Chapter – 5: Arithmetic Progressions
Que 1. How many two digits numbers are divisible by 3?
Solution: First two digit number divisible by 3 = 12
Last two digit number divisible by 3 = 99
An arithmetic progression (A.P) is a progression in which the difference between two consecutive terms is constant.
A.P = 12,15,18,…,99
here
First term (a) = 12
Common difference (d) = 3
Let us consider there are n numbers then
an = 99
a + (n – 1)d = 99
12 + (n – 1)3 = 99
12 + 3n – 3 = 99
n = 29+1
n = 30
∴ Two digit numbers divisible by 3 = 30.
Que 2. In an AP, if the common difference (d) = -4 and the seventh term (a7) is 4, then find the the first term.
Solution: we know,an = a + (n−1)d
so, a7 = a + (7−1)d
or, 4 = a + 6 (−4)
or, 24 + 4 = a
∴ a = 28
Que 3. Which term of AP 8,14,20,26…will be 72 more than its 41st term?
Solution:
Given :
8, 14, 20, 26, ..
first term , a = 8 , common difference , d = 14 – 8 = 6
nth term = 72 + a41
a + (n -1)d = 72 + a41
[nth term = a + (n -1)d]
8 + (n – 1) 6 = 72 + a + (41 – 1) d
8 + 6n – 6 = 72 + 8 + 40 × 6
2 + 6n = 72 + 8 + 240
2 + 6n = 80 + 240
2 + 6n = 320
6n = 320 – 2
6n = 318
n = 318/6
n = 53
Hence, 53rd term of the A.P 8, 14, 20, 26, … will be 72 more than its 41st term.
Que 4. Write the nth term of A.P. 1/m, (1+m)/m, (1+2m)/m, ……..?
Solution: We know that,
![798641c9d3](https://www.selfstudys.com/ckeditor/plugins/imageuploader/uploads/798641c9d3.png)
Que 5. Find the middle term of 6, 13, 20… 216?
Solution: The given AP is 6,13,20,……..,216.
First term, a = 6
Common difference d = 13-6 = 7
Suppose these are n terms in the given AP. Then,
![79872a96c9](https://www.selfstudys.com/ckeditor/plugins/imageuploader/uploads/79872a96c9.png)
Hence, the middle term of the given AP is 111.
Que 6. The 8th term of AP is zero. Prove that 38th term is triple of 18th term.
Solution: T8 = a + 7d = 0
⇒ a = −7d
T38 = a + 37d
= −7d + 37d
= 30d
T18 = a + 17d
= −7d + 17d
= 10d.
Clearly T38=3(T18).
Que 7. Find the sum of all two digit positive odd numbers?
Solution: All two-digit odd positive numbers are 11, 13, 15, 17, …., 99. which are in AP with a = 11, d = 2, l = 99
Let the number of terms be n.
an = 99
a + (n − 1)d = 99
11 + (n − 1) × 2 = 99
n = 45
Sum of n terms is given by
Sn = n/2(a + l)
Sn = 45/2(11 + 99) = 2475
Therefore, the sum of all two-digit odd positive numbers is 2475.
Que 8. Three numbers are in AP and their sum is 21, find the middle number ?
Solution: Let the numbers be,
a − d, a, a + d
Given,
a − d + a + a + d = 21
3a = 21
∴ a = 7
Hence the middle number.
Que 9. If 7 times the 7th term of an AP is equal to 11 times its 11th term. Then find its 18th term.
Solution: Let a and d be first term and common difference respectively
According to the question,
7 times the 7th term of an AP is equal to 11 times its 11th term
⇒ 7a7 = 11a11
As we know, nth term of an AP is
an = a + (n – 1)d
where a = first term
an is nth term
d is the common difference
we have,
7(a + 6d) = 11(a + 10d)
⇒ 7a + 42d = 11a + 110d
⇒ 4a + 68d = 0
⇒ a + 17d = 0
⇒ a18 = 0
18th term of AP is 0
Que 10. The consecutive terms of an AP are 2, x, 26, find the value of x?
Solution: x = 14
Step-by-step explanation:
2, x, 26 are in AP.
Hence,
x – 2 = 26 – x
2x = 28
x = 14
Que 11. For what value of p is 2p+1, 13, 5p-3, are 3 consecutive terms of an AP?
Solution: Given 2p+1, 13 , 5p-3 are in AP.
To find: value of P
If a,b,c are in AP then we know that
2b = ( a+c)
Here a = 2p + 1
b = 13
c = 5p – 3
2 (13) = (2p+1+5p-3)
26 = 7p-2
26 + 2 = 7p
28 = 7p
p = 28/7 = 4
Therefore the value of P is 4.
Que 12. Which term of the AP: 3, 8, 13, 18… is 78?
Solution: The formula for nth term of an AP is aₙ = a + (n – 1) d
The given arithmetic progression is 3, 8, 13, 18, …
First term: a = 3
Second term: a + d = 8
Common difference: d = 8 – 3 = 5
an = a + (n – 1) d
an = 78, n = ?
3 + (n – 1) 5 = 78
5(n – 1) = 78 – 3
n – 1 = 15
n = 16
78 is the 16th term of the given AP.
Que 13. Write the 5th term from the end of the AP 3,5,7,9 .. 201?
Solution:
Given :
A.P. 3, 5, 7, 9, ….,201.
Let ‘l’ be the last term of an A.P
Here, l = 201, a = 3, a2 = 5
d = a2 – a
d = 5 – 3 = 2
nth term from the end = l – (n – 1)d
5th term from the end = l – (n – 1)d
5th term from the end = 201 – (5 – 1) × 2
5th term from the end = 201 – 4 × 2
5th term from the end = 201 – 8
5th term from the end = 193
Hence, the 5th term from the end of the A.P is 193.
Que 14. From the given AP: 8, 10, 12… Find the sum of its last 10 terms if it has 60 terms?
Solution: A.P is 8,10,12………………….
Then first term is 8 and common difference is 2.
an = a + (n−1)d.
Then a60 = 8 + (60−1)2 = 8 + 59 × 2 = 126
Last 10 term =a51, a52………………a60
Then a51 = 8 + (51 − 1)2 = 108
So total of last 10 terms = 10/2 [108 × 2 + (10 − 1)2] = 216 + 81 = 234 × 5 = 1170
Que 15. Find the number of terms of an AP 5, 9, 13 …185.
Solution:
Given:
l = 185
a = 5
d = 4
To find:
n = ?
Solution:
l = a + (n – 1)d
185 = 5 + (n – 1)4
185 = 5 + 4n -4
185 = 1 + 4n
185 – 1 = 4n
184 = 4n
184/4 = n
46 = n