PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions
PSEB 10th Class Maths Solutions Chapter 5 Arithmetic Progressions
Ex 5.1
Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why ?
(i) The taxi fare after each km when the fare is 15 for the first km and 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs 150 for the first metre and rises by 50 for each msubsequent metre.
(iv) The amount of money in the account every year when 10000 is deposited at compound interest at 8% per annum.
Solution:
(i) Let Tn denotes the taxi fare in nth km.
According to question,
T1 = 15 km;
T2 = 15 + 8 = 23;
T3 = 23 + 8 = 31
Now, T3 – T2 = 31 – 23 = 8
T2 – T1 = 23 – 15 = 8
Here, T3 – T2 = T2 – T1 = 8
∴ given situation form an AP.
(ii) Let Tn denotes amount of air present in a cylinder.
According to question,
(iii) Let Tn denotes cost of digging a well for the nth metre,
According to question,
T1 = ₹ 150; T2 = (150 + 50) = ₹ 200;
T3 = ₹ (200 + 5o) = 250 and so on
Now, T3 – T2 = ₹ (250 – 200) = 50
T2 – T1 = ₹ (200 – 150) = 50
Here, T3 – T2 = T2– T1 = 50
∴ given situation form an A.P.
(iv) Let Tn denotes amount of money in the nth year.
According to question
Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(1) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = 1/2
(w) a = -1.25, d = -0.25
Solution:
(i) Given that first term = a = 10
and common difference = d = 10
∴ T1 = a = 10;
T2 = a + d = 10 + 10 = 20;
T3 = a + 2d
= 10 + 2 × 10 = 10 + 20 = 30;
T4 = a + 3d = 10 + 3 × 10
= 10 + 30 = 40
Hence, first four terms of an A.P. are 10, 20, 30, 40………….
(ii) Given that first term = a = -2
and common iifference = d = 0
∴ T1 = a = -2;
T2 = a + d = -2 + 0 = -2
T3 = a + 2d = -2 + 2 × 0 = -2
T4 = a + 3d = -2 + 3 × 0 = -2
Hence, first four terms of an A.P. are -2, -2, -2, -2,…………….
(iii) Given that first term = a = 4
and common difference = d = -3
∴ T1 = a = 4;
T2= a + d = 4 – 3 = 1
T3 = a + 2d = 4 + 2(-3) = 4 – 6 = -2
T4 = a + 3d = 4 + 3(-3) = 4 – 9 = -5
Hence, first four terms of an A.P. are 4, 1, -2, -5, ……….
(v) Given that first term = a = – 1.25
and common difference = d = – 0.25
∴ T1 = a = – 1.25;
T2 = a + d = – 1.25 – 0.25 = -1.50
T3 = a + 2d = – 1.25 + 2(- 0.25)
= – 1.25 – 0.50 = – 1.75
T4 = a + 3d = – 1.25 + 3(- 0.25)
= – 1.25 – 0.75 = – 2
Hence, first four terms of an A.P. are – 1.25, – 1.50, – 1.75, – 2, ……………..
Question 3.
For the following APs, wilte the first term and the common difference:
(i) 3, 1, -1, -3, …………
(ii) 5, -1, 3, 7, ………….
(iii) 1/3,5/3,9/3,13/3, …………..
(iv) 0.6, 1.7, 2.8, 3.9, ………..
Solution:
(i) Given A.P., is 3, 1, -1, -3, ………
Here T1 = 3, T2 = 1,
T3 = -1, T4 = -3
First term = T1 = 3
Now, T2 – T1 = 1 – 3 = – 2
T3 – T2 = – 1 – 1 = -2
T4 – T3 = -3 + 1 = -2
∴ T2 – T1 = T3 – T2 = T4 – T3 = – 2
Hence, common difference = – 2 and first term = 3.
(ii) Given A.P. is – 5, – 1, 3, 7, ………….
Here T1 = – 5, T2 = – 1,
T3 = 3, T4 = 7
First term T1 = -5
Now, T2 – T1 = -1 + 5 = 4
T3– T2 = 3 + 1 = 4
T4 – T3 = 7 – 3 = 4
∴ T2 – T1 = T3 – T2 = T4 – T3 = 4
Hence, common difference = 4 and first term = – 5.
(iv) Given A.P. is 0.6, 1.7, 2.8, 3.9,…
Here, T1 = 0.6, T2 = 1.7, T3 = 2.8, T4 = 3.9
First term = T1 = 0.6
Now, T2 – T1 = 1.7 – 0.6 = 1.1
T3 – T2 = 2.8 – 1.7 = 1.1
T4 – T3 = 3.9 – 2.8 = 1.1
Hence, common difference = 1.1 and first term = 0.6.
Question 4.
WhIch of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16
(ii) 2, 5/2, 3, 7/2, ………
(iii) – 1.2, – 3.2, – 5.2, – 7.2, ………….
(iv) – 10, – 6, – 2, 2, ………….
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, ……….
(vi) 0.2, 0.22, 0.222, 0.2222, ………….
(vii) 0, -4, -8, -12, …………..
(viii) −1/2, −1/2, −1/2, −1/2, ………..
(ix) 1, 3, 9, 27 …………….
(x) a, 2a, 3a, 4a, ………………
(xi) a, a2, a3, a4, ……………….
(xii) √2, √8, √18, √32, …………
(xiii) √3, √6, √9, √12, ……………..
(xiv) 12, 32, 52, 72, ………..
(xv) 12, 52, 72, 73, ………….
Solution:
(i) Given terms are 2, 4, 8, 16 ………………
Here T1 = 2, T2 = 4, T3 = 8, T4 = 16
T2 – T1 = 4 – 2 = 2
T3 – T2 = 8 – 4 = 4
∵ T2 – T1 ≠ T3 – T2
Hence, given terms do not form an A.P.
