ML Aggarwal ICSE Solutions Class 10 Math 9th Chapter Arithmetic and Geometric Progression
Class 10 Chapter 9 Arithmetic and Geometric ProgressionChapter 9 – Arithmetic and Geometric Progression
Exercise 9.1
(1) For the following A.P.S, write the first term a and the common differenced:
(i) 3, 1, -1, -3
Solution:
Given A.P. is 3, 1, -1, -3, —
First term = a = 3
Common difference (d) = 1 – 3 = -2
= -1 – 1 = -2
= -3+1 = -2
d = -2
(ii) 1/3, 5/3, 9/3, 13/3, —
Solution:
Given A.P. is 1/3, 5/3, 9/3, 13/3, ….
First term (a) = 1/3
Common difference (d) = 5/3 – 1/3 = 4/3
= 9/3 – 5/3 = 4/3
= 13/3 – 9/3 = 4/3
d = 4/3
(iii) -3.2, -3, -2.8, -2.6, ….
Solution:
Given A.P. is -3.2, -3, -2.8, -2.6, …
First term (a) = -3.2
Common difference (d) = -3+3.2 = 0.2
= -2.8 + 3 = 0.2
= -2.6 + 2.8 = 0.2
d = 0.2
(2) Write first four terms of the A.P, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10
Solution:
First term (a) = 1
Common difference (d) = 10
2nd term = a + d = 10 + 10 = 20
3rd term = a + 2d = 10 + 20 = 30
4th term = a + 3d = 10 + 30 = 40
5th term = a + 4d = 10 + 40 = 50
Hence, first four terms are 10, 20, 30, 40, 50, …..
(ii) a = -2, d = 0
Solution:
First term (a) = -2
Common difference (d) = 0
2nd term = a + d = -2 + 0 = -2
3rd term = a + 2d = -2 + 0 = -2
4th term = a + 3d = -2 + 0 = -2
Thus, the required first four terms are -2, -2, -2, -2
(iii) a = 4, d = -3
Solution:
First term (a) = 4
Common difference (d) = -3
2nd term = a + d = 4 – 3 = 1
3rd term = a + 2d = 4 – 6 = -2
4th term = a + 3d = 4 – 9 = -5
Thus, first four terms are found to be 4, 1, -2, -5
(iv) a = 1/2, d = -1/6
Solution:
First term (a) = 1/2
Common difference (d) = -1/6
2nd term = a + d = 1/2 – 1/6 = 3-1/6 = 1/3
3rd term = a + 2d = 1/2 – 1/3 = 1/6
4th term = a + 3d = 1/2 – 1/2 = 0
Thus, the required first four terms are 1/2, 1/3, 1/6, 0
(3) Which of the following lists of numbers forms an A.P.? If they form an A.P., find the common difference ‘d’ and write the next three terms:
(i) 4, 10, 16, 22, ….
Solution:
Given numbers are 4, 10, 16, 22, …..
For common difference:
10 – 4 = 6
16 – 10 = 6
22 – 16 = 6
d = 6
Here, common difference is same,
Thus given sequence is A.P.
Then, next three terms are:
(22 + 6) = 28
(28 + 6) = 34
(34 + 6) = 40
(ii) -2, 2, -2, 2, ….
Solution:
Given numbers are
-2, 2, -2, 2, …
For common difference:
2 + 2 = 4
-2 – 2 = -4
2 + 2 = 4
d ≠ same
Here, the common difference is not same for given numbers.
Hence, the given sequence is not an arithmetic progression.
(iii) 2, 4, 8, 16,….
Solution:
Given numbers are
2, 4, 8, 16, ….
For common difference:
4 – 2 = 2
8 – 4 = 4
16 – 8 = 8
d ≠ same
Here, the common difference is not same for given numbers.
Hence, the given sequence is not an arithmetic progression.
(iv) -10, -6, -2, 2, …
Solution:
Given numbers are
-10, -6, -2, 2 ….
For common difference:
-6 + 10 = 4
-2 + 6 = 4
2 + 2 = 4
d = 4
Here, the common difference is same for given numbers.
Hence, the given sequence is an arithmetic progression.
Exercise 9.2
(1) Find the A.P. whose nth term is (7 – 3n), also find the 20th term.
Solution:
Given that, nth term of an A.P. = an (7 – 3n)
Now, to find the terms of an A.P. put n = 1, 2, 3 ….
n = 1, => a1 = (7 – 3) = 4
n = 2 => a2 = (7 – 6) = 1
n = 3 => a3 = (7 – 9) = -2
Thus, the required A.P is 4, 1, -2, ….
20th term => a20 = (7 – 60) = – 53
(2) Find the indicated term in each of the following A.P
(i) 1, 6, 11, 16, …., a20
Solution:
Here, first term (a) = 1
Common difference = 5
Then, an (n – 1) d
a20 = a + (20 – 1) d
a20 = 1 + 19 (5)
a20 = 1 + 95
a20 = 96
(ii) -4, -7, -10, -13, —- a25, an
Solution:
Here, first term (a) = -4
Common difference (d) = -3
Then, an = a (n – 1) d
Now, a25 = -4 + (25 – 1 (-3)
= -4 + 24 (-3)
= -4 – 72 = -76
a25 = -76
And an = -4 – (n – 1) 3
= -4 – 3n + 3
an = -3n – 1
an = -1 – 3n
(3) Find the nth term and the 12th term of the list of the numbers:
5, 2, -1 – 4, ……
Solution:
Given sequence is 5, 2, -1, -4, …..
