Exercise 9.1
1. For the following A.P.s, write the first term ‘a’ and the common difference ‘d’:
(i) 3,1,– 1,– 3,…
Solution:-
We have given that:
From the question,
The first term a = 3
Then, difference d = 1 – 3 = – 2
– 1 – 1 = – 2
– 3 – (-1 ) = – 3 + 1 = -2
Hence, common difference d = – 2
(ii) 1/3,5/3,9/3,13/3,….
Solution:-
We have given that:
From the question,
The first term a = 1/3
Then, difference d = 5/3 – 1/3 = (5 – 1)/3 = 4/3
9/3 – 5/3 = (9 – 5)/3 = 4/3
13/3 – 9/3 = (13 – 9)/3 = 4/3
Hence, common difference d = 4/3
(iii) -3.2,-3,-2.8,-2.6,….
Solution:-
We have given that:
From the question,
The first term a = -3.2
Then, difference d = -3 – (-3.2) = -3 + 3.2 = 0.2
-2.8 – (-3) = -2.8 + 3 = 0.2
-2.6 – (-2.8) = -2.6 + 2.8 = 0.2
Hence, common difference 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:-
We have given that:
From the question,
First term a = 10
Common difference d = 10
Then the first four terms are = 10 + 10 = 20
20 + 10 = 30
30 + 10 = 40
Hence, first four terms are 10,20,30 and 40.
(ii) a = -2,d = 0
Solution:-
We have given that:
From the question,
First term a = -2
Common difference d = 0
Then the first four terms are = -2 + 0 = -2
-2 + 0 = -2
-2 + 0 = -2
Hence, first four terms are -2,-2,-2 and -2.
(iii) a = 4,d = -3
Solution:-
We have given that:
From the question,
First term a = 4
Common difference d = -3
Then the first four terms are = 4 + (-3) = 4 – 3 = 1
1 + (-3) = 1 – 3 = – 2
-2 + (-3) = -2 – 3 = – 5
Hence, first four terms are 4,1,-2 and -5.
(iv) a = ½,d = -1/6
Solution:-
We have given that:
From the question,
First term a = ½
Common difference d = -1/6
Then the first four terms are = ½ + (-1/6) = ½ – 1/6 = (3 – 1)/6 = 2/6 = 1/3
1/3 + (-1/6) = 1/3 – 1/6 = (2 – 1)/6 = 1/6
1/6 + (-1/6) = 1/6 – 1/6 = 0
Hence, first four terms are ½,1/3,1/6 and 0.
3. Which of the following lists of numbers form 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:-
We have given that:
From the question,
First term a = 4
Then, difference d = 10 – 4 = 6
16 – 10 = 6
22 – 16 = 6
So, common difference d = 6
Hence, the numbers are form A.P.
(ii) -2,2,-2,2,…
Solution:-
We have given that:
From the question,
First term a = -2
Then, difference d = -2 – 2 = – 4
-2 – 2 = -4
2 – (-2) = 2 + 2 = 4
Therefore, common difference d is not same in the given numbers.
Hence, the numbers are not form A.P.
(iii) 2,4,8,16,…
Solution:-
We have given that:
From the question,
First term a = 2
Then, difference d = 4 – 2 = 2
8 – 4 = 4
16 – 8 = 8
Hence, common difference d is not same in the given numbers.
So, the numbers are not form A.P.
(iv) 2,5/2,3,7/2,…
Solution:-
We have given that:
From the question,
First term a = 2
Then, difference d = 5/2 – 2 = (5 – 4)/2 = ½
3 – 5/2 = (6 – 5)/2 = ½
7/2 – 3 = (7 – 6)/2 = ½
Hence, common difference d = ½
So, the given values d = ½ are not roots of the equation.
Hence, the numbers are form A.P.
(v) – 10,-6,-2,2,…
Solution:-
We have given that:
From the question,
First term a = -10
Then, difference d = -6 – (- 10) = – 6 + 10 = 4
-2 – (-6) = – 2 + 6 = 4
2 – (-2) = 2 + 2 = 4
Therefore, common difference d = 4
So, the given values d=4 are not roots of the equation.
Hence, the numbers are form A.P.
(vi) 12,32,52,72,…
Solution:-
We have given that:
From the question,
First term a = 12 = 1
Then, difference d = 32 – 12 = 9 – 1 = 8
52 – 32= 25 – 9 = 16
72 – 52 = 49 – 25 = 24
So, common difference d is not same in the given numbers.
Hence, the numbers are not form A.P.
(vii) 1,3,9,27,…
Solution:-
We have given that:
From the question,
First term a = 1 = 1
Then, difference d = 3 – 1 = 2
9 – 3 = 6
27 – 9 = 18
Therefore, common difference d is not same in the given numbers.
Hence, the numbers are not form A.P.
(viii) √2,√8,√18,√32,…
Solution:-
We have given that:
Given numbers can be written as, √2,2√2,3√2,4√2,…
From the question,
First term a =√( 2)
Then, difference d = 2√2 – √2 = √2
3√2 – 2√2 = √2
4√2 – 3√2 = √2
Therefore, common difference d = √2
Hence, the given values d = √2 are not roots of the equation.
Hence, the numbers are form A.P.
(ix) a,2a,3a,4a,…
Given, From the question,
First term a = a
Then, difference d = 2a – a = a
3a – 2a = a
4a – 3a = a
Therefore, common difference d = a
Hence, the given values d=a are not roots of the equation.
Hence, the numbers are form A.P.
(x) a,2a + 1,3a + 2,4a + 3,…
Given, From the question,
First term a = a
Then, difference d = (2a + 1) – a = 2a + 1 – a = a + 1
(3a + 2) – (2a + 1) = 3a + 2 – 2a – 1 = a + 1
(4a + 3) – (3a + 2) = 4a + 3 – 3a – 2 = a + 1
Therefore, common difference d = a +1
Hence, the given values d = a +1 are not roots of the equation.
Hence, the numbers are form A.P.
Exercise 9.2
1. Find the A.P. whose nth term is 7 – 3K. Also find the 20th term.
Solution:-
We have given that:
From the question,
2. Find the indicated terms in each of following A.P.s:
(i) 1,6,11,16,…; a20
Solution:-
We have given that:
From the question,
The first term a = 1
Then, difference d = 6 – 1 = 5
11 – 6 = 5
16 – 11 = 5
Therefore, common difference d = 5
From the formula, an = a + (n – 1)d
So, a20 = a + (20 – 1)d
= 1 + (20 – 1)5
= 1 + (19)5
= 1 + 95
= 96
Therefore, a20 = 96
Hence, the given values a20 = 96 are not roots of the equation.
(ii) -4,-7,-10,-13,…,a25,an
Solution: -
We have given that:
From the question,
The first term a = -4
Then, difference d = -7 – (-4) = – 7 + 4 = -3
-10 – (-7) = -10 + 7 = -3
-13 – (-10) = -13 + 10 = -3
Therefore, common difference d = -3
From the formula, an = a + (n – 1)d
So, a25 = a + (25 – 1)d
= -4 + (25 – 1)(-3)
= -4 + (24)-3
= – 4 – 72
= -76
Therefore, a25 = -76
Now, an = a + (n – 1)d
an = -4 + (n – 1)-3
= -4 – 3n + 3
= -1 – 3n
= 8 – 3n
So, T12 = a + (12 – 1)d
= 5 + (12 – 1)(-3)
= 5 + (11)-3
= 5 – 33
= – 28
Hence, the 12th term is -28
4. Find the 8th term of the A.P. whose first term is 7 and common difference is 3.
Solution:-
We have given that:
From the question,
The first term a = 7
Then, common difference d = 3
Tn = a + (n – 1)d
So, T10 = a + (10 – 1)d
0 = -18 + (10 – 1)d
0 = -18 + 9d
9d = 18
d = 18/9
d = 2
Therefore, common difference d = 2
6. Which term of the A.P.
(i) 3,8,13,18,… is 78?