(iii) Given terms are – 1.2, – 3.2, – 5.2, – 7.2, …………
Here T1 = – 1.2, T2 = – 3.2,
T3 = – 5.2, T4 = – 7.2
T2 – T1 = – 3.2 + 1.2 = – 2
T3 – T2 = – 5.2 + 3.2 = – 2
T 4 – T3 = – 7.2 + 5.2 = – 2
∵ T2 – T1 = T3 – T2 = T4 – T3 = – 2
∴ Common difference = d = – 2
Now, T5 = a + 4d = – 1.2 + 4(-2) = – 1.2 – 8 = – 9.2
T6 = a + 5d = – 1.2 + 5(-2) = – 1.2 – 10 = – 11.2
T7 = a + 6d = – 1.2 + 6(-2) = -1.2 – 12 = – 13.2
(iv) Given terms are – 10, – 6, – 2, 2, ………..
Here T1 = – 10,T2 = – 6
T3 = – 2, T4=2 .
T2 – T1 = – 6 + 10 = 4
T3 – T2 = – 2 + 6 =4
T4 – T3 = 2 + 2 = 4
∵ T2 – T1=T3 – T2 = T4 – T3 = 4 .
∴ Common difference = d = 4
Now, T5 = a + 4d = – 10 + 4(4) = – 10 + 16 = 6
T6 = a + 5d = – 10 + 5(4) = – 10 + 20 = 10
T7 = a + 6d = – 10 + 6(4) = – 10 + 24 = 14.
(v) Given terms are 3, 3 + √2, 3 + 2√2, 3 + 3√2, …………
Here T1 = 3, T2 = 3 + √2,
T3 = 3 + 2√2, T4= 3 + 3√2
T2 – T1 = 3 + √2 – 3 = √2
T3 – T2 = 3 + 2√2 – (3 + √2)
= 3 + 2√2 – 3 – √2 = √2
T4 – T3 = 3 + 3√2 – (3 + 2√2)
= 3 + 3√2 – 3 – 2√2 = √2
∵ T2 -T1 = T3 – T2 = T4 – T3 = √2
∴ Common difference = d = √2
Now, T5 = a + 4d = 3 + 4(√2) = 3 + 4√2
T6 = a + 5d = 3 + 5√2
T7 = a + 6d = 3 + 6√2
(vi) Given terms are 0.2, 0.22, 0.222, 0.2222, …………..
Here Here T1 = 0.2, T2 = 0.22,
T3 = 0.222, T4 = 0.2222.
T2 – T1 = 0.22 – 0.2 = 0.02
T3 – T2 = 0.222 – 0.22 = 0.002
∵ T2 – T1 ≠ T3 – T2
∴ given terms do not form an A.P.
(vii) Given terms are 0, -4, -8, -12
Here T1 = 0, T2 = -4,
T3 = -8, T4 = -12
T2 – T1 = – 4 – 0 = -4
T3 – T2= – 8 + 4 = -4
T4 – T3= – 12 + 8 = -4.
T2 – T1 = T3 – T2 = T4 – T3
∴ Common difference = d = -4
Now, T5= a + 4d = 0 + 4(-4) = -16
T6 = a + 5d = 0 + 5(-4) = -20
T7 = a + 6d = 0 + 6(-4) = -24.
(ix) Given terms are 1, 3, 9, 27
T1 = 1, T2 = 3, T3 = 9, T4 = 27
T2 – T1 = 3 1 = 2
T3 – T2 = 9 – 3 = 6.
∵ T2 – T1 ≠ T3 – T2
∴ Given terms do not form an A.P.
(x) Given terms are a, 2a, 3a, 4a, …
T1 = a, T2 = 2a, T3 = 3a, T4 = 4a
T2 – T1 = 2a – a = a
T3 – T2 = 3a – 2a = a
T4 – T3 = 4a – 3a = a
∵ T2 – T1 = T3 – T2 = T4 – T3 = a
∴ Common difference = d = a
Now T5 = a + 4d = a + 4(a) = a + 4a = 5a
T6 = a + 5d = a + 5a = 6a
T7 = a + 6d = a + 6a = 7a
(xi) Given terms are a, a2, a3, a4, …………
T1 = a, T2 = a2, T3 = a3, T4 = a4
T2 – T1 = a2 – a
T3 – T2 = a3 – a2
∵ T2 – T1 ≠ T3 – T2
∴ Given terms do not form an A.P.
(xii) Given terms are √2, √8, √18, √32, …………
T1 = √2, T2 = √8, T3 = √18, T4 = √32
or T1 = √2, T2 = 2√2 T3 = 3√2, T4 = 4√2
T2 – T1 = 2√2 – √2 = √2
T3 – T = 3√2 – 2√2 = √2
T4 – T3 = 4√2 – 3√2 = √2
∵ T2 – T1 = T3 – T2 = T4 – T3= √2
∴ Common difference = d = √2
Now, T5 = a + 4d = √2 + 4√2 = 5√2
T6 = a + 5d = √2 + 5√2 = 6√2
T7 = a + 6d = √2 + 6√2 = 7√2
(xiii) Given terms are √3, √6, √9, √12, ……………..
T1 = √3, T2= √6, T3= √9, T4= √12
or T1 = √3, T2 = √6, T3 = 3, T4 = 2√3
T4 – T1 = √6 – √3
T3 – T2 = 3 – √6
∵ T2 – T1 ≠ T3 – T2
∴Given terms do not form an A.P.
(xiv) Given terms are 12, 32, 52, 72, ………..
T1 = 12, T2 = 32, T3 = 52, T4 = 72
or T1 = 1, T2 = 9, T3 = 25, T4 = 49
T4 – T1 = 9 – 1 = 8
T3 – T2 = 25 – 9 = 16
∵ T2 – T1 ≠ T3 – T2
∴ Given terms do not form an A.P.