Here, a = 5
Common difference => d = -3
Then, nth term: an = a + (n – 1) d
an = 5 + (n – 1) (-3)
= 5 – 3n + 3
= 8 – 3n
an = 8 – 3n is the required nth term of an given A.P.
Now, 12th term: a12 = 8 – 3 (12)
= 8 – 36
a12 = 28
(4) (i) If the common difference of an A.P. is -3 and 18th term is (-5), then find the first term.
Solution:
Given that, common difference (d) = -3
18th term => a18 = -5
We have, an = a + (n – 1)d
a18 = a + (18 – 1) d = a + 17 d
-5 = a + 17 (-3)
-5 = a – 51
a = -5 + 51 = 46
a = 46 => First term
(ii) If the first term of an A.P. is – 18 and it’s 10th term is 3000, then find its common difference.
Solution:
Given that, first term (a) = -18, a10 = 0
We have, an = a + (n – 1) d
a10 = -18 + (10 – 1) d
0 = -18 + 9d
18 = 9d
d = 2 => Common difference
(5) Which term of th A.P.
(i) 3, 8, 13, 18, ….. is 78?
Solution:
Given that, first term (a) = 3
Common difference (d) = 5
Now, an = a + (n – 1) d
78 = 3 + (n – 1) 5
75 = 5n – 5
80 = 5n
n = 16
Thus, 78 is the 16th term of given A.P.
(ii) 18, 15 1/2, 13, …. is -47?
Solution:
Given that, first term (a) = 18
Common difference (d) = – 5/2
Now, an = a + (n – 1) d
-47 = 18 + (n – 1) (-5/2)
-65 = -5n/2 + 5/2
-130 = -5n + 5
-135 = -5n
n = +27
Thus, -47 is the 27th term of given A.P.
(6) (i) Check whether (-150) is a term of A.P. 11, 8, 5, 2, ….
Solution:
Given term (a) = 11 and common difference (d) = -3
We have, an = a + (n – 1) d
-150 = 11 + (n – 1) (-3)
-161 = -3n + 3
-158 = -3n
= n = 52.6
But, here n = 52.6 is not a natural number.
Hence, -150 is not the term of given A.P.
(ii) Find whether 55 is a term of the A.P. 7, 10, 13, …. or not. If yes, find which term form it is.
Solution:
Given A.P. is 7, 10, 13, ….
First term (a) = 7,
Common difference (d) = 3
We have, an = a + (n – 1) d
55 = 7 + (n – 1) 3
48 = 3n – 3
51 = 3n
n = 17
Here, n = 17 is the natural number.
Hence, 55 is the term of given A.P.
Also, 55 is the 17th term of given A.P.
(7) Find the 20th term form the last term of the A.P. 3, 8, 13, —, 253
Solution:
Given, A.P. is 3, 8, 13, —-, 253
First term => (a) = 3
Common difference => d = 5
We have, an = a + (n – 1) d
253 = 3 + (n – 1) 5
250 = 5n – 5
255 = 5n
n = 51
253 is the 51st term of given A.P.
Let us consider, ‘p’ is the 20th term from the last term .
Then, P = L – (n – 1) d
P = 253 – (20 – 1) 5
= 253 – 19 × 5
= 253 – 95
P = 158
Thus, 158 is the 20th term from the last term
(8) Find the sum of the two middle most terms of the A.P -4/3, -1, -2/3, …, 4 1/3
Solution:
Given A.P. is – 4/3, -1, -2/3, …., 4 1/3
Here, first term (a) = -4/3
Common difference (d) = -1 – (-4/3) = -1 + 4/3 = 1/3
d = 1/3
We have, an = a + (n – 1)
13/3 = -4/3 + (n – 1) 1/3
13/3 + 4/3 = n/3 – 1/3
17/3 = (n-1)/3
17 = n – 1
n = 18
Thus, middle term is 18/2 & (18/2) + 1 à 9th & 10th term
Then, a9 + a10 = a + 8d + a + 9d
a9 + a10 = 2a + 17d
= 2 (-4/3) + 17 (1/3)
= -8/3 + 17/3
= 9/3
a9 + a10 = 3 is the required answer.
(Q9) Which term of the A.P. 53, 48, 43, —-, is the first negative term?
Solution:
Given, A.P. is 53, 48, 43, ….
Here, first term –> a = 53
Common difference –> d = -5
We have, an = a + (n – 1) d
= 53 + (n – 1) (-5)
= 53 – 5n + 5
= 58 – 5n
= 58 = 5n
n = 11.6 = 12th
Thus, 12th term is the first negative term of the A.P.
(Q10) Determine the A.P. whose third term is 16 & the 7th term exceeds the 5th term by 12
Solution:
Given that, a3 = 16
a7 – a5 = 12
Let us consider, a be the first term & d be the common difference.