Solution:-
According to the question,
Let us assume 78 as nth term.
Now, from the question
The first term a = 3
Then, difference d = 8 – 3 = 5
13 – 8 = 5
18 – 13 = 5
Therefore, common difference d = 5
Tn = a + (n – 1)d
So, 78 = a + (n – 1)d
78 = 3 +(n-1)5
78 = 3 + 5n – 5
78= -2 + 5n
5n = 78 + 2
5n = 80
n = 80/5
n = 16
Therefore, 78 is 16th term.
(ii) 7,13,19,… is 205 ?
Solution:-
According to the question,
Let us assume 205 as nth term.
From the question,
The first term a = 7
Then, difference d = 13 – 7 = 6
19 – 13 = 6
Therefore, common difference d = 6
Tn = a + (n – 1)d
So, 205 = a + (n – 1)d
205 = 7 + (n – 1)6
205 = 7 + 6n – 6
205 = 1 + 6n
6n = 205 – 1
6n = 204
n = 204/6
n = 34
Therefore, 205 is 34th term.
(iii) 18,15½,13,… is – 47 ?
Solution:-
We have given that:
Convert mixed fraction into improper fraction = 15½ = 31/2
Let us assume -47 as nth term.
From the question,
The first term a = 18
Then, difference d = 31/2 – 18 = (31 – 36)/2 = -5/2
13 – 31/2 = (26 – 31)/2 = -5/2
Therefore, common difference d = -5/2
Tn = a + (n – 1)d
So, -47 = a + (n – 1)d
-47 = 18 + (n – 1)(-5/2)
-47 = 18 – 5/2n + 5/2
-47 – 18 = -5/2n + 5/2
-65 = -5/2n + 5/2
-65 – 5/2 = – 5/2n
(-130 – 5)/2 = -5/2n
-135/2 = -5/2n
n = (-135/2) × (-2/5)
n = -135/-5
n = 27
Hence, -47 is 27th term.
7. (i) Check whether – 150 is a term of the A.P. 11,8,5,2,…
Solution:-
We have given that:
From the question it is given that,
The first term a = 11
Then, difference d = 8 – 11 = -3
5 – 8 = -3
2 – 5 = -3
Then, common difference d = – 3
Let us assume -150 as nth term,
Tn = a + (n – 1)d
So, -150 = 11 + (n – 1)(-3)
-150 = 11 – 3n + 3
-150 = 14 – 3n
3n = 150 + 14
3n = 164
n = 164/3
Hence, the given values n=164/3 are not roots of the equation.
Therefore, – 150 is not a term of the A.P. 11,8,5,2,…
(ii) Find whether 55 is a term of the A.P. 7,10,13,… or not. If yes, find which term is it.
Solution:-
As per given information,
The first term a = 7
Then, difference d = 10 – 7 = 3
13 – 10 = 3
Then, common difference d = 3
Let us assume 55 as nth term,
Tn = a + (n – 1)d
So, 55 = 7 + (n – 1)3
55 = 7 + 3n – 3
55 = 4 + 3n
3n = 55 – 4
3n = 51
n = 51/3
n = 17
Therefore, 55 is 17th term of the A.P. 7,10,13,…
(iii) Is 0 a term of the A.P. 31,28,25,…? Justify your answer.
Solution:-
According to question,
It is given that
The first term a = 31
Then, difference d = 28 – 31 = -3
25 – 28 = -3
Then, common difference d = – 3
Let us assume 0 as nth term,
Tn = a + (n – 1)d
So, 0 = 31 + (n – 1)(-3)
0 = 31 – 3n + 3
0 = 34 – 3n
3n = 34
n = 34/3
Hence, the given values n = 34/3 are not roots of the equation.
Therefore, 0 is not a term of the A.P. 31, 28, 25, …
8. (i) Find the 20th term from the last term of the A.P. 3,8,13,…,253.
Solution:-
We have given that:
Let us assume 253 as nth term.
From the question,
The first term a = 3
Then, difference d = 8 – 3 = 5
13 – 8 = 5
Therefore, common difference d = 5
Tn = a + (n – 1)d
So, 253 = a + (n – 1)d
253 = 3 + (n – 1)5
253 = 3 + 5n – 5
253 = -2 + 5n
5n = 253 + 2
5n = 255
n = 255/5
n = 51
Therefore, 253 is 51th term.
Now, assume ‘P’ be the 20th term from the last.
Then, P = L – (n – 1)d
= 253 – (20 – 1) 5
= 253 – (19) 5
= 253 – 95
P = 158
Hence, the given values p=158 are not roots of the equation.
Therefore, 158 is the 20th term from the last.
(ii) Find the 12th from the end of the A.P. – 2,– 4,– 6,…,– 100.
Solution:-
As per given information,
Let us assume -100 as nth term.
From the question,
The first term a = -2
Then, difference d = – 4 – (-2) = – 4 + 2 = -2
-6 – (-4) = -6 + 4 = – 2
Therefore, common difference d = -2
Tn = a + (n – 1)d
So, – 100 = a + (n – 1)d
– 100 = -2 + (n – 1)(-2)
– 100 = -2 – 2n + 2
– 100 = -2n
n = -100/-2
n = 50
Therefore,-100 is 50th term.
Now, assume ‘P’ be the 12th term from the last.
Then, P = L – (n – 1)d
= -100 – (12 – 1) (-2)
= -100 – (11) (-2)
= -100 + 22
P = – 78
Therefore, -78 is the 12th term from the last of the A.P. – 2,– 4,– 6,…
Then, a9 + a10 = a + 8d + a + 9d
= 2a + 17d
Now substitute the value of ‘a’ and ‘d’.
= 2(-4/3) + 17(1/3)
= -8/3 + 17/3
= (-8 + 17)/3
= 9/3
= 3
Therefore, the sum of the two middle most terms of the A.P is 3.