(xv) Given terms are 12, 52, 72, 73
T1 = 12, T2 = 52, T3 = 72, T4 = 73
or T1 = 1, T2 = 25, T3 = 49, T4 = 73
T2 – T1 = 25 – 1 = 24
T3 – T2 =49 – 24= 24
T4 – T3 = 73 – 49 = 24
∵ T2 – T1 = T3 – T2 = T4 – T3 = 24
∴ Common difference = d = 24
T5 = a + 4d = 1 + 4(24) = 1 + 96 = 97
T6 = a + 5d = 1 + 5(24) = 1 + 120 = 121
T7 = a + 6d = 1 +6(24) = 1 + 144 = 145
Ex 5.2
Question 1
Fill in the blanks in the following table, given that a is the first term, d the common difference and a the nth term of the
AP:
Solution:
(i) Here a = 7, d = 3, n = 8
∵ an = a + (n – 1)d
∴ a8 = 7 + (8 – 1)3
= 7 + 21 = 28.
(ii) Here a = – 18, n = 10, an = 0
∵ an = a + (n – 1)d
∴ a10 = – 18 + (10 – 1)d
or 0 = – 18 + 9d .
or 9d = 18
d = 18/2 = 2.
(iii) Here d = – 3, n = 18, an = – 5
∵ an = a + (n – 1)d
∴ a18 = a + (18 – 1)(-3)
or -5 = a – 51
or a = – 5 + 51 = 46.
(iv) Here a = – 18.9, d = 2.5 an = 3.6
∵ an = a + (n – 1)d
∴ 3.6 = – 18.9 + (n – 1) 2.5
or 3.6 + 18.9 = (n – 1) 2.5
or (n – 1) 2.5 = 22.5
or n – 1 = 22.5/2.5
or n = 9 + 1 = 10.
(v) Here a = 3.5, d = 0, n = 105
∵ an = a + (n – 1) d
∴ an = 3.5 + (105 – 1) 0
an = 3.5 + 0 = 3.5.
Question 2.
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
Solution:
(i) Given A.P. is 10, 7, 4 ……………
T1 = 10, T2 = 7, T3 = 4
T2 – T1 = 7 – 10 = – 3
T3 – T2 = 4 – 7 = – 3
∵ T2 – T1 = T3 – T2 = – 3 = d(say)
∵ Tn = a + (n – 1) d
Now, T30 = 10 + (30 – 1)(-3)
= 10 – 87 = – 77
∴ Correct choice is (C).
Question 3.
In the following APs, find the missing terms in the boxes:
(i) 2, 
, 26
(ii), 13, , 3
(iii) 5,, , 9
(iv) – 4, , , , , 6
(v) , 38, , , , – 22
Solution:
Let a be the first term and ‘d’ be the common difference of given A.P.
(i) Here T1 = a = 2
and T3 = a + 2d = 26
or 2 + 2d = 26
or 2d = 26 – 2 = 24
or d = 12
∴ Missing term = T2 = a + d = 2 + 12 = 14.
(ii) Here, T2 = a + d = 13 ……………(1)
and T4 = a + 3d = 3 …………….(3)
Now, (2) – (1) gives
Substitute this value of d in (1), we get
a – 5 = 13
a = 13 + 5 = 18.
∴ T1 = a = 18
T3 = a + 2d = 18 + 2(-5)
= 18 – 10 = 8.
(iv) Here T1 = a = —
T6 = a + 5d = 6
or -4 + 5d = 6
or 5d = 6 + 4
or 5d = 10
or d = 10/2 = 2
Now, T2 = a + d = -4 + 2 = -2
T3 = a + 2d = – 4 + 2(2)
= – 4 + 4 = 0
T4 = a + 3d = – 4 + 3(2)
= – 4 + 6 = 2
T5 = a + 4d = – 4 + 4(2)
= – 4 + 8 = 4
(v) Here T2 = a + d = 38 ………….(1)
and T6 = a + 5d = -22 ……………(2)
Now, (2) – (1) gives
Substitute this value of d in (1), we get
a + (-15) = 38
a = 38 + 15 = 53
∴ T1 = a = 53
T3 = a + 2d = 53 + 2(-15) = 53 – 30 = 23.
T4 = a + 3d = 53 + 3(-15) = 53 – 45 = 8
T5 = a + 4d = 53 + 4(-15) = 53 – 60 = – 7.
Question 4
Which term of the A.P. 3, 8, 13, 18, …………… is 78?
Solution:
Given A.P. is 3, 8, 13, 18, ………….
T1 = 3, T2 = 8, T3 = 13, T4 = 18
T2 – T1 =8 – 3=5
T3 – T2= 13 – 8=5
T2 – T1 = T3 – T,= 5 = d (say)
Using, Tn = a + (n – I) d
or 78 = 3 + (n – 1) 5
or 5(n – 1) = 78 – 3 = 75
or n – 1 = 15
or n = 15 + 1 = 16
Hence, 16th term of given AP. is 78.
Question 5.
Find the number of terms in each of the following APs:
(i) 7, 13, 19,…, 205
(ii) 18, 15 1/2, 13, ………….., – 47
Solution:
(i) Given A.P. is 7, 13, 19, …………..
T1 = 7, T2 = 13, T3 = 19
T2 – T1 = 13 – 7 = 6
T3 – T2 = 19 – 13 = 6
T2 – T1 = T3 – T2 = 6 = d(say)
Using formula, Tn = a + (n – 1) d
205 = 7 + (n – 1) 6
or (n – 1) 6 = 205 – 7 = 198
or (n – 1) = 198/6
or n – 1 = 33
n = 33 + 1 = 34
Hence, 34th term of an AP. is 205.