Then, an = a + (n – 1) d
a3 = a + 2d
6 = a + 2d —- (1)
a5 = a + 4d
a7 = a + 6d
a7 – a5 = a + 6d – a – 4d
12 = 2d
d = 6 put in (1)
16 = a + 12
4 = a
Here, first term is a = 4 and common difference (d) = 6
Then, required A.P. is a, a + d, a + 2d, a + 3d, —-
I.e. 4, 10, 16, 22, —–
(Q11) Find the 20th term of the A.P. whose 7th term is 24 less than the 11th term, first term being 12.
Solution:
Given that, first term (a) = 12 and a7 = a11
Then, a11 – a7 = 24 ∵ an = a + (n – 1) d
(a + 10d) – (a + 6d) = 24
4d = 24
d = 6
a = 12
Then 20th term of A.P => a20 = 12 + 19 (6)
= 12 + 114
a20 = 126
Thus, 20th term of an A.P. is found to be 126.
(Q12) Find the 31st term of an A.P. whose 11th term is 38 and 6th term is 73
Solution:
Given that, a11 = 38
a11 = a + 10d
38 = a + 10d —- (1)
a6 = 73
a6 = a + 5d
73 = a + 5d —– (2)
(1) – (2) => (73 – 38) = (5d – 10d)
+35 = -5d
d = -7 Put I (1)
38 = a – 70
70 + 38 = a
a = 108
Thus, 31st term of an A.P. is given by
a31 = a + 30d
a31 = 108 – 210
a31 = -178
Thus, 31s term of given A.P. is found to be (-178)
(Q13) If the seventh term of an A.P. is 1/9 and its ninth term is 1/7, find its 63rd term.
Solution:
Given that, a7 = 1/9 and a9 = 1/7
a7 = a + 6d —– (1) an a9 = a + 8d —- (2)
(1) – (2) => (1/9 – 1/7) = (a + 6d) – (a + 8d) = -2d
(-2/63) = -2d
d = 1/63 put in (1)
1/9 = a + 6/63
7 = 63a + 6
1 = 63a
a = 1/63
Thus, 63rd term of an A.P. is given by
a63 = a + (62) d
= 1/63 + 62/63
a63 = 63/63 = 1
Thus, 63rd term of given A.P. is found to be 1
(Q15) If 8th term of an A.P. is 3000, prove that its 38th term is tripal of its 18th term.
Solution:
Given that, a8 = 0
Let us, consider a is the first term & d is common
Difference of given A.P.
Then, a8 = a + 7d
=> a + 7d = 0
a = -7d
Now, a18 = a + 17d
a18 = -7d + 17d
a18 = 10d —– (1)
a38 = a + 37d
a38 = -d (7) + 37d
a38 = 30d —– (2)
From (1) & (2) = a38 = 30d = 3 (10d)
a38 = 3 (a18) Hence Proved
(Q16) Which of the A.P. 3, 10, 17, —– will be 84 more than its 13th term?
Solution:
Given that, A.P. is 3, 10, 17, ….
First term => a = 3 common difference (d) = 7
Now, a13 = a + 12d = 3 + 84 = 87
a13 = 87
Given that, an = a13 + 84
= 87 + 84
an = 171
a + (n – 1) d = 171
3 + (n – 1) 7 = 171
7n – 7 = 168
7n = 175
n = 25
Thus, 25th term of given A.P. is 84 more than its 13th term.
(Q18) If the numbers (n – 2), (4n – 1) and (5n + 2) are in A.P. find the value of n.
Solution:
Given that, (n – 2), (4n – 1) and (5n + 2) are the numbers which are in A.P.
That mean common difference should be same.
Thus, (4n – 1) – (n – 2) = (5n + 2) – (4n – 1)
3n + 1 = n + 3
2n = 2
n = 1 is the required value
(Q19) The sum of three numbers in A.P. is 3 and their product is -35. Find the numbers.
Solution:
Given that, the sum of three numbers in A.P. is 3
Also, their product = -35
Let us, consider the numbers (a – d), a and (a + d) are in A.P.
=> a – d + a + a + d = 3
3a = 3
a = 1
And, (a – d) a (a + d) = -35
a (a2 – d2) = -35
(a3 – ad2) = -35
(1 – d2) = -35
1 + 35 = d2
=> d2 = 36 => d = ±6
If d = – 6 then the required three numbers are –
a – d = 1 + 6 = 17
a = 1
a + d = 1 – 6 = -5
Hence, the three numbers 7, 1, -5, …… are in A.P.
(Q20) The sum of three numbers in A.P. is 30 & the ratio of first number to the third number is 3:7. Find the numbers.
Solution:
Given that, the sum of three numbers in A.P. is 30. And the ratio of first number to third number is 3:7.
Let us, consider the three numbers be (a – d), a & (a + d).
=> (a – d) + a + (a + d) = 30
3a = 30
a = 10
Again, 3:7 = (a – d) : (a + d)
3/7 = (a-d)/(a+d)
= 3 (a + d) = 7 (a – d)
3a + 3d = 7a – 7d
= 7d + 3d = 7a – 3a
10d = 4a
40 = 10d
d = 4
Thus, the required numbers are: a = 10
a + d = 10 + 4 = 14, a – d = 10 = 10 – 4 = 6
Thus, the required three numbers are 6, 10, 14 , …. in A.P.
(Q21) The sum of the first three terms of an A.P. is 33. If the product of the first and the third terms exceeds the second term by 29, find the A.P.
Solution:
Given that, the sum of the first three term of an A.P. is 33.