10. Which term of the A.P. 53,48,43,… is the first negative term ?
Solution:-
As per given information,
The first term a = 53
Then, difference d = 48 – 53 = -5
= 43 – 48 = -5
Therefore, common difference d = -5
Tn = a + (n – 1)d
= 53 + (n – 1)(-5)
= 53 – 5n + 5
= 58 – 5n
5n = 58
n= 58/5
n = 11.6 ≈ 12
11. Determine the A.P. whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20.
Solution:-
As per given information:
From the question,
T5 = 19
T8 – T13 = 20
We know that, Tn = a + (n – 1)d
So, T5 = a + 4d = 19 … [equation (i)]
T13 – T8 = (a + 12d) – (a + 7d) … [equation (ii)]
20 = a + 12d – a – 7d
20 = 5d
d = 20/5
d = 4
Now, substitute value of d in equation (i) we get,
Then, T5 = a + 4d
19 = a + 4(4)
a = 19 – 16
a = 3
Therefore, A.P. is 3 + 4 = 7,7 + 4 = 11,11 + 4 = 15
Hence, the four term of A.P. is 3,7,11,15, …
As per given information:
From the question,
T5 = 19
T8 – T13 = 20
We know that, Tn = a + (n – 1)d
So, T5 = a + 4d = 19 … [equation (i)]
T13 – T8 = (a + 12d) – (a + 7d) … [equation (ii)]
20 = a + 12d – a – 7d
20 = 5d
d = 20/5
d = 4
Now, substitute value of d in equation (i) we get,
Then, T5 = a + 4d
19 = a + 4(4)
a = 19 – 16
a = 3
Therefore, A.P. is 3 + 4 = 7,7 + 4 = 11,11 + 4 = 15
Hence, the four term of A.P. is 3,7,11,15, …
14. Find the 31st term of an A.P. whose 11th term is 38 and 6th term is 73.
Solution: -
As We have given that:
T11 = 38
T6 = 73
Let us assume ‘a’ be the first term and ‘d’ be the common difference,
So, T11 = a + 10d = 38 equation (i)
T6 = a + 5d = 73 equation (ii)
Subtracting both equation (i) and equation (i), we get
(a + 10d) – (a + 5d) = 73 – 38
a + 10d – a – 5d = 35
5d = 35
d = 35/5
d = 7
Now, substitute the value of d in equation (i) to find out a, we get
a + 10d = 38
a + 10(7) = 38
a + 70 = 38
a = 38 – 70
a = -32
Therefore, T31 = a + 30d
= -32 + 30(7)
= -32 + 210
= 178
∴ 31st term is 178.
a = 1/9 – 6/63
a = (7 – 6)/63
a = 1/63
Therefore, T63 = a + 62d
= 1/63 + 62(1/63)
= 1/63 + 62/63
= (1 + 62)/63
= 63/63
= 1
Hence, 63rd term of an A.P. is 1.
(ii) The sum of 5th and 7th terms of an A.P. is 52 and the 10th term is 46. Find the A.P.
Solution:-
According to the question
a5 + a7 = 52
(a + 4d) + (a + 6d) = 52
a + 4d + a + 6d = 52
2a + 10d = 52
Divide both the side by 2 we get,
a + 5d = 26 … (i)
Given, a10 = a + 9d = 46
a + 9d = 46 … (ii)
Now subtracting equation (i) from equation (ii), we get
(a + 9d) – (a + 5d) = 46 – 26
a + 9d – a – 5d = 20
4d = 20
d = 20/4
d = 5
Then, Substitute the value of d in equation (i) to find out a, we get
a + 5d = 26
a + 5(5) = 26
a + 25 = 26
a = 26 – 25
a = 1
Then, a2 = a + d
= 1 + 5 = 6
a3 = a2 + d
= 6 + 5
= 11
a4 = a3 + d
= 11 + 5
= 16
Therefore, 1,6,11,16,… are A.P.
17. If 8th term of an A.P. is zero, prove that its 38th term is triple of its 18th term.
Solution:-
We have given that:
T8 = 0
We have to prove that, 38th term is triple of its 18th term = T38 = 3T18
T8 = a + 7d = 0
T8 = a = -7d
T38 = a + 37d
= -7d + 37d
= 30d
Take, T18 = a + 17d
Substitute the value of a and d,
T18 = -7d + 17d
T18 = 10d
Hence, the given values
T18 = 10d are not roots of the equation.
By comparing results of T38 and T18,38th term is triple of its 18th term.
18. Which term of the A.P.3,10,17,… will be 84 more than its 13th term?
Solution:-
According to the question,
We have given that:
First term a = 3
Common difference d = 10 – 3 = 7
Then, T13 = a + 12d
= 3 + 12(7)
= 3 + 84
= 87
Let us assume that, nth term is 84 more than its 13th term
So, Tn = 84 + 87
= 171
We know that, Tn = a + (n – 1)d = 171
3 + (n – 1)7 = 171
3 + 7n – 7 = 171
7n – 4 = 171
7n = 171 + 4
7n = 175
n = 175/7
n = 25
Hence, the given values n = 25 are not roots of the equation.
20. (i) How many two digit numbers are divisible by 3 ?
Solution:-
We have given that:
The two digits numbers divisible by 3 are, 12,15,18,21,24,…..,99.
The above numbers are A.P.
So, first number a = 12
Common difference d = 15 – 12 = 3
Then, last number is 99
We know that, Tn (last number) = a + (n – 1)d
99 = 12 + (n – 1)3
99 = 12 + 3n – 3
99 = 9 + 3n
99 – 9 = 3n
3n = 90
n = 90/3
n = 30
Hence, the given values n = 30 are not roots of the equation.
Therefore, 30 two digits number are divisible by 3.
(ii) Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
Solution:-
We have given that:
The natural numbers which are divisible by both 2 and 5 are 110,120,130,140,….,999
The above numbers are A.P.
So, first number a = 110
Common difference d = 120 – 110 = 10
Then, last number is 999
We know that,Tn (last number) = a + (n – 1)d
999 = 110 + (n – 1)10
999 = 110 + 10n – 10
999 = 100 + 10n
999 – 100 = 10n
10n = 888
n = 888/10
n = 88
Hence, the given values n=88 are not roots of the equation.
The number of natural numbers which are divisible by both 2 and 5 are 88.
(iii) How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?
Solution:-
The numbers which are lie between 10 and 300, when divisible by 4 leave a remainder 3 are 11,15,19,23,27,….299
The above numbers are A.P.
So, first number a = 11
Common difference d = 15 – 11 = 4
Then, last number is 299
We know that, Tn (last number) = a + (n – 1)d
299 = 11 + (n – 1)4
299 = 11 + 4n – 4
299 = 7 + 4n
299 – 7 = 4n
4n = 292
n = 292/4
n = 73
The total which are lie between 10 and 300, when divisible by 4 leave a remainder 3 are 73.
22. The sum of three numbers in A.P. is 3 and their product is – 35. Find the numbers.
Solution:-
We have given that:
The sum of three numbers in A.P.= 3
Given, Their product = -35
Let us assume the 3 numbers which are in A.P. are, a – d,a,a + d
Now adding 3 numbers = a – d + a + a + d = 3
3a = 3
a = 3/3
a = 1
From the question, product of 3 numbers is – 35
So, (a – d) × (a) × (a + d) = – 35
(1 – d) × (1) × (1 + d) = – 35
12 – d2 = – 35
d2 = 35 + 1
d2 = 36
d = √36
d = ±6
Therefore, the numbers are (a – d) = 1 – 6 = – 5
a = 1
(a + d) = 1 + 6 = 7
If d = – 6
The numbers are (a – d) = 1 – (-6) = 1 + 6 = 7
a = 1
(a + d) = 1 + (-6) = 1 – 6 = -5
Therefore, the numbers -5,1,7,… and 7,1,-5,… are in A.P.
23. The sum of three numbers in A.P. is 30 and the ratio of first number to the third number is 3∶ 7. Find the numbers.
Solution:-
We have given that:, sum of three numbers in A.P.= 30
The ratio of first number to the third number is 3: 7
Let us assume the 3 numbers which are in A.P. are, a – d,a,a + d
Now adding 3 numbers = a – d + a + a + d = 30
3a = 30
a = 30/3
a = 10
Given ratio 3∶ 7 = a – d∶ a + d
3/7 = (a – d)/(a + d)
(a – d)7 = 3(a + d)
7a – 7d = 3a + 3d
7a – 3a = 7d + 3d
4a = 10d
4(10) = 10d
40 = 10d
d = 40/10
d = 4
Therefore, the numbers are a – d = 10 – 4 = 6
a = 10
a + d = 10 + 4 = 14
So, 6,10,14, … are in A.P.