Question 6.
Is – 150 a term of 11, 8, 5, 2….? why?
Solution:
Given sequence is 11, 8, 5, 2, ………..
T1 = 11, T2 = 8, T3 = 5, T4 = 2
T2 – T1 = 8 – 11 = – 3
T3 – T2 = 5 – 8 = – 3
T4 – T3 = 2 – 5 = – 3
T2 – T1 = T3 – T2 = T4 – T3 = – 3 = d (say).
Let – 150 be any term of given A.P.
then Tn = – 150
a+(n – 1)d = – 150
or 11 +(n – 1)(- 3) = – 150
or (n – 1)( – 3) = – 150 – 11 = – 161
Question 7.
Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Solution:
Let ‘a’ and 4d’ be the first term and common difference of given A.P.
Given that T11 = 38
a +(11 – 1) d = 38
[∵ Tn = a + (n – 1) d]
a + 10 d = 38
and T16 = 73
a + (16 – 1) d = 73
[∵ Tn = a + (n – 1) d]
a + 15 d = 73
Now, (2) – (1) gives
Substitute this value of d in (1), we get
a + 10 (7) = 38
or a + 70 = 38
or a = 38 – 70 = – 32
Now, T31 = a + (31 – 1) d
= – 32 + 30 (7) = – 32 + 210 = 178.
Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 291h term.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Given that, T3 = 12
a + (3 – 1) d = 12
∵ Tn = a + (n – 1) d
or a + 2d = 12 ………………(1)
and Last term = T50 = 106
a + (50 – 1) d = 106
∵ Tn = a + (n – 1) d
a + 49 d = 106 ……………(2)
Now, (2) – (1) gives
Substitute this value of d in (1), we get
a + 2(2) = 12
or a + 4 = 12
or a + 12 – 4 = 8
Now, T29 = a + (29 – 1) d
= 8 + 28 (2) = 8 + 56 = 64.
Question 9.
If the 3rd and 9th tenus of an A.P. are 4 and – 8 respectively, which term of this A.P. is zero.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given AP.
Given that: T3 = 4
a + (3 – 1) d = 4
∵ Tn = a + (n – 1) d
a + 2d = 4 …………..(1)
and T9 = – 8
a + (9 – 1) d = 8
and T9 = – 8
a + (9 – 1)d = 8
∵ Tn = a + (n – 1) d
or a + 8d = – 8
Now, (2) – (1) gives
Substitute this value of d in (1), we get
a + 2(- 2) = 4
or a – 4 = 4
or a = 4 + 4 = 8
Now, Tn = 0 (Given)
a + (n – 1) d = 0
or 8 + (n – 1)(- 2)=0
or -2 (n – 1) = – 8
or n – 1 = 4
or n = 4 + 1 = 5
Hence, 5th term of an AP. is zero.
Question 10.
The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Now, T17 = a + (17 – 1) d = a + 16 d
and T10 = a + (10 – 1) d = a + 9 d
According to question
T17 – T10 = 7
(a + 16 d) – (a + 9 d) = 7
or a + 16 d – a – 9 d = 7
7 d = 7
or d = 7/7 = 1
Hence, common difference is 1.
Question 11.
Which term of the A.P. 3, 15, 27, 39, …………. will be 132 more than its 54th term?
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Given A.P. is 3, 15, 27, 39, …
T1 = 3, T2 = 15, T3 = 27, T4 = 39
T2 – T1 = 15 – 3 = 12
T3 – T2 = 27 – 15 = 12
:. d=T2 – T1 = T3 – T2 =12
Now, T54 = a + (54 – 1) d
= 3 + 53 (12) = 3 + 636 = 639
According to question
T = T54 + 132
a + (n – 1)d = 639 + 132
3 + (n – 1)(12) = 771
(n – 1) 12 = 771 – 3 = 768
or n – 1 = 768/12 = 64
or n = 64 + 1 = 65
Hence, 65th term of A.P. is 132 more than its 54th term.
Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of first AP.
Also, ‘A’ and ‘d’ be the first term and common difference of second A.P.
According to question
[T100 of second A.P.] – [T100 of first A.P.] = 100
or[A +(100 – 1)d] – [a +(100 – 1)d] = 100
or A + 99d – a – 99d = 100
or A – a = 100
Now, [T1000 of second A.P.] – [T1000 of first A.P.]
= [A + (1000 – 1) d) – (a + (1000 – 1) d]
= A + 999 d – a – 999 d
= A – a = 100 [Using (I)].
Question 13.
How many three-digits numbers are divisible by 7?
Solution:
Three digits numbers which divisible by 7 are 105, 112, 119 , 994
Here a = T1 = 105, T2 = 112, T3 = 119 and Tn = 994
T2 – T1 = 112 – 105=7
T3 – T2 = 119 – 112=7
∴ d = T2 – T1 = T3 – T2 = 7
Given that, Tn = 994
a + (n – 1) d = 994
or 105 + (n – 1) 7 = 994
or (n – 1) 7 = 994 – 105
or (n – 1) 7 = 889
or n – 1 = 889/7 = 127
or n = 127 + 1 = 128.
Hence, 128 terms of three digit number are divisible by 7.
Question 14.
How many multiples of 411e between 10 and 250?
Solution:
Multiples of 4 lie between 10 and 250 are 12, 16, 20, 24, … 248
Here a = T1 = 12, T2 = 16, T3 = 20 and Tn = 248
T2 – T1 = 16 – 12 = 4
T3 – T2 = 20 – 16 = 4
∴ d = T2 – T1 = T3 – T2 = 4
Given that, Tn = 248
a + (n – 1) d = 248
or 12 + (n – 1)4 = 248
or 4(n – 1) = 248 – 12 = 236
or n – 1 = 236/4 = 59
or n = 59 + 1 = 60
Hence, there are 60 terms which are multiples of 4 lies between 10 and 250.
