than the 54th term.
an = a54 + 132
an = 639 + 1325) Now we will find the value of n.
an = 771 --------- equation 3
an = a + (n – 1) d
771 = 3 + 12 (n – 1)
12 (n – 1) = 771 - 3
12 (n – 1) = 768
(n – 1) = 768/12
(n – 1) = 64
n = 64 + 1
n = 65 --------- equation 4
6) So, 65th term is 132 more than 54th term.
7) Second method: our term is 132 more than the 54th term. As the difference is 12,
132 will give us 132/12 = 11. So our term is 54 + 11 = 65. So, 65th term is 132 more than 54th 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?
Solution:
1) Let the first terms of two APs be "x" and "y", and the common difference be "d".
2) The 100th term first AP of which the first term is x:
an = a + (n – 1) d
a100 = x + (100 - 1)(d)
a100 = x + 99d --------- equation 1
3) The 100th term of the second AP of which the first term is y:
an = a + (n – 1) d
a100 = y + (100 - 1)(d)
a100 = y + 99d --------- equation 2
4) As the difference between their 100th term 100, we have,
(x + 99d) - (y + 99d) = 100
(x - y) = 100 --------- equation 3
5) The 1000th term first AP of which the first term is x:
an = a + (n – 1) d
a1000 = x + (1000 - 1)(d)
a1000 = x + 999d --------- equation 4
6) The 1000th term second AP of which the first term is y:
an = a + (n – 1) d
a1000 = y + (1000 - 1)(d)
a1000 = y + 999d --------- equation 5
7) Now we find the difference between their 1000th terms,
The difference = (x + 999d) - (y + 999d)
The difference = (x - y) --------- equation 6
8) From equations 3 and 6, we have,
The difference = 100
9) So, the difference between their 1000th terms is 100.
Q13. How many three-digit numbers are divisible by 7?
Solution:
1) Here the first 3-digit number which is divisible by 7 is a = 105 and the common
difference d = 7.
2) The greatest 3-digit number is 999. When 7 divides 999, the remainder is 5.
So 999-5 divisible by 7. i.e 994 is divisible by 7. So here an = 994.
3) Using the formula an = a + (n – 1) d, we have,
an = a + (n – 1) d
994 = 105 + 7 (n – 1)
7 (n – 1) = 994 - 105
7 (n – 1) = 889
(n – 1) = 889/7
(n – 1) = 127
4) So, 128 3-digit numbers are divisible by 7.n = 127 + 1n = 128
Q14. How many multiples of 4 lie between 10 and 250?
Solution:
1) Here the first number between 10 and 250 which is divisible by 4 is a = 12 and
the common difference d = 4.
2) When 4 divides 250, the remainder 2. So 250-2 is divisible by 4. i.e. 248 is
divisible by 4. So here an = 248.
3) Using the formula an = a + (n – 1) d, we have,
(n – 1) = 59an = a + (n – 1) d248 = 12 + 4 (n – 1)4 (n – 1) = 248 - 12
4 (n – 1) = 236
(n – 1) = 236/4
n = 59 + 1
n = 604) So, there are 60 numbers that are divisible by 4 between 10 and 250.
Q15. For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
Solution:
1) For first AP, a1 = a = 63, a2 = 65, a3 = 67.
2) Here.
d = a2 - a1
d = 65 - 63
d = 2
2) The nth term first AP will be,
an = a + (n – 1) d
an = 63 + (n - 1)(2)
an = 63 + 2(n - 1) --------- equation 1
3) For first AP, a1 = a = 3, a2 = 10, a3 = 17.
4) Here.
d = a2 - a1
d = 10 - 3
d = 7
5) The nth term second AP will be,
an = a + (n – 1) d
an = 3 + (n - 1)(7)
an = 3 + 7(n - 1) --------- equation 2
6) As nth term of both the APs are the same, from equations 1 and 2 we have,
63 + 2(n - 1) = 3 + 7(n - 1)
7(n - 1) - 2(n - 1) = 63 - 3
7n - 7 - 2n + 2 = 607) Therefore, the 13th terms of both these A.P.s are equal.
5n - 5 = 60
5n = 60 + 5
5n = 65
n = 65/5
n = 13
Q16. Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
1) Let the first term of an AP be "a" and the common difference be "d".
2) Using the formula an = a + (n – 1) d,
an = a + (n – 1) d
a3 = a + (3 – 1) d
16 = a + 2d
a + 2d = 16 --------- equation 1
3) Now we will find the 5th term.
an = a + (n – 1) d
a5 = a + (5 – 1) d
a5 = a + 4d --------- equation 2
4) Now we will find the 7th term.
an = a + (n – 1) d
a7 = a + (7 – 1) d
a7 = a + 6d --------- equation 3
5) As the 7th term exceeds the 5th term by 12, we have,
a7 = a5 + 12 --------- equation 4
6) So, from equations 2, 3, and 4, we have,
a7 = a5 + 12
a + 6d = a + 4d + 12
a + 6d - a - 4d = 12
6d - 4d = 12
2d = 12
d = 12/2
d = 6 --------- equation 5
7) Put d = 6 from equation 5 in equation 1, and we get.
a + 2d = 16
a + 2(6) = 16
a + 12 = 16
a = 16 - 12
a = 4 --------- equation 6
8) So, the AP will be 4, 10, 16, 22...
Q17. Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
Solution:
1) Here,
a1 = a = 3, a2 = 8, a3 = 13, an = 253.
2) Here.
d = a2 - a1
d = 8 - 3
d = 5 --------- equation 1
3) Using the formula an = a + (n – 1) d,
an = a + (n – 1) d
253 = 3 + (n - 1)(5)
253 = 3 + 5(n - 1)
5(n - 1) = 253 - 3
5(n - 1)/2 = 65
5(n - 1) = 2(65)
(n - 1) = 2(65)/5
(n - 1) = 2(13)
(n - 1) = 26
n = 26 + 1n = 27
6) The number of terms in the given AP is 27.
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