Therefore, the numbers are (a – d) = 11 – 9 = 2
a = 11
(a + d) = 11 + 9 = 20
If d = – 9
The numbers are (a – d) = 1 – (-9) = 11 + 9 = 20
a = 11
(a + d) = 11 + (-9) = 11 – 9 = 2
Therefore, the numbers 2,11,20,… and 20,11,2,… are in A.P.
25. Justify whether it is true to say that the following are the nth terms of an A.P.
(i) 2n – 3
(ii) n² + 1
Solution:-
(i) 2n – 3
We have given that:
nth term is 2n – 3
So, Tn = 2n – 3
Now, we start giving values, 1, 2, 3, … in the place of n, we get,
(2 × 1) – 3 = 2 – 3 = – 1
(2 × 2) – 3 = 4 – 3 = 1
(2 × 3) – 3 = 6 – 3 = 3
(2 × 4) – 3 = 8 – 3 = 5
From the above results, -1,1,3,5, …. Are in A.P.
So, first term a = – 1
Common difference d = 1 – (-1) = 1 + 1 = 2
= 3 – 2 = 2
(ii) n2 + 1
We have given that:
nth term is n2 + 1
So, Tn = n2 + 1
Now, we start giving values, 1,2,3, … in the place of n, we get,
12 + 1 = 1 + 1 = 2
22 + 1 = 4 + 1 = 5
32 + 1 = 9 + 1 = 10
42 + 1 = 16 + 1 = 17
From the above results,2,5,10,17, ….
So, first term a = 2
Common difference d = 5 – 2 = 3
= 10 – 5 = 5
The common difference is not same.
Therefore, 2,5,10,17,… are not in A.P.
Exercise 9.3
1. Find the sum of the following A.P.s :
(i) 2,7,12, … to 10 terms
Solution:-
We have given that:
From the question,
First term a = 2
Then, d = 7 – 2 = 5
12 – 7 = 5
So, common difference d = 5
n = 10
S10 = n/2(2a + (n – 1)d)
= 10/2 ((2 × 2) + (10 – 1)5)
= 5(4 + 45)
= 5(49)
= 245
(ii) 1/15,1/12,1/10, … to 11 terms
Solution:-
We have given that:
From the question,
First term a = 1/15
Then, d = 1/12 – 1/15
= (5 – 4)/60
= 1/60
So, common difference d = 1/60
n = 11
S11 = 11/2(2a + (n – 1)d)
= 11/2 ((2 × (1/15)) + (11 – 1)(1/60))
= 11/2 ((2/15) + (10/60))
= 11/2 (2/15 + 1/6)
= 11/2 (4 + 5)/30
= 11/2 (9/30)
= 11/2(3/10)
= 33/20
3. Find the sums given below :
(i) 34 + 32 + 30 + … + 10
Solution:-
We have given that:
From the question,
First term a = 34,
Difference d = 32 – 34 = -2
So, common difference d = – 2
Last term Tn = 10
We know that, Tn = a + (n – 1)d
10 = 34 + (n – 1)(-2)
-24 = – 2(n – 1)
-24 = – 2n + 2
2n = 24 + 2
2n = 26
n = 26/2
n = 13
Sn = n/2(a + 1)
= 13/2 (34 + 10)
= 13/2 (44)
= 13 (22)
= 286
Hence, the sum is 286
(ii) – 5 + ( – 8) + ( – 11) + … + ( – 230)
Solution:-
We have given that:
From the question,
First term a = -5,
Difference d = -8 – (-5) = -8 + 5 = -3
So, common difference d = – 3
Last term Tn = -230
We know that, Tn = a + (n – 1)d
-230 = -5 + (n – 1)(-3)
-230 = – 5 – 3n + 3
-230 = – 2 – 3n
3n = 230 -2
3n = 228
n = 228/3
n = 76
Therefore, Sn = n/2 (a + l)
= 76/2 (-5 + (-230))
= 38 (-5 – 230)
= 38 (235)
= – 8930
Hence, the sum is -8930
4. In an A.P. (with usual notations) :
(i) given a = 5,d = 3,an = 50, find n and Sn
Solution:-
We have given that:
From the question,
First term a = 5
Then common difference d = 3
an = 50,
We know that, 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
So, Sn = (n/2)(2a + (n – 1)d)
= (16/2) ((2 × 5) + (16 – 1) × 3)
= 8(10 + 45)
= 8(55)
= 440
Hence, the Sn is 440 and n=16
(ii) given a = 7,a13 = 35, find d and S13
Solution:-
We have given that:
From the question,
First term a = 7
a13 = 35,
We know that, a_n = a + (n – 1)d
35 = 7 + (13 – 1)d
35 = 7 + 13d – d
35 = 7 + 12d
12d = 35 – 7
12d = 28
d = 28/12 … [divide by 4]
d = 7/3
So, S13 = (n/2)(2a + (n – 1)d)
= (13/2) ((2 × 7) + ((13 – 1) × (7/3))
= (13/2) ((14 + (12 × 7/3))
= (13/2) (14 + 28)
= (13/2) (42)
= 13 × 21
= 273
Hence, d is 7/3 and S13 is 273
(iii) given d = 5,S9 = 75, find a and a_9.
Solution:-
We have given that:
From the question,
Common difference d = 5
S9 = 75
We know that, an = a + (n – 1)d
a9 = a + (9 – 1)5
a9 = a + 45 – 5
a9 = a + 40 … [equation (i)]
Then, S9 = (n/2) (2a + (n – 1)d)
75 = (9/2) (2a + (9 – 1)5)
75 = (9/2) (2a + (8)5)
(75 × 2)/9 = 2a + 40
150/9 = 2a + 40
2a = 150/9 – 40
2a = 50/3 – 40
2a = (50 – 120)/3
2a = -70/3
a = -70/(3 × 2)
a = – 35/3
Hence, a=-35/3
Now, substitute the value of a in equation (i),we get
a9 =a + 40
= -35/3 + 40
= (-35 + 120)/3
= 85/3
Hence, a9 = 85/3
(iv) given a = 8,an = 62,Sn = 210, find n and d
Solution:-
We have given that:
From the question,
First term a = 8,
an = 62 and Sn = 210
We know that, an = a + (n – 1)d
62 = 8 + (n – 1)d
(n – 1)d = 62 – 8
(n – 1)d = 54 … [equation (i)]
Then, S_n = (n/2) (2a + (n – 1)d)
210 = (n/2) ((2 × 8) + 54) … [from equation (i) (n – 1)d = 54]
210 = (n/2) (16 + 54)
420 = n(70)
n = 420/70
n = 6
Hence, the value of n is 6
Now, substitute the value of n in equation (i),
(n – 1)d = 54
(6 – 1)d = 54
5d = 54
d = 54/5
Therefore, d = 54/5
(v) given a = 3,n = 8,S = 192, find d.