Question 15.
For what value of n, are the n terms of two A.P.s 63, 65, 67, …………. and 3, 10, 17, …………….. equal?
Solution:
Given A.P. is 63, 65, 67, ……………..
Here a = T1 = 63, T2 = 65, T3 = 67
T2 – T1 = 65 – 63 = 2
T3 – T2 = 67 – 65 = 2
∴ d = T2 – T1 = T3 – T2 = 2
and second A.P. is 3, 10, 17, …
Here a = T1 = 3, T2 = 10, T3 = 17
T2 – T1 = 10 – 3 = 7
T3 – T2 = 17 – 10 = 7
According to question.
[nth term of first A.P.] = [nth term of second A.P.]
63 + (n – 1)2 = 3 + (n – 1) 7
or 63 + 2n – 2 = 3 + 7n – 7
or 61 + 2n = 7n – 4
or 2n – 7n = – 4 – 61
– 5n = – 65
n = 65/5 = 13.
Question 16.
Determine the AP. whose third term is 16 and 7 term exceeds the by 12.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Given that T3 = 16
a + (3 – 1) d = 16
a + 2d = 16
According to question
T7 – T5 = 12
[a + (7 – 1) d] – [a + (5 -1) d] = 12
a + 6 d – a – 44 = 12
2d = 12
d = 12/2 = 6
Substitute this value of d in (1), we get
a + 2(6) = 16
a = 16 – 12 = 4 .
Hence, given A.P. are 4, 10, 16, 22, 28, ………….
Question 17.
Find the 20th term from the last term of the AP: 3, 8, 13, ………., 253.
Solution:
Given A.P. is 3, 8, 13, …………., 253
Here, a = T1 = 3, T2 = 8, T3 =13 and Tn = 253
T2 – T1 = 8 – 3 = 5
T3 – T2 = 13 – 8 = 5
∴ d = T2 – T1 = T3 – T1 = 5
Now, Tn = 253
3 + (n – 1)5 = 253
∵ Tn = a + (n – 1) d
(n – 1) 5 = 250
n-1 = 250/5 = 50
n = 50 + 1 = 51
20th term from the end of AP = (Total number of terms) – 20 + 1
= 51 – 20 + 1 = 32nd term
∴ 20th term from the end of AP
= 32nd term from the starting
= 3 + (32 – 1)5
∵ Tn = a + (n – 1)d
= 3 + 31 × 5
= 3 + 155 = 158.
Question 18.
The sum of the 4th and 8th term of an AP is 24 and the sum of the 6’ and 10th terms is 44. FInd the first three terms of the A.P.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
According to 1st condition
T4 + T8 = 24
a + (4 – 1) d + a + (8 – 1) d = 24
∵ Tn = a + (n – 1) d
or 2a + 3d + 7d = 24
2a + 10d = 24
a + 5d = 12 …………(1)
According to 2nd condition
T6 + T10 = 44
a + (6 – 1) d + a +(10 – 1) d = 44
∵ Tn = a + (n – 1) d
2a + 5d + 9d = 44
2a + 14d = 44
a + 7d = 22
Now (2) – (1) gives
Substitute this value of d in (I). we get
a + 5(5) = 12
a + 25 = 12
a = 12 – 25 = -13
T1 = a = -13
T2 = a + d = 13 + 5 = -8
T2 = a + 2d = – 13 + 2(5) = – 13 + 10 = -3
Hence, given A.P. is -13, -8, -3, ……………
Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
Solution:
Subba Rao’s starting salary = ₹ 5000
Annual increment = ₹ 200
Let ‘n’ denotes number of years.
∴ first term = a = ₹ 5000
Common diflerence = d = ₹ 200
and Tn = ₹ 7000
5000 + (n – 1) 200 = 7000
∵ Tn = a + (n – 1) d
(n – 1) 200 = 7000 – 5000
or (n – 1) 200 = 2000
or n – 1 = 2000/200 = 10
or n = 10 + 1 = 11
Now, in case of year the sequence are 1995. 1996, 1997, 1998, ……………
Here a = 1995, d = 1 and n = 11
Let Tn denotes the required year.
∴ Tn = 1995 + (11 – 1) 1
= 1995 + 10 = 2005
Hence, in 2005, Subba Rao’s salary becomes 7000.
Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly saving by ₹ 1.75. If in the nth week, her weekly saving becomes ₹ 20.75, find n.
Solution:
Amount saved in first week = ₹ 5
Increment in saving every week = ₹ 1.75
It is clear that, it form an A.P. whose terms are
T1 = 5, d = 1.75
∴ T2 = 5 + 1.75 = 6.75
T3 = 6.75 + 1.75 = 8.50
Also. Tn = 20. 75 (Given)
5 + (n – 1) 1.75 = 20.75
∵ Tn = a + (n – 1) d
or (n – 1) 1.75 = 20.75 – 5
or (n – 1) 1.75 = 15.75
or (n – 1) = 1575/100×100/175
or n – 1 = 9
or n = 9 + 1 = 10
Hence, in 10th week, Ramkali’s saving becomes ₹ 20.75.
Ex 5.3
Question 1.
Find the sum of the following APs:
(i) 2, 7, 12, … to 10 terms.
(ii) – 37, – 33, – 29, ………….. to 12 terms.
(iii) 0.6, 1.7, 2.8, … to 100 terms.
(iv) 1/5, 1/12, 1/10, ………… to 11 terms.
Solution:
(i) Given AP. is 2, 7, 12, …
Here a = 2, d = 7 – 2 = 5 and n = 10
Sn = n/2 [2a + (n – 1) d]
∴ S10 = 10/2 [2 × 2 + (10 – 1)5]
= 5 [4 + 45] = 245.