Let us consider the three numbers be (a – d), a, (a + d).
= a – d + a + a + d = 33
3a = 33 => a = 11
Also, given that product of first & third term exceeds the second term by 29
= (a – d) (a + d) = a + 29
a2 – d2 = 11 + 29
112 – d2 = 40
121 – 40 = d2
d2 = 81
d = ±9
If d = 9,
If d = 9, then the required numbers are
a – d = 11 – 9 = 2, a = 11, a + d = 11 + 9 = 20
If d = -9, then the required numbers are
(a – d) = 11 – (-9) = 11 + 9 = 20, a = 11, a + d = 11.9 = 2
Hence, the required three numbers are 2, 11, 20, —- and 20, 11, 2, …… which are in A.P.
Exercise 9.3
(Q1) Find the sum of the following A.P.:
(i) 2, 7, 12, … to 10 terms
Solution:
First term (a) = 2
Common difference (d) = 5
S10 = n/2 [2a + (n – 1) d]
= 10/2 [4+9×5]
= 5 (4 + 45)
S10 = 5 (49)
S10 = 245
(ii) 1/15, 1/12, 1/10, ….. to 11 terms
Solution:
First term (a) = 1/15
Common difference (d) = 1/12 – 1/15
= (15-12)/15×12
d = 1/60
S11 = 11/2 [2a + (n – 1) d]
= 11/2 [2× 1/15 + 10×1/60]
= 11/2 (2/15 + 1/6)
= 11/2 (4 + 5) =/30
= 11/2 (9/30)
= 11/2 (3/10)
S11 = 33/20
(Q2) Find the sum given below:
(i) 34 + 32 + 30 + ….. + 10
Solution:
Here, first term (a) = 34
Common difference (d) = -2
Last term (1) = 10
Then, an = a + (n – 1) d
10 = 34 + (n – 1) (-2)
-24 = -2 (n – 1)
-24 = -2n + 2
2n = 24 + 2
2n = 26
n = 13
Thus, 5n = n/2 (a + 1)
= 13/2 (34 + 10)
= 13/2 (44)
= 13 (22)
5n = 1286
(ii) – 5 + (-8) + (-11) + ….. + (-230)
Solution:
First term (a) = -5
Common difference (d) = -3
Last term (l) = -230
We have, an = a + (n – 1) d
-230 = -5 + (n – 1) (-3)
-230 = -5 – 3n + 3
– 230 = -2 – 3n
3n = 230 – 2
3n = 228
n = 76
Thus, 5n = n/2 (a + l)
= 76/2 (-5-230)
= 38 (-5 – 230)
= 38 (-235)
5n = -8930
(Q3) In an A.P. (with usual rotations):
(i) Given a = 5, d = 3, an = 50 find n and 5n.
Solution:
an = a + (n – 1) d
50 = 5 + (n – 1) 3
50 = 5 + 3n – 3
50 = 2 + 3n
3n = 50 – 2
3n = 48
n = 48/3
n = 16
Thus,
5n = n/2 [2a + (n – 1) d]
= 16/2 [2×5+15×3]
= 8 (10 + 45)
5n = 8 × 55
5n = 440
(ii) Given, a = 7, a13 = 35, find d and 513
Solution:
We have, an = a + (n – 1) d
35 = 7 + (13 – 1) d
35 = 7 + 13d – d
35 = 7 + 12d
12d = 28
d = 7/3
Thus, 513 = n = 2 [2a + (n – 1) d]
= 13/2 [2×7+12×7/3]
= 13/2 [14 + 28] = 13/2 × 42 = 13×21
513 = 273
(Q4) (i) The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
Given that, first term (a) = 5, last term (l) = 45
Sum = 400
an = a + (n – 1) d
45 = 5 + (n – 1) d
= (n – 1) d = 45 – 5 = 40
= (n – 1) d = 40 — (1)
Now, Sn = n/2 [2a + (n – 1) d]
400 = n/2 [2×5+40] from (1)
800 = n (10 + 40)
800 = 50n
n = 16 put in (1)
(16 – 1) d = 40
d = 40/15 = 8/3
d = 8/3
Thus, the number terms are 16.
And the common differences is 8/3
(Q5) The first and last terms of an A.P. are 17 and 350 —-. If the common difference is 9, how many terms are there and what is their sum?
Solution:
Given that, first term (a) = 17, last term (l) = 350, Common difference (d) = 9,
an = a + (n – 1) d
350 = 17 + (n – 1) × 9
350 – 17 = 9n – 9
333 + 9 = 9n
9n = 342
n = 342/9
n = 38
Thus, Sn = n/2 [2a + (n – 1) d]
38/2 [2×17+37d]
= 19 [34 + 37 × 9]
= 19 (34 + 333)
= 19 × 367
Sn = 6973
(Q6) Solve for x: 1 + 4 + 7 + 10 + …. + x = 287
Solution:
Given that, first term (a) = 1, Common difference (d) = 3
(Q8) Find the sum of first 22 terms of an A.P. in which d = 7 and a22 is 149.
Solution:
Given that, the sum of first 22 terms of A.P. whose d = 7 and a22 = 149
a22 = (n – 1) d
149 = a + 21d
149 = a + 21 × 7
149 = a + 147
a = 2
Thus, S22 = 22/2 [2a + (n – 1) d]
= 11 [2×2 + 21×7]
= 11 [4 + 21×7]
= 11 (4 + 147)
S22 = 11 × 151 = 1661
Thus, the sum of first 22 terms is found to be 1661.