Solution:-
We have given that:
First term a = 3
n = 8
S = 192
We know that, Sn = (n/2) (2a + (n – 1)d)
192 = (8/2) ((2 × 3) + (8 – 1)d)
192 = 4 (6 + 7d)
192/4 = 6 + 7d
48 = 6 + 7d
48 – 6 = 7d
42 = 7d
d = 42/7
d = 6
Hence, the given values d=6 are not roots of the equation.
Therefore, common difference d is 6.
5.
(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:-
We have given that:
First term a = 5
Last term = 45
Then, sum = 400
We know that, last term = a + (n – 1)d
45 = 5 + (n – 1)d
(n – 1)d = 45 – 5
(n – 1)d = 40 … [equation (i)]
So, Sn = (n/2) (2a + (n – 1)d)
400 = (n/2) ((2 × 5) + 40) … [from equation (i) (n – 1)d = 40]
800 = n(10 + 40)
800 = 50n
n = 800/50
n = 16
Hence, the given values n=16 are not roots of the equation.
(ii) The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.
Solution:-
We have given that:
First term a = 15
Therefore, sum of first n terms of an A.P. is given by,
Sn = (n/2) (2a + (n – 1)d)
S15 = (15/2)(2a + (15 – 1)d)
750 = (15/2) (2a + 14d)
(750 × 2)/15 = 2a + 14d
100 = 2a + 14d
Dividing both the side by 2 we get,
50 = a + 7d
Now, substitute the value a,
50 = 15 + 7d
7d = 50 – 15
7d = 35
d = 35/7
d = 5
So, 20th term a20 = a + 19d
= 15 + 19(5)
= 15 + 95
= 110
Therefore, its 20th term is 110
6. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:-
We have given that:
First term a = 17
Last term (l) = 350
Common difference d = 9
We know that, l = Tn = a + (n – 1)d
350 = 17 + (n – 1) × 9
350 – 17 = 9n – 9
333 + 9 = 9n
342 = 9n
n = 342/9
n = 38
So, Sn = (n/2) (2a + (n – 1)d)
= (38/2) ((2 × 17) + (38 – 1)d)
= 19(34 + (37 × 9))
= 19(34 + 333)
= 19 × 367
= 6973
Therefore, n = 38 and Sn = 6973
7. Solve for x∶ 1 + 4 + 7 + 10 + … + x = 287.
Solution:-
We have given that:
From the question,
First term a = 1
Difference d = 4 – 1 = 3
n = x
x = a = (n – 1)d
x – 1 = (n – 1)d
Sn = (n/2) (2a + (n – 1)d)
287 = (n/2) ((2 × 1)+ (n – 1)3)
= n (2 + 3n – 3)
574 = n(2 + 3n – 3)
574 = 2n + 3n2 – 3n
574 = – n + 3n2
3n2 – n – 574 = 0
3n2 – 42n + 41 – 574 = 0
3n(n – 14) + 41(n – 14) = 0
(n – 14) (3n + 41) = 0
If n – 14 = 0
n = 14
or 3n + 41 = 0
3n = -41
n = -41/3
We have to take positive number so n = 14
Then, = a + (n – 1)d
= 1 + (14 – 1) 3
= 1 + (13)3
= 1 + 39
= 40
Therefore, x = 40
Hence, the given values x = 40 are not roots of the equation.
8. (i) How many terms of the A.P.25,22,19, … are needed to give the sum 116 ? Also find the last term.
Solution:-
We have given that:
First term a = 25
Common difference d = 22 – 25 = – 3
Sum = 116
Sn = (n/2) (2a + (n – 1)d)
116 = (n/2) (2a + (n – 1)d)
By cross multiplication,
232 = n ((2 × 25) + (n – 1) (-3))
232 = n (50 – 3n + 3)
232 = n (53 – 3n)
232 = 53n – 3n2
3n2 – 53n + 232 = 0
32n2 – 24n – 29n + 232 = 0
3n (n – 8) – 29 (n – 8) = 0
(n – 8) (3n – 29) = 0
If n – 8 = 0
n = 8
or 3n – 29 = 0
3n = 29
n = 29/3
not possible to take fraction,
So, n = 8
Then, T = a + (n – 1)d
= 25 +(8 – 1) (-3)
= 25 + 7 (-3)
= 25 – 21
= 4
(ii) How many terms of the A.P.24,21,18, … must be taken so that the sum is 78 ? Explain the double answer.
Solution:-
We have given that:
First term a = 24
Common difference d = 21 – 24 = – 3
Sum = 78
Sn = (n/2) (2a + (n – 1)d)
78 = (n/2) (2a + (n – 1)d)
By cross multiplication,
156 = n ((2 × 24) + (n – 1) (-3))
156 = n (48 – 3n + 3)
156 = n (51 – 3n)
156 = 51n – 3n2
3n2 – 51n + 156 = 0
3n2 – 12n – 39n + 156 = 0
3n (n – 4) – 39 (n – 4) = 0
(n – 4) (3n – 39) = 0
If n – 4 = 0
n = 4
or 3n – 39 = 0
3n = 39
n = 39/3
n = 13
Now we have to consider both values
So, n = 4
Then, T = a + (n – 1)d
= 24 +(4 – 1) (-3)
= 24 + 3 (-3)
= 24 – 9
= 15
n = 13
Then, T = a + (n – 1)d
= 24 +(13 – 1) (-3)
= 24 + 12 (-3)
= 24 – 36
= -12
So, (12 + 9 + 6 + 3 + 0 + (-3)+ (-6) + (-9) + (-12)) = 0
Hence, the sum of 5th term to 13th term = 0
9. Find the sum of first 22 terms, of an A.P. in which d = 7 and a22 is 149.
Solution:-
We have given that:
Common difference d = 7
a22 = 149
n = 22
we know that,
a22 = (n – 1)d
149 = a + (22 – 1)7
149 = a + (22)7
149 = a + 147
a = 149 – 147
a = 2
So, S22 = (n/2) (2a + (n – 1)d)
= (22/2) ((2 × 2) + (22 – 1)7)
= 11(4 + (21)7)
= 11 (4 + 147)
= 11 (151)
= 1661
(ii) If the third term of an A.P. is 1 and 6th term is – 11, find the sum of its first 32 terms.
Solution:-
We have given that:
T3 = 1,T6 = – 11 and n = 32
As we know that,
T3 = a + 2d = 1 … [equation (i)]
T6 = a + 5d = – 11 … [equation (ii)]
Now, subtracting equation (ii) from equation (i), we get,
(a + 2d) – (a + 5d) = 1 – (-11)
a + 2d – a – 5d = 1 + 11
-3d = 12
d = -12/3
d = -4
Now, substitute value of d in equation (i), we get
a + 2d = 1
a + 2(-4) = 1
a – 8 = 1
a = 8 + 1
a = 9
S32 = (n/2) (2a + (n – 1)d)
= (32/2) (2(9) + (32 – 1)(-4))
= 16 (18 + (31)(-4))
= 16 (18 – 124)
= 16 (-106)
= – 1696
Therefore, the sum of its first 32 terms is – 1696.