(ii) Given A.P. is – 37, – 33, – 29…
Here a = -37, d = – 33 + 37 = 4 and n = 12
Sn = n/2 [2a + (n – 1) d]
∴ S12 = [2(-37) + (12 – 1)4]
= 6 [- 74 + 44] = – 180.
Question 3.
In an AP:
(i) given a = 5, d = 3, an = 50 find n and Sn.
(ii) given a = 7. a13 = 35, find d and S13.
(iii) given a12 = 37, d = 3. find a and S12.
(iv) given a3 = 15, S10 = 125, find d and a10.
(y) given d = 5, S9 = 75, find a and a9.
(vi) given a = 2, d = 8, Sn = 90, find n and an.
(vii) given a = 8, an = 62, Sn = 210, find n and d.
(viii)given an = 4, d = 2, Sn = -14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms. Find a.
Solution:
(i) Given a = 5, d = 3, an = 50
an = 50
a + (n – 1) d = 50
or 5 + (n – 1) 3 = 50
or 3 (n – 1) = 50 – 5 = 45
or n – 1 = 45/3 = 15
or n = 15 + 1 = 16
Now, Sn = latex]\frac{n}{2}[/latex] [a + l]
= latex]\frac{16}{2}[/latex] [5 + 50] = 8 × 55 = 440.
(iv)Given a3 = 15, S10 = 125
∵ a3 = 15
a + (n – 1) d = 15
a + (3 – 1) d = 15
or a + 2d = 15 …………….(1)
∵ S10 = 125
∵ [Sn = latex]\frac{n}{2}[/latex] [2a + (n – 1)d]
10/2 [2a + (10 – 1) d] = 125
or 5[2a + 9d] = 125
Note this
or 2a + 9d = 125/5 = 25
or 2a + 9d = 25 …………(2)
From (1), a = 15 – 24 …………..(3)
Substitute this value of (a) in (2i, we c1
2(15 – 2d) + 9d = 25
or 30 – 4d + 9d = 25
5d = 25 – 30
or d = −5/5 = -1
Substitute this value of d in (3), we get
a = 15 – 2(- 1)
a = 15 + 2 = 17
Now, a10 = 17 + (10 – 1)(- 1)
∵ Tn = a + (n – 1) d = 17 – 9 = 8.
(viii) Given an = 4, d = 2, Sn = – 14
∵ an = 4
a + (n – 1) d = 4
or a + (n – 1)2 = 4
or a + 2n – 2 = 4
or a = 6 – 2n ……………(1)
and Sn = – 14
n/2 [a + an] = – 14
or n/2 [6 – 2n +4] = – 14 (Using (1))
or n/2 [10 – 2n] = – 14
or 5n – n2 + 14 = 0
or n2 – 5n – 14 = 0
S = – 5
P = 1 × – 14 = – 14
or n2 – 7n + 2n – 14 = 0
or n(n – 7) + 2 (n – 7) = 0
or (n – 7) (n + 2) = 0
either n – 7 = 0
or n + 2 = 0
n = 7 or n = -2
∵ n cannot be negative so reject n = – 2
∴ n = 7
Substitute this value of n in (1), we get
a = 6 – 2 × 7
a = 6 – 14 = – 8.
Question 6.
The first and last terms of an AP are 17 and 350 respectively. 1f the common difference is 9, how many terms are there
and what is their sum?
Solution:
Given that a = T1 = 17;
l = an = 350 and d = 9
∵ l = an = 350
a + (n – 1) d = 350
17 + (n – 1) 9 = 350
or 9 (n – 1) = 350 – 17 = 333
or n – 1 = 333/9 = 37
n = 37 + 1 = 38
Now, S38 = n/2 [a + l]
= 38/2 (17 + 350]
= 19 × 367 = 6973.
Hence, sum of 38 terms of given A.P. arc 6973.
Question 7.
Find the sum of first 22 terms of an AP. in which d = 7 and 22nd term is 149.
Solution:
Given that d = 7; T22 = 149 and n = 22
∵ T22 = 149
a + (n – 1) d = 149
or a + (22 – 1) 7 = 149
or a + 147 = 149
or a = 149 – 147 = 2
Now, S22 = [a + T22]
= 22/2 [2+ 149] = 11 × 151 = 1661
Hence, sum of first 22 temis of given A.P. is 1661.
Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Let ‘a’ and ‘d’ be fïrst term and common difference
Given that T2 = 14; T3 = 18 and n = 51
∵ T2 = 14
a + (n – 1) d = 14
a + (2 – 1)d = 14
or a + d = 14
a = 14 – d …………….(1)
and T3 = 18 (Given)
a + (n – 1) d = 18
a + (3 – 1) d = 18
or a + 2d = 18
or 14 – d + 2d = 18
or d = 18 – 14 = 4
or d = 4
Substitute this value of d in (1), we get
Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
According to 1st condition
Question 10.
Show that a1, a2, ……., an, … form an AP where is defined as below.
(i) an = 3 + 4n
(ii) an = 9 – 5n
Also find the sum of the first 15 terms In each case.
Solution:
(i) Given that an = 3 + 4n ………..( 1)
Putting the different values of n in(1), we get
a1 = 3 + 4(1) = 7;
a2 = 3 + 4 (2) = 11
a3 = 3 + 4 (3) = 15, …………
Now, a2 – a1 = 11 – 7
and a3 – a2 = 15 – 11 = 4
a2 – a1 = 11 – 7 = 4
and a3 – a2 = 4 = d(say)
∵ given sequence form an A.P.
Here a = 7, d = 4 and n = 15
∴ S15 = n/2 [2a + (n – 1)d]
= 15/2 [2(7) + (15 – 1) 4]
= 15/2 [14 + 56] = 15/2 × 70
= 15 × 35 = 525.