(Q9) In an A.P. the fourth & sixth term are 8 & 14 respectively find
(i) First term
(ii) Common difference
(iii) Sum of first 20 term
Solution:
Given that, a4 = 8 and a6 = 14
=> a + 3d = 8 —- (1) and a + 5d = 14 —– (2)
=> 3d – 5d = 8 – 14
-2d = -6
d = 3 put in (1) => a + 9 = 8
a = -1
Thus, (i) first term (a) = 1
(ii) Common difference (d) = 3
(iii) Then the sum of first 20 terms is given by
S20 = 20/2 [2a + 19 × d]
= 10 [2 × (-1) + 19 × 3]
= 10 [-2 + 57]
= 10 × 55
S20 = 550 is the required sum.
(Q10) Find the sum of first 51 terms of the A.P. whose second and third are 14 and 18 respectively.
Solution:
Given that, the sum of first 51 terms of an A.P. where a2 = 14, a3 = 18
Then, Common difference = a3 – a2 = 18 – 14 = 4
d = 4
Now, first term = 14 – 4 = 10 and n = 51
a = 10
We have, Sn = n/2 [2a + (n – 1) d]
= 51/2 [2 × 10 + 50 × 4]
= 51/2 [20 + 200]
S51 = 51/2 × 220 = 5610
S51 = 5610 is the required sum of first 51 terms.
(11) If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.
Solution:
Given that S6 = 36 and S16 = 256
We have, Sn = n/2 [2a + (n – 1) d]
Then, S6 = 6/2 [2a + 5d]
36 = 3 (2a + 5d)
12 = 2a + 5d —- (1)
S16 = 16/2 [2a + 15d]
256 = 8 [2a + 15d]
32 = (2a + 15d) —– (2)
(2) — (1) => 32 – 12 = (2a + 15d) – (2a + 5d)
20 = 10d
d = 2 put in (1)
12 ≠ 2a + 10
2 = 2a
a = 1
Thus, the sum of first ten terms is given by,
S10 = n/2 [2a + (n – 1) d] = 5 [2 + 9 × 2]
S10 = 5 (2 + 18)
S10 = 5 (20) = 100
Thus, the sum of first ten terms is found to be 100.
(Q12) Show that: a1, a2, a3, ….. form an A.P. where an is defined as (an = 3 + 4n). Also find the sum of the first 15 terms.
Solution:
Given that, an = 3 + 4n —– (1)
Put n = 1 in (1) => a1 = 3 + 4 = 7
n = 2 => a2 = 3 + 8 = 11
n = 3 => a3 = 3 + 12 = 15
n = 4 => a4 = 3 + 16 = 19
Thus, the required numbers are found to be 7, 11, 15, 19 …..
First term => a = 7 and common difference (d)
d = 11 – 7 = 4
d = 15 – 11 = 4
d = 19 – 15 = 4
d = 4
Thus, 7, 11, 15, 19, …… (i.e. a1, a2, a3, …. forms an A.P.
Now, The sum of first 15 terms is given by
S15 = 15/2 [2 × 7 + 14 × 4]
= 15/2 [14 + 14 × 4] = 15/2 × 14 (1 + 4)
= 105 × 5
S15 = 825 is the required sum of first 15 terms.
(Q13) Sum of first 6 terms of an arithmetic progression is 42 and the ratio of the 10th and 30th term is 1:3. Calculate first and 13th term.
Solution:
Given that, S6 = 42
We have, Sn = n/2 [2a + (n – 1) d]
Thus, S6 = 3 [2a + 5d]
42 = 3 (2a + 5d)
14 = 2a + 5d —- (1)
Put a = d in (1)
14 = 2a + 5a
14 = 7a
a = 2 Put in (1)
14 = 2a + 5d
14 = 4 + 5d
10 = 5d
d = 2
a10 : a30 = 1:3
an = a + (n – 1) d
a10 = a + 9d & a30 = a + 29d
a10/a30 = (a+9d)/(a+29d) = 1/3
=> 3 (a + 9d) = (a + 29d)
3a – a + 27d – 29d = 0
2a – 2d = 0
a = d
Thus, first term => a = 2
And a13 = a + (n – 1) d
= 2 + 12 × 2
= 2 + 24
a13 = 26 is the required term
(Q14) In an A.P. the sum of first n terms is (6n – n2). Find 25th term.
Solution:
Given that, Sn = 6n – n2
Then, S (n – 1) = 6 (n – 1) – (n – 1)2
= 6n – 6 – (n2 + 2n + 1)
= 6n – 6 – n2 + 2n + 1
= – n2 + 8n – 7
Sn-1 = -n2 + 8n – 7
We know that, an = Sn – Sn-1
a25 = S25 – S24
= (6×25 – 252) – (6×24 – 242)
= 25 (6 – 25) – 24 (6 – 24)
= 25 (-19) – 24 (-18)
= – 475 + 432
a25 = -43 is the required 25th term.