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:-
As we have given that:
S6 = 36
S16 = 256
We know that,
Sn = (n/2) (2a + (n – 1)d)
S6 = (6/2) (2a + (6 – 1)d) = 36
3 (2a + 5d) = 36
Divide both the side by 3,
2a + 5d = 12 … [equation (i)]
Now, S16 = (16/2) (2a + (16 – 1)d) = 256
8 (2a + 15d) = 256
Divide both the side by 8,
2a + 15d = 32 … [equation (ii)]
Then, subtract equation (ii) from equation (i) we get,
(2a + 5d) – (2a + 15d) = 12 – 32
2a + 5d – 2a – 15d = -20
-10d = -20
d = -20/-10
d = 2
Substitute the value of d in equation (i) to find a, we get
2a + 5d = 12
2a + 5(2) = 12
2a + 10 = 12
2a = 12 – 10
2a = 2
a = 2/2
a = 1
So, S10 = (n/2) (2a + (n – 1)d)
= (10/2) ((2 × 1) + (10 – 1)2)
= 5 (2 + 18)
= 5 (20)
= 100
Therefore, the sum of first 10 terms is 100.
12. Show that a1 ,a2 ,a3 , … form an A.P. where an is defined as an = 3 + 4n. Also find the sum of first 15 terms.
Solution:-
As we have given that:
nth term is 3 + 4n
So, an = 3 + 4n
Now, we start giving values, 1, 2, 3, … in the place of n, we get,
a1 = 3 + (4 × 1) = 3 + 4 = 7
a2 = 3 + (4 × 2) = 3 + 8 = 11
a3 = 3 + (4 × 3) = 3 + 12 = 15
a4 = 3 + (4 × 4) = 3 + 16 = 19
So, The numbers are 7,11,15,19, ….
Then, first term a = 7, common difference d = 11 – 7 = 4
We know that,
S15 = (n/2) (2a + (n – 1)d)
= (15/2) ((2 × 7) + (15 – 1) × 4)
= (15/2) (14 + (14 × 4))
= (15/2) (14 + 56)
= (15/2) × 70
= 525
Therefore, the sum of first 15 terms is 525.
Therefore, the S20 is -780.
Hence, the given values S20 =-780 are not roots of the equation.
(ii) Find the common difference of an A.P. whose first term is 5 and the sum of first four terms is half the sum of next four terms.
Solution:-
We have given that:
First term a = 5
And also it is given that, the sum of first four terms is half the sum of next four terms,
By cross multiplication,
2(4a + 6d) = (4a + 22d)
2 ((4 × 5) + 6d) = ((4 × 5) + 22d) … [given a = 5]
2(20 + 6d) = (20 + 22d)
40 + 12d = 20 + 22d
40 – 20 = 22d – 12d
20 = 10d
d = 20/10
d = 2
Hence, the given values d=2 are not roots of the equation.
Therefore, the common difference d is 2.
Exercise 9.4
(ii) Find the next term of the list of numbers 3/16,-3/8,¾,-3/2,…
Solution:-
According to the question
We have given that:
First term a = 3/16
Then, r = (-3/8) ÷ (3/16)
r = (-3/8) × (16/3)
r = (-3 × 16)/(8 × 3)
r = (-1 × 2)/ (1 × 1)
r = -2
Therefore, next term = -3/2 × (-2) = 6/2 = 3
(vii) Find the 6th term from the end of the list of numbers 3,– 6,12,– 24,…,12288.
Solution:-
We have given that:
Last term = 12288
First term a = 3,
Then, r = (-6) ÷ (3)
r = (-6) × (1/3)
r = (-6 × 1)/(1 × 3)
r = (-2 × 1)/(1 × 1)
r = -2
3. Which term of the G.P.
(i) 2,2√2,4,… is 128?
Solution:-
We have given that:
Last term = 128
First term a = 2,
Then, r = (2√2) ÷ (2)
r = (2√2)/2
r = √2
8. Find the value of x such that,
(i) -2/7,x,-7/2 are three consecutive terms of a G.P.
Solution:-
According to the question,
x2 = -2/7 × -7/2
x2 = 1
x = ± 1
Therefore, x = 1 or x = – 1
Hence, the given values x = 1 are not roots of the equation.
(ii) x + 9,x – 6 and 4 are three consecutive terms of a G.P.
Solution:-
According to the question,
(x – 6)2 = (x + 9) × 4
x2 – 12x + 36 = 4x + 36
x2 – 12x – 4x + 36 – 36 = 0
x2 – 16x = 0
x(x – 16) = 0
Either let us take x – 16 = 16
Or x = 0
So, x = 0,16
(iii) x,x + 3,x + 9 are first three terms of a G.P. Find the value of x.
Solution:-
We have given that:
From the question,
(x + 3)2 = x(x + 9)
x2 + 6x + 9 = x2 + 9x
9 = 9x – 6x
9 = 3x
X = 9/3
X = 3
Hence, the given values x = 3 are not roots of the equation.
11. If a,b,c are in G.P., then show that a²,b²,c² are also in G.P.
Solution:-
We have given that:
a,b,c are in G.P.
We have to show that a²,b²,c² are also in G.P
Then,
b² = ac … (i)
Therefore, a²,b²,c² will be in G.P.
if (b²)² = a² x c²
(ac)² = a²c² … [from eq. (i)]
a²c² = a²c²
Therefore, it is proved that a²,b²,c² are also in G.P.
12. If a,b,c are in A.P., then show that 3a,3b,3c are in G.P.
Solution:-
We have given that:
a,b and c are in A.P.
So, 2b = a + c
We have to show that 3a,3b,3c are also in G.P.
If (3b)2 = 3a × 3c
32b = 3a + c
Now, comparing the results we get,
2b = a + c
Therefore, 3a,3b,3c are in G.P
14. If a,a2+ 2 and a3 + 10 are in G.P., then find the values(s) of a.
Solution:-
We have given that:
From the question,
(a2 + 2)2 = a(a3 + 10)
a4 + 4 = a4 + 10a
4a2 – 10a + 4 = 0
2a2 – 5a + 2 = 0
2a2 – a – 4a + 2 = 0
a(2a – 1) – 2(2a – 1) = 0
(2a – 1) (a – 2) = 0
Then, 2a – 1 = 0
a = ½
a – 2 = 0
a = 2
Hence, the given values a=2 are not roots of the equation.
Therefore, a = 2 or a= ½
18. 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:-
According to the question,
We have given that
The sum of first three terms of a G.P. is 39/10
The product of first three terms of a G.P. is 1
Let us assume that a be the first term and ‘r’ be the common ratio,
And also assume that, three terms of the G.P. is a/r,a,ar,
The sum of three terms = (a/r) + a + ar = 39/10
Take out ‘a’ as common then, we get
a(1/r + 1 + r) = 39/10 … [equation (i)]
Now, product of three terms = (a/r) × a × ar = 1
20. Three numbers form an increasing G.P. If the middle term is doubled, then the new numbers are in A.P. Find the common ratio of the G.P.
Solution:-
We have given that:
Three numbers form an increasing G.P.
Let us assume the three numbers a/r,a, are
Then, double the middle term we get,
a/r,2a,ar will be in A.P.
So, 2(2a) = a/r + ar
4a = a(1/r + r)
4 = 1/r + r
By cross multiplication,
4r = 1 + r2
r2 – 4r + 1 = 0
21. Three numbers whose sum is 70 are in GP. If each of the extremes is multiplied by 4 and the mean by 5, the numbers will be in A.P. Find the numbers.
Solution:-
We have given that:
Three numbers are in G.P. whose sum is 70.
Let us assume the three number be a/r,a,ar
Then, sum = (a/r) + a + ar = 70
Take out a as common,
a((1/r) + 1 + r) = 70 … [equation (i)]
Now, multiplying the extremes by 4 and mean by 5,
Then, (a/r) × 4 = 4a/r
(a × 5) = 5a
(ar × 4) = 4ar
4a/r,5a,4ar
Therefore, these are in A.P.