(ii) Given that an = 9 – 5n ……………….(1)
Putting the different values of n is (1), we get
a1 = 9 – 5(1) = 4;
a2 = 9 – 5(2) = -1;
a3 = 9 – 5(3) = -6.
Now, a2 – a1 = – 1 – 4 = – 5
and a3 – a2 = – 6 + 1 = – 5
∵ a – a1 = a3 – a2 = – 5 = d (say)
∴ given sequence form an A.P.
Here a = 4, d = – 5 and n = 15
∴ S15 = n/2 [2a + (n – 1) d]
= 15/2 [2(4) + (15 – 1) (-5)]
= 15/2 [8 – 70]
= 15/2 (-62) = – 465
Question 11.
If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1) ? What is the sum of two terms? What is the second term ? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
Given that, sum of n terms of an A.P. are
Sn = 4n – n2
Putting n = 1 in (1), we get
S1 = 4(1) – (1)2 = 4 – 1
S1 = 3
∴ a = T1 = S1 = 3
Putting n = 2, in (1), we get
S2 = 4(2) – (2)2 = 8 – 4
S2 = 4
or T1 + T2 = 4
or 3 + T2 = 4
or T2 = 4 – 3 = 1
Putting n = 3 in (1), we get
S3 =4(3) – (3)2 = 12 – 9
S3 = 3
or S2 + T3 = 3
or 4+ T3 = 3
or T3 = 3 – 4 = – 1
Now, d = T2 – T1
d = 1 – 3 = -2
∴ T10 = a + (n – 1) d
= 3 + (10 – 1) (- 2)
T10 = 3 – 18 = – 15
and Tn = a + (n – 1) d
= 3 + (n – 1) (- 2)
= 3 – 2n + 2
Tn = 5 – 2n.
Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
Positive integers divisible by 6 are 6, 12, 18, 24, 30, 36 42, …
Here a = T1 = 6, T2 = 12, T3 = 18, T4 = 24
T2 – T1 = 12 – 6 = 6
T3 – T2 = 18 – 12 = 6
T4 – T3 = 24 – 18 = 6
∵ T2 – T1 = T3 – T2 = T4 – T3 = 6 = d (say)
Using formula, Sn = n/2 [2a + (n – 1) d]
S40 = 40/2 [2(6) + (40 – 1)6].
= 20 [12 + 234]
= 20 (246) = 4920
Hence, sum of first 40 positive integers divisible by 6 is 4920.
Question 13.
Find the sum of first 15 multiples of 8.
Solution:
Multiples of 8 are 8, 16, 24, 32, 40, 48, …………..
Here a = T1 = 8; T2 = 16; T3 = 24 ; T4 = 32
T2 – T1 = 16 – 8= 8
T3 – T2 = 24 – 16=8
T2 – T1 = T3 – T2 = 8 = d(say)
̪sing formula. Sn = [2a + (n Р1) d}
S15 = 15/2 [2(8) + (15 – 1) 8]
= 15/2 [16 + 112]
= 15/2 × 128 = 960
Hence, sum of first 15 multiples of 8 is 960.
Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
Odd numbers between 0 and 50 are 1, 3, 5, 7, 9, …………, 49
Here a = T1 = 1; T2 = 3; T3 = 5 ; T4 = 7
and l = Tn = 49
T2 – T1 = 3 – 1 = 2
T3 – T2 = 5 – 3 = 2
∵ T2 – T1 = T3 – T2 = 2 = d (say)
Also, l = Tn = 49
a + (n – 1) d = 49
1 + (n – 1) 2 = 49
or 2 (n – 1) = 49 – 1 = 48
n – 1 = 48/2 = 24
n = 24 + 1 = 25.
Using formula, Sn = n/2 [2a + (n – 1) d]
S25 = 25/2 [2(1) + (25 – 1)2]
= 25/2 [2 + 48]
= 25/2 × 50 = 625
Hence, sum of the odd numbers between 0 and 50 are 625.
Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for
the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day, How much money the contractor has to pay as penalty, if he hasdelayed the work by 30 days?
Solution:
Penalty (cost) for delay of one, two, third day are ₹ 200, ₹ 250, ₹ 300
Now, penalty increase with next day with a difference of ₹ 50.
∴ Required A.P. are ₹ 200, ₹ 250, ₹ 300, ₹ 350, …
Here a = T1 = 200; d = ₹ 50 and n = 30
Amount of penalty gives after 30 days
= S30
= n/2 [2a + (n – 1) d]
= 30/2 [2(2oo) + (30 – 1) 50)
= 15 [400 + 1450]
= 15 (1850) = 27750
Hence, ₹ 27350 pay as penalty by the contractor if he has delayed the work 30 days.
Question 16.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If
each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let amount of prize given to 1st student = ₹ x
Amount of prize given to 2nd student = ₹ (x – 20)
Amount of prize given to 3rd student = ₹ [x – 20 – 20] = ₹ (x – 40)
and so on.
∴ Required sequence are ₹ x, ₹ (x – 20), ₹ (x – 40), … which form on A.P. with
a = ₹ x, d = – 20 and n = 7
Using formula, Sn = n/2 [2a + (n – 1) d]
S7 = 7/2 [2(x) + (7 – 1) (- 20)]
S7 = 7/2 [2x – 120] = 7 (x – 60)
According to question,
7 (x – 60) = 700
x – 60 = 7 = 100
x – 60 = 7 = 100
x = 100 + 60
x = 160
Hence, 7 prizes are ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, ₹ 40.
Question 17.
In a school, student thought of planting trees in and around the school to reduce air pollution. It was decided that
number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g. – a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
Number of trees planted by three sections of class I = 3 × 1 = 3
Number of trees planted by three sections of class II = 3 × 2 = 6
Number of trees planted by three sections of class III = 3 × 3 = 9
…………………………………………………………………..