(Q15) If Sn denote the sum of first n terms of an A.P. prove that S30 = 3 (S20 – S10)
Solution:
We have, Sn = n/2 [2a + (n – 1) d]
Given that, S30 = 3 (S20 – S10) —– (1)
S20 = 10 [2a + 19d] = 20a + 190d
S10 = 5 (2a + 9d) = 10a + 45d
=> LHS of (1) = 3 (S20 – S10)
S30 = 3 (S20 – S10)
= 3 [20a + 190d – (10a + 45d)]
= 3 [20a + 190d – 10a – 45d]
= 3 [10a + 145d]
S30 = 30a + 145 —- (2)
Now, S30 = 15 (2a + 29d) = 30a + 435d — (3)
From (2) and (3) => S30 = 3 (S20 – S10) (Hence Proved)
(Q16) (i) Find the sum of first 100 positive integers.
Solution:
Let us, consider the sum of the first 1000 positive integers is (1 + 2 + 3 + 4 + ….. + 1000)
Here, a = 1 and common difference d = 1
Thus, we have Sn = n/2 [2a + (n – 1) d]
S1000 = 500 [2 + 999 × 1]
= 500 [2 + 999] = 500 × 1001
S1000 = 500,500
This is the required sum of first 1000 positive integers.
(ii) Find the sum of first 15 multiples of 8
Solution:
The multiples of 8 are: 8, 16, 24, 40, ….. & given n = 15
Here, a = 8, d = 8
Thus, S15 = 15/2 [2 × 8 + 14 × 8]
S15 = 15 × 4 [2 + 14] = 60 [16] = 960
S15 = 960
This is the required sum of first 15 multiples of 8.
(Q17) (i) Find the sum of two digit natural numbers which are divisible by 4.
Solution:
The two digit natural numbers which are divisible by 4 starts from 12, 16, 20, …. is an arithmetic progression.
Here, a = 12 and d = 4
Thus, the last two digit number which is divisible by 4 is 96
Hence, last term = 96
Hence, an = a + (n – 1) d
96 = 12 + (n – 1) 4 = 12 + 4n – 4 = 8 + 4n
4n = 88
n = 22
Hence, 96 is the 22nd term here.
So, we have to find the sum of 22 terms here.
Thus, S22 = 22/2 [2 × 12 + (22 – 1) × 4)
= 11 [24 + 21 × 4]
= 11 [24 + 84] = 11 (108)
S22 = 1188 is the required sum of two digit natural numbers which are divisible by 4.
(ii) Find the sum of all natural numbers less than 100 which are divisible by 6.
Solution:
The natural numbers divisible by 6 start from 6, 12, 18, 24, …. and the last number less than 100 which is divisible by 6 is found to be 98.
Hence, here a = 6, d = 6 and last term = 98
Then, an = 6 + (n – 1) 6
98 = 6 + 6n – 6
98 = 6n
n = 13
98 is the 13th term. Hence, we have to find the sum of first 13th terms which are divisible by 6.
S13 = 13/2 [2×6 + 12×6]
= 13 × 3 (2 + 12)
= 49 × 14
S13 = 546
This is the required sum.
Exercise 9.4
(1) (i) Find the next term of the list of numbers 1/6, 1/3, 2/3, ….
Solution:
Given list numbers are 1/6, 1/3, 2/3, …..
Here, a = 1/6, common ratio => (1/3)/(1/6) = 6/3 = 2
Here, common ratio r = 2 is same (1/3)/(1/3) = 2
r = 2
Hence, the list of numbers given
1/6, 1/3, 2/3, ….. is a geometric progression.
Hence, the next term will be à 2/3 × 2 = 4/3
(ii) Find the next term of the list of numbers 3/16, -3/8, 3/4, -3/2,
Solution:
Given list of numbers is 3/16, -3/8, 3/4, -3/2 …..
Here, a = 3/6 and common ratio
(iii) Find the 10th and nth terms of the list of numbers 5, 25, 125, ….
Solution:
Given list of numbers is 5, 25, 125, ….
Here, r = 5
Then, an = arn-1
a10 = 5 (5)10-1 = 5 × 59 = 510
a10 = 510 is the 10th term
And nth term is give by
an = 5 (5)n-1 = 5n
an = 5n
(2) (i) Which term of the GP. (i) 2, 2√2, 4, …..
Solution:
Here, a = 2 and common ratio =>
2√2/2 = √2
4/2√2 = √2×√2/√2 = √2
r = √2
We have, an = arn-1
128 = 2 (√2)n-1
(√2)n-1 = 64
(√2)n-1 =26
= (√2)n-1 = (√2)12
Comparing exponential on both sides,
We get, n – 1 = 12
n = 13
Hence, 128 is the 13th term of given geometric progression.
(Q3) Determine the 12th term G.P., whose 8th term is 192 and common ratio is 2.
Solution:
Given that, term of G.P. => a8 = 192 and common ratio (r) = 2
We have, an = arn-1
a8 = a (2)8-1 = a (2)7 => 192 = a (128)
=> a = 192/128 —- (1)
Now, a12 = a (2)12-1 = a211
= a = a12/211 —- (2) from (1) & (2)
a12/211 = 192/27
a12 = 192/27 × 211 = 192×211×2-7
= 192 × 24-7
= 192 × 24
= 192 × 16
a12 = 3072
Thus, the required 12th term of G.P. is found to be 3072.
(Q4) In a G.P. the third term is 24 and 6th term is 192 find the 10th term.