So, 2(5a) = (4a/r) + 4ar
10a = 4a(1/r) + r
Divide both the side by 2 we get,
(10/2)a = (4/2)a (1/r) + r
5a = 2a((1/r) + r)
5r = 2 + 2r2
2r2 – 5r + 2 = 0
2r2 – r – 4r + 2 = 0
r(2r – 1) – 2(2r – 1) = 0
(2r – 1) (r – 2) = 0
So, 2r – 1 = 0
2r = 1
r = ½
r – 2 = 0
r = 2
Now substitute the value r in equation (i), we get
a((1/2) + 1 + 2) = 70
a ((½) + 3) = 70
a ((1 + 6)/2) = 70
a (7/2) = 70
a = 70 × (2/7)
a = 10 × 2
a = 20
Then,
r = 2,a = 20
= (a/r),a,ar
= (20/2),20,(20 × 2)
= 10,20,40
Then,
r = ½,a = 20
= (a/r),a,ar
= (20/½),20,(20 × ½)
= (20 × 2),20,10
= 40,20,10
Therefore, the numbers are 40, 20, 10
Substitute the value of a
Then, 2b = a + a
2b = 2a
Therefore, b = a … [equation (iv)]
By comparing equation (iii) and equation (iv),
a = b = c
3. Find the sum of the series 81 – 27 + 9 … – 1/27
Solution: -
According to the question:
First term a = 81
r = -27/81
= -1/3
Last term l = -1/27
Sn = (a – lr)/(l – r)
= [81 + ((1/27) × (-1/3)]/[1 + (1/3)]
= [(81 – (1/81))]/(4/3)
= (6561 – 1)/[81 × (4/3)]
= (6560 × 3)/(81 × 4)
= 1640/27
4. The nth term of a G.P. is 128 and the sum of its n terms is 255. If its common ratio is 2, then find its first term.
Solution:-
According to the question:
5. If the sum of first six terms of any G.P. is equal to 9 times the sum of the first three terms, then find the common ratio of the G.P.
Solution:-
We have given that:
the sum of first six terms of any G.P. is equal to 9 times the sum of the first three terms,
(iii) 0.5 + 0.55 + 0.555 + …
Solution: -
We have given that:
Consider the given numbers 0.5 + 0.55 + 0.555 + … n terms
Take out 5 as common we get,
= 5(0.1 + 0.11 + 0.111 + … n terms)
= 5/9 (0.9 + 0.99 + 0.999 + … n terms)
= 5/9 ((1 – 0.1) + (1 – 0.01) + (1 – 0.001) + … n terms)
= 5/9 (1 + 1 + 1 + … n terms – (0.1 + 0.01 + 0.001 + ….n terms))
Chapter – Test
1. Write the first four terms of the A.P. when its first term is – 5 and the common difference is – 3.
Solution:-
We have given that:
First term a = – 5
Common difference d = -3
Then the first four terms are = – 5 + (-3) = -5 – 3 = – 8
-8 + (-3) = – 8 – 3 = -11
– 11 + (-3) = – 11 – 3 = – 14
Therefore, first four terms are -5,-8,-11 and -14.
2. Verify that each of the following lists of numbers is an A.P., and the write its next three terms:
(i) 0,¼,½,¾,…
Solution:-
We have given that:
First term a = 0
Common difference = ¼ – 0 = ¼
So, next three numbers are ¾ + ¼ = 4/4 = 1
1 + ¼ = (4 + 1)/4 = 5/4
5/4 + ¼ = 6/4 = 3/2
Therefore, the next three term are 1,5/4 and 3/2.
(ii) 5,14/3,13/3,4,…
Solution:-
We have given that:
First term a = 5
Common difference = 14/3 – 5 = (14 – 15)/3 = -1/3
So, next three numbers are 4 + (-1/3) = (12 – 1)/3 = 11/3
11/3 + (-1/3) = (11 – 1)/3 = 10/3
10/3 + (-1/3) = (10 – 1)/3 = 9/3 = 3
Therefore, the next three term are 11/3,10/3 and 3.
4. Show that the list of numbers 9,12,15,18,… form an A.P. Find its 16th term and the n^th.
Solution:-
We have given that:
The first term a = 9
Then, difference d = 12 – 9 = 3
15 – 12 = 3
18 – 15 = 3
Therefore, common difference d = 3
From the formula, an = a + (n – 1)d
Tn = a + (n – 1)d
= 9 + (n – 1)3
= 9 + 3n – 3
= 6 + 3n
So, T16 = a + (n – 1)d
= 9 + (16 – 1)3
= 9 + (15)(3)
= 9 + 45
= 54
(3d + 16d)/2 = 19
(19/2)d = 19
d = (19 × 2)/19
d = 2
To find out the value of a substitute the value of d in equation (i)
a + 8d = 19
a + (8 × 2) = 19
a + 16 = 19
a = 19 – 16
a = 3
Hence, the given values a = 3 are not roots of the equation.
Therefore, A.P.is 3,5,7,9,…
9. If the 3rd and the 9th terms of an A.P. are 4 and – 8 respectively, then which term of this A.P. is zero?
Solution:-
We have given that:
a3 = 4
a9 = – 8
We know that, a3 = a + 2d = 4 … [equation (i)]
a9 = a + 8d = -8 … [equation (ii)]
Now, subtracting equation (i) from equation (ii)
(a + 8d) – (a + 2d) = -8 – 4
a + 8d – a – 2d = -12
6d = -12
d = -12/6
d = -2
To find out the value of a substitute the value of d in equation (i)
a + 2d = 4
a + (2 × (-2)) = 4
a – 4 = 4
a = 4 + 4
a = 8
let us assume nth term be zero, then
a + (n – 1)d = 0
8 + (n – 1)(-2) = 0
-2n + 2 = -8
-2n = -8 – 2
-2n = -10
n = -10/-2
n = 5
Hence, the given values n = 5 are not roots of the equation.
Therefore, 0 will be the fifth term.
10. Which term of the list of numbers 5,2,– 1,– 4,… is – 55?
Solution:-
We have given that:
First term a = 5
nth term = -55
Common difference d = 2 – 5 = – 3
We know that, a_n = a + (n – 1)d
– 55 = 5 + (n – 1)(-3)
-55 – 5 = – 3n + 3
-60 – 3 = -3n
-63 = -3n
n = -63/-3
n = 21
Hence, the given values n = 21 are not roots of the equation.
Therefore, -55 is the 21st term.
83 < 3n
83/3 < n
28 < n
Therefore, 28th is the first negative term.
13. If the pth term of an A.P. is q and the qth term is p, show that its nth term is (p + q – n)
Solution:-
We have given that:
pth term = q
qth term = p
We have to show that, nth term is (p + q – n)
We know that, an = a + (n – 1)d
So, pth term = a + (p -1)d = q … [equation (i)]
qth term = a + (q – 1)d = p … [equation (ii)]
Now subtracting equation (ii) from equation (i), we get
q – p = (a + (p – 1)d) – (a + (q – 1)d)
q – p = (a + pd – d) – (a +qd – d)
q – p = a + pd – d – a – qd + d
q – p = pd – qd
q – p = d (p – q)
d = (q – p)/(p – q)
d = -(p – q)/(p – q)
d = – 1
Substitute the value of d in equation (i), we get
a + (p -1)(-1) = q
a – p + 1= q
a = q + p – 1
Then, nth term = a + (n – 1)d
= (p + q – 1) + (n – 1) (- 1)
= (p + q – 1) – n + 1
= p + q – 1 – n + 1
= p + q – n
900 = 9n
n = 900/9
n = 100
Hence, the given values n = 100 are not roots of the equation.