…………………………………………………………………..
……………………………………………………………………
Number of trees planted by three sections of class XII = 3 × 12 = 36
:. Required A.P. are 3, 6, 9, …………., 36
Here a = T1 = 3; T2 = 6; T3 = 9
and l = Tn = 36; n = 12
d = T2 – T1 = 6 – 3 = 3
Total number of trees planted by students
= S12
= n/2 [a + l]
= 12/2 [3+ 36]
= 6 × 39 = 234
Hence, 234 trees will be planted by students to reduce air pollution.
Question 18.
A spiral is made up of successive semicircics, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …. as shown in Fig. What is the total length of such a spiral made up of thirteen consecutive semicircies? (Take Ï€ = 22/7)
[Hint: Length of successive semicircies is ‘l1’ ‘l2’ ‘l3‘ ‘l4‘ … wIth centres at A, B, A, B, …, respectively.]
Hence, total length of a spiral made up of thirteen consecutive semi circies is 143 cm
Question 19.
200 logs are stacked in the following manner : 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig). In how many rows are the 200 logs placed and how many logs are in the top row?
Solution:
Number of logs in the bottom (1st row) = 20
Number of logs in the 2nd row = 19
Number of log in the 3rd row = 18 and so on.
∴ Number of logs in the each steps form an
Here a = T1 = 20; T2 = 19; T3 = 18…
d = T2 – T1 = 19 – 20 = – 1
Let S, denotes the number logs.
Using formula, Sn = n/2 [2a + (n – 1) d]
= n/2 [2 (20) + (n – 1) (-1)]
∴ Sn = n/2 [40 – n + 1]
= n/2 [41 – n]
According to question,
n/2 [41 – n] = 200
or 41 – n2 = 400
or – n2 + 41n – 400 = 0
or n2 – 41n + 400 = 0
S = – 41, P = 400
or n2 – 16n – 25n + 400 = 0
or n (n – 16) – 25 (n – 16)=0
or (n – 16) (n – 25) = 0
Either n – 16 = 0 or n – 25 = 0
Either n = 16 or n = 25
∴ n = 16, 25.
Case I:
When n = 25
T25 = a + (n – 1) d
= 20 + (25 – 1)(- 1)
= 20 – 24 = – 4;
which is impossible
∴ n = 25 rejected.
Case II. When n = 16
T16 = a + (n – 1) d
= 20 + (16 – 1)(- 1)
= 20 – 15 = 5
Hence, there are 16 row and 5 logs are in the top row.
Question 20.
In a potato race a bucket ¡s placed at the starting point which Ls 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig.)
Each competitor starts from the bucket, picks up the nearest potato, runs back with It, drops It in the bucket, runs back to pick up the next potato, runs to the bucket to drop it In, and the continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
(Hint : To pick up the first potato and second potato, the distance run is [2 × 5 + 2 × (5 + 3)] m)
Solution:
Distance covered to pick up the Ist potato = 2(5) m = 10 m
Distance between successive potato = 3 m
distance covered to pick up the 2nd potato = 2(5 + 3) m = 16 m
Distance covered to pick up the 3rd potato = 2 (5 + 3 + 3) m = 22 m
and this process go on. h is clear that this situation becomes an A.P. as 10 m, 16 m, 22 m, 28 m, …
Here a = T1 = 10; T2 = 16; T3 = 22, …
d = T2 – T1 = 16 – 10 = 6 and n = 10
∴ Total distance the competitor has to run = S10
= n/2 [2a + (n – 1) d]
= 10/2 [2(10) + (10 – 1) 6]
= 5 [20 + 54]
= 5 × 74 = 370
Hence, 370 m is the total distance run by a competitor.
Ex 5.4
Question 1.
Which term of the A.P. 121, 117, 113, …………. is its first negative term?
Solution:
Given A.P is 121, 117, 113, …
Here a = T1 = 121 ;T2 = 117; T3 = 113
d = T2 – T1 = 117 – 121 = – 4
Using formula, Tn = a + (n – 1) d
Tn = 121 + (n – 1) (- 4)
= 121 – 4n + 4
= 125 – 4n.
According to question :—
Tn < 0
or 125 – 4n < 0
or 125 < 4n or 4n > 125.
or n > 125/4
or n > 31 1/4.
But n must be integer, for first negative term.
∴ n = 32.
Hence, 32nd term be the first negative term of given A.P.
Question 2.
The sum of the third and the seventh term of an A.P. is 6 and their product ¡s 8. Find the sum of first sixteer
terms of an A.P.
Solution:
Let ‘a’ and ‘d’ be the first term and common diftèrence of given A.P.
According to 1st condition
T3 + T7 = 6
[a + (3 – 1)d] + [a + (7 – 1) d] = 6
∵ [Tn = a + (n – 1) d]
or a + 2d + a + 6d = 6
or 2a + 8d = 6
or a + 4d = 3 …………….(1)
According to 2nd condition
T3 (T7) = 8
[a + (3 – 1) d] [a + (7 – 1)d] = 8
∵ [Tn = a + (n – 1) d]
or (a + 2d) (a + 6d) = 8
or [3 – 4d + 2d] [3 – 4d + 6d] = 8
[Using (1), a = 3 – 4d]
or (3 – 2d) (3 + 2d) = 8
or 9 – 4d2 = 8
or 4d2 = 98
or d2 = 1/4
d = ± 1/2
Question 3.
A ladder has rungs 25 cm apart (see fig.) The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2 latex]\frac{1}{2}[/latex] m apart, what is the length of the wood required for the rungs?
Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it and find this value of x.
[Hint: Sx – 1 = S49 – S1]
Solution:
Let ‘x’ denotes the number of any house.
Here a = T1 = 1 ;d = 1
According to question,