Solution:
Given that, 3rd term of G.P. is 24 => a3 = 24
6th term of G.P. is 192 => a6 = 192
We have, an = arn-1
a3 = ar2 => 24 = ar2 => a = 24/r2 —- (1)
And a6 = ar5 => 192 = ar5 => a = 192/r5 —- (2)
From (1) & (2) => 24/r2 = 192/r5
192/24 = r5/r2 = r3 = 8 = r3
r = 2 => a = 6
Thus, 10th term of G.P. is given by, a10 = 6 (2)9
a10 = 3072 is the required 10th term of G.P.
(Q5) Find the number of terms of a G.P. whose terms is 3/4, common ratio is 2 and last term is 384.
Solution:
Given that, first term (a) = 3/4, common ratio (r) = 2 and last term = 384
We have, an = arn-1
384 = (3/4) 2n-1
384×4/3 = 2n-1
512 = 2n-1
=> 29 = 2n-1
9 = n – 1
n = 10
Thus, the required total no. of terms are found to be 10.
(6) Find the value of ‘x’ such that
(i) (x + 9), (x – 6) and 4 are three consecutive terms of a G.P.
Solution:
Given that, (x + 9), (x – 6) and 4 are three consecutive terms of G.P.
= x – 6/x+9 = 4/x-6 = constant = r
= (x – 6)2 = 4x + 36
x2 – 12x + 36 = 4x + 36
x2 – 16x = 0
x (x – 16) = 0
x = 0 or x = 16 are the require/x+d values of x.
(ii) x, (x + 3), (x + 9) are first three terms of G.P. find x.
Solution:
Given that, x, (x + 3), (x + 9) are first three terms of G.P.
=> Common ratio = x+3/x = x+9/x+3
= (x + 3)2 = x (x + 9)
x2 + 6x + 9 = x2 + 9x
-3x + 9 = 0 => -3x = -9
x = 9/3 = 3
x = 3 is the required value of x.
(7) If the fourth, seventh and tenth of G.P. are x, y, z respectively then prove that x, y, z are in G.P.
Solution:
Given that, x, y, z are 4th, 7th & 10th term of G.P. respectively.
Thus, an = arn-1
=> a4 = ar3, a7 = ar6, a10 = ar9
We know that, the geometric mean of G.P. is b2 = ac => b = √ac
=> ar2 = √ar3 × ar9
ar6 = √a2r12 = √(ar6)2 = ar6
y2 = xz
Thus, x, y, z are in G.P. (Hence Proved)
(Q8) The 5th, 8th & 11th terms of a G.P. are p, q and S respectively, show that q2 = ps
Solution:
Given that, 5th 8th and 11th terms of a G.P. are p, q and S respectively.
We have, an = arn-1
= a5 = a1r4, a8 = a1r7, a11 = a1r10
p = a1r4, q = a1r7, s = a1r10
Now, q2 = a12 r14
And ps = (a1r4) (a1r10) = a12 r14 (2)
From (1) & (2) = q2 = ps (Hence proved)
(Q9) If a, (a2 + 2) and (a3 + 10) are in G.P. then find the values of a
Solution:
Given that, a, (a2 + 2) and (a3 + 10) are in G.P.
=> Common ratio => (a2 + 2)/a = (a3+10)/(a2+2)
= (a2 + 2)2 = a (a3 + 10)
= a4 + 4a2 + 4 = a4 + 10
4a2 + 4 – 10a = 0
4a2 – 8a – 2a + 4 = 0
4a (a – 2) – 2 (a – 2) = 0
(4a – 2) (a – 2) = 0
a = 2 or a = 1/2 are the required values of a.
(Q11) The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.
Solution:
Given that, the sum of first three terms of G.P. is 39/10
Their product = 1
Let us consider, a/r, a and ar be the first three terms which are in G.P.
= a/r + a + ar = 39/10 —- (1)
and (a/r) (a) (ar) = 1
∵ From given condition
a3 = 1
=> a = 1 put in (1)
1/r + 1 + r 39/10 => 1 + r + r2 = 39/10 r
=> 10 + 10r + 10r2 – 39r = 0
10r2 – 29r + 10 = 0
10r2 – 25r – 4r + 10 = 0
5r (2r – 5) – 2 (2r – 5) = 0
(5r – 2) (2r – 5) = 0
=> r = 2/5 or 5/2
Thus, when common ratio r = 2/5 then required terms are a/r, a, ar => 5/2, 1, 2/5
And, when common ratio r = 5/2 then required terms are a/r, a, ar => 2/5, 1, 5/2
Exercise 9.5
(1) Find the sum of
(i) 20 terms of the series 2 + 6 + 18 +, …
Solution:
Given, series is 2 + 6 + 18 + ….
Here, first term => a = 2
Common ratio r = 6/2 = 3
Common ratio ∵ r = 3
As r>1 => Sn = a (rn-1)/(r-1)
Sn = 2 (320-1)/3-1 = 320-1
S20 = 320-1 is the required sum of 20 terms for given GP
(ii) 10 terms of series 1 + √3 + 3 +, —-
Solution:
(Q2) Find the sum of the series 81 – 27 + 9 …. – 1/27
Solution:
Given, series is 81 – 27 + 9 ….. – 1/27
(3) The nth term of a G.P. is 128 and the sum of its n term is 255. If its common ratio is 2, then find its first term.