Therefore, there are 100 three digits numbers.
15. The sum of three numbers in A.P. is – 3 and the product is 8. Find the numbers.
Solution:-
We have given that:
The sum of three numbers in A.P.= – 3
The product of three numbers in A.P.= 8
Let us assume the 3 numbers which are in A.P. are, a – d,a,a + d
Now adding 3 numbers = a – d + a + a + d = – 3
3a = -3
a = -3/3
a = -1
From the question, product of 3 numbers is – 35
16. The angles of a quadrilateral are in A.P. If the greatest angle is double of the smallest angle, find all the four angles.
Solution:-
We have given that:
The angles of a quadrilateral are in A.P.
Greatest angle is double of the smallest angle
Let us assume the greatest angle of the quadrilateral is a + 3d,
Then, the other angles are a + d,a – d,a – 3d
So, a – 3d is the smallest
Therefore, a + 3d = 2(a – 3d)
a + 3d = 2a – 6d
6d + 3d = 2a – a
9d = a … [equation (i)]
19. Find the sum : 18 + 15½ + 13 + … + (-49½)
Solution:-
We have given that:
First term a = 18
Common difference d = 15½ – 18
= 31/2 – 18
= (31 – 36)/2
= -5/2
Last term = -49½ = -99/2
We know that, a_n = a + (n – 1)d
-99/2 = 18 + (n – 1)(-5/2)
(-99/2) – (18/1) = (n – 1)(-5/2)
(-99 – 36)/2 = (-5/2)(n – 1)
(-135/2) = (-5/2) (n – 1)
(-135/2) × (-2/5) = n – 1
-135/-5 = n – 1
27 = n – 1
n = 27 + 1
n = 28
Then, Sn = (n/2) [2a + (n – 1)d]
(ii) Solve the equation 2 + 5 + 8 + … + x = 155
Solution:-
We have given that:
First term a = 2
Last term = x
Common difference d = 5 – 2 = 3
Then, sum of the terms = 155
L = a + (n – 1)d
x = 2 + (n – 1)3
x = 2 + 3n – 3
x = 3n – 1 … [equation (i)]
We know that, Sn = (n/2) [2a + (n – 1)d]
155 = (n/2) [(2 × 2) + (n – 1) × 3]
155 × 2 = n[4 + 3n – 3]
310 = n(3n + 1)
310 = 3n2 + n
3n2 + n – 310 = 0
3n2 – 30n + 31n – 310 = 0
3n(n – 10) + 31(n – 10) = 0
(n – 10) (3n + 31) = 0
So, n – 10 = 0
n = 10
or 3n + 31 = 0
n = -31/3
negative is not possible.
Therefore, n = 10
Now, substitute the value of n in equation (i),
x = 3n – 1
= (3 × 10) – 1
= 30 – 1
= 29
21. If the third term of an A.P. is 5 and the ratio of its 6th term to the 10th term is 7∶ 13, then find the sum of first 20 terms of this A.P.
Solution:-
We have given that:
The third term of an A.P.a_3 = 5
The ratio of its 6th term to the 10th term a6 : a10 = 7∶ 13
We know that, an = a + (n – 1)d
a3 = a + (3 – 1)d = 5
= a + 2d = 5 … [equation (i)]
Then, a6 /a10 = 7/13
(a + 5d)/(a + 9d) = 7/13
By cross multiplication we get,
13(a + 5d) = 7(a + 9d)
13a + 75d = 7a + 63d
13a – 7a + 65d – 63d = 0
6a + 2d = 0
Divide by 2 on both side we get,
3a + d = 0
d = -3a … [equation (ii)]
Substitute the value of d in equation (i),
a + 2(-3a) = 5
a – 6a = 5
-5a = 5
a = -5/5
a = -1
Now substitute the value of a in equation (ii),
d = -3(-1)
d = 3
Then, sum of first 20 terms,
= (n/2) [2a + (n – 1)d]
= (20/2)[(2 × (-3)) + (2- – 1)3]
= 10[-2 + 57]
= 10 × 55
= 550
22. In an A.P., the first term is 2 and the last term is 29. If the sum of the terms is 155, then find the common difference of the A.P.
Solution:-
We have given that:
First term a = 2
Last term = 29
The sum of terms = 155
We know that, last term = an = a + (n – 1)d
29 = 2 + (n – 1)d
29 – 2 = d(n – 1)
27 = d(n – 1) … (i)
Then, Sn = (n/2)[2a + (n – 1)d]
155 = (n/2)[(2 ×2) + 27]
155 = (n/2)[4 + 27]
155 = (31/2)n
n = (155 × 2)/31
n = 10
d(n – 1) = 27
d(10 – 1) = 27
d(9) = 27
d = 27/9
d = 3
Hence, the common difference, d is 3
23. The sum of first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term.
Solution:-
We have given that:
First term a = 10
The sum of first 14 terms of an A.P.= 1505
25th term = ?
We know that, Sn = (n/2) [2a + (n – 1)d]
S14 = (n/2) [2a + (n – 1)d]
1505 = (14/2) [(2 × 10) + (14 – 1)d]
1505 = 7[20 + 13d]
1505/7 = 20 + 13d
215 = 20 + 13d
13d = 215 – 20
13d = 195
d = 195/13
d = 15
Then, an = a + (n – 1)d
a25 = 10 + (25 – 1)(15)
= 10 + (24)15
= 10 + 360
= 370
= – 150 – 550
= -700
S20 = (20/2) [2a + 19d]
-700 = 10(2a + 19d)
-700 = 20a + 190d … [equation (ii)]
Now, multiplying equation (i) by 2 we get,
20a + 90d = – 300 … [equation (iii)]
Subtract equation (iii) from equation (ii),
(20a + 190d) – (20a + 90d) = -700 – (-300)
20a + 190d – 20a – 90d = -700 + 300
100d = -400
d = -400/100
d = -4
Substitute the value of d in equation (i) we get,
10a + 45(-4) = -150
10a – 180 = – 150
10a = -150 + 180
10a = 30
a = 30/10
a = 3
a2 = 3 + (-4) = 3 – 4 = -1
a3 = -1 + (-4) = -1 – 4 = -5
a3 = -5 + (-4) = -5 – 4 = -9
Therefore, A.P. is 3,-1,-5,-9,…
r = 3
To find out a, consider ar3 = 54
a(3)3 = 54
a = 54/27
a = 2
Therefore, a = 2,r = 3
Hence, the given values a = 2,r = 3 are not roots of the equation.
So, G.P.is 2,6,18,54,…
31. Find the sum of first n terms of the series: 3 + 33 + 333 + …
Solution:-
We have given that:
Consider the given numbers 3 + 33 + 333 + … n terms
Take out 3 as common we get,
= 3 (1 + 11 + 111 + … n terms)
= 3/9 (9 + 99 + 999 + … n terms)
= 3/9 ((10 – 1) + (100 – 1) + (1000 – 1) + … n terms)
= 3/9 (10 + 100 + 100 + … n terms – (1 + 1 + 1 + ….n terms))