Solution:
Given that, nth term of a G.P. is 128 => an = 128
And Sn = 255 and r = 2
We have an = arn-1
255 × 2n-1 = 128 (2n-1)
255 × 2n-1 = 128 × 2n – 128
(255 × 2n)/2 = 128 × 2n – 128
255 × 2n = 256 × 2n – 256
256 × 2n – 255 × 2n = 256
After × 2n – 255 × 2n = 256
After simplification, 2n = 256
2n = 28, n = 8
∴ 128 = a27
128 = a × 128
a = 1
Thus, the first term is found to be a = 1
(Q4) (i) How many terms of the G.P. 3, 32, 33, …. are needed to give the sum 120?
Solution:
Given, G.P. is 3, 32, 33, ….
Here, first term (a) = 3, common ratio (r) = 3
Thus, we can write,
= a (rn – 1)/(r-1) = 3(3n-1)/(3-1) = 120
= 3 (3n – 1) = 240
3n – 1 = 80
= 3n = 81 = 34
= n = 4
Thus, the sum of first 4 terms of given G.P. is found to be 120.
(ii) How many terms of the G.P. 1, 4, 16, …. must be taken to have their sum equal to 341?
Solution:
Given G.P. is 1, 4, 16, …..
Here, first term (a) = 1, common ratio r = 4
Thus, we can write,
∴ Sn = a (rn-1)/(r-1) = 341 => 1(4n-1)/(4-1) = 341
4n – 1 = 341 × 3
4n = 1024
4n = 45
= n = 5
Thus, the sum of first 5 terms of given G.P. is found to be 341.
(Q6) The 2nd and 5th terms of G.P. are -1/2 and 1/16 respectively find the sum of the series upto 8 terms.
Solution:
Given that, a2 = -1/2 and a5 = 1/16
Let us, consider, a —> first term and
r —> common ratio of G.P.
Then, a2 = ar2-1 = ar = 1/2
a5 = ar5-1 = ar4 = 1/16
Now, ar4/ar = r3 = 1/16/-1/2 = -2/16
r3 = -1/8
r3 = (-1/2)3
=> r = -1/2
Thus, ar = -1/2
a (-1/2) = -1/2
a = 1
and
Thus, the sum of given series upto 8 terms is found to be 85/128.
(Q7) The first term of a G.P. is 27 and 8th term is 1/81. Find the sum of its first 10 terms.
Solution:
Given that, a = 27, a8 = 1/81
Let us, consider ‘a’ be first term & r is the common ratio.
Then, a8 = arn-1 = ar8-1 = ar7 = 1/81
=> 27r7 = 1/81 => r7 = 1/81×27 = 1/2187 = 1/37
∴ r = 1/3 < 1
Thus,
(Q8) Find the first term of the G.P. whose common ratio is 3, last term is 486 and the sum of whose terms is 728.
Solution:
Given that, common ratio (r) = 3
Last term (l) = 486
Sum of the terms = 728
We have, Sn = a(rn-1)/(r-1) ∵ r > 1
728 = a (3n-1)/(3-1)
=> a (3n – 1) = 728×2
a (3n – 1) = 1456 —- (1)
But, last term = arn-1
486 = a × 3n-1
486 = a (3n/3)
486 × 3 = a3n
1458 = a3n —– (2)
From (1) => a (3n-1) = 1456
3n a – a = 1456
1458 – a = 1456 ∵ from (2)
a = 2 is the required first term of G.P.
(Q9) In a G.P. the first term is 7, the last term is 448 and the sum is 889 – find the common ratio.
Solution:
Given that, first term (a) = 7, Last term = 448
Sum of terms = 889
We know that, a, ar, ar2, ar3, —- arn are the terms in G.P.
Let, r be the common ratio here,
Then, an = arn-1 = 448
Sn = a(rn-1)/(r-1) = 889
=> Sn = arn-1r-a/r-1 = 889
448r – 7 = 889r – 88
882 = 441r
r = 2 is the required common ratio.
(Q10) Find the third term of a G.P. whose common ratio is 3 and the sum of whose first seven terms is 2186.
Solution:
Given that, common ratio (r) = 3,
Sum of first seven terms = 2186 = S7
We have, Sn = a(rn-1)/(r-1)
S7 = a(3n-1)/(3-1) = 2186
= a (3n-1)/2 = 2186
= a (37 – 1) = 4372
a (2187 – 1) = 4372
= a = 2 is the required first term of G.P.
(Q11) If the first term of a G.P. is 5 and the sum of first three terms is 31/5, find the common ratio.
Solution:
Given that, a = 5, S3 = 31/5
We have, Sn = a(rn-1)/(r-1) => S3 = (a(r3-1)/(r-1)
31/5 = S(r3-1)(r-1)
31/25 = r3-1/r-1 = (r2+r+1)/(r-1) = 31/25
= r2 + r + 1 = 31/25
25r2 + 25r + 25 = 31
25r2 + 25r – 6 = 0
25r2 + 30r – 5r – 6 = 0
5r (5r + 6) – 1 (5r + 6) = 0
(5r + 6) (5r – 1) = 0
5r – 1 = 0 => r = 1/5
and 5r + 6 = 0
r = -6/5 are the required common ratios.