## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A

Other Exercises

Question 1.
Which of the following sequences are in arithmetic progression?
(i) 2, 6, 10, 14, …….
(ii) 15, 12, 9, 6,…….
(iii) 5, 9, 12, 18, ……
(iv) $$\frac { 1 }{ 2 } ,\frac { 1 }{ 3 } ,\frac { 1 }{ 4 } ,\frac { 1 }{ 5 }$$,…..
Solution:
(i) 2, 6, 10, 14,
Here, d = 6 – 2 = 4
10 – 6 = 4
14 – 10 = 4
∴ In each case (d) is same.
∴It is an arithmetic progression.

∴d is not the same
∴It is not an arithmetic progression

Question 2.
The nth term of a sequence is (2n – 3), find its fifteenth term.
Solution:
Tn = 2n – 3
T15 = 2 x 15 – 3
= 30 – 3
= 27

Question 3.
If the pth term of an A.P. is (2p + 3); find the A.P.
Solution:
Tp = 2p + 3
T1 = 2 x 1 + 3 = 2 + 3 = 5
T2 = 2 x 2 + 3 = 4 + 3 = 7
T3 = 2 x 3 + 3 = 6 + 3 = 9
∴A.P. is 5, 7, 9,…..

Question 4.
Find the 24th term of the sequence : 12, 10, 8, 6,…….
Solution:
A.P. is 12, 10, 8, 6,……
Here a = 12, d = 10 – 12 = – 2
Tn = a + (n – 1)d
T24 = 12 + (24 – 1) x ( – 2)
= 12 + 23 x ( – 2)
= 12 – 46
= – 34

Question 5.
Find the 30th term of the sequence :
$$\\ \frac { 1 }{ 2 }$$, 1, $$\\ \frac { 3 }{ 2 }$$,….
Solution:
$$\\ \frac { 1 }{ 2 }$$, 1, $$\\ \frac { 3 }{ 2 }$$,….
Here a = $$\\ \frac { 1 }{ 2 }$$, d = $$1- \frac { 1 }{ 2 }$$
$$\\ \frac { 1 }{ 2 }$$

Question 6.
Find the 100th term of the sequence :
√3, 2√3, 3√3,….
Solution:
√3, 2√3, 3√3,….
Here a = √3, d = 2√3 – √3 = √3

Question 7.
Find the 50th term of the sequence :
$$\\ \frac { 1 }{ n }$$, $$\\ \frac { n+1 }{ n }$$, $$\\ \frac { 2n+1 }{ n }$$,……
Solution:
$$\\ \frac { 1 }{ n }$$, $$\\ \frac { n+1 }{ n }$$, $$\\ \frac { 2n+1 }{ n }$$,……
=>$$\\ \frac { 1 }{ n }$$, $$1+ \frac { 1 }{ n }$$, $$2+ \frac { 1 }{ n }$$,…..

Question 8.
Is 402 a term of the sequence :
8, 13, 18, 23,…?
Solution:
In sequence 8, 13, 18, 23,…..
a = 8 and d = 13 – 8 = 5
Let 402 be the nth term, then
402 = a + (n – 1)d = 8 + (n – 1) x 5
$$\\ \frac { 402-8 }{ 5 }$$ = n – 1
=> $$\\ \frac { 394 }{ 5 }$$ = n – 1
394 is not exactly divisible by 5,
∴ 402 is not its term.

Question 9.
Find the common difference and 99th term of the arithmetic progression :
$$7 \frac { 3 }{ 4 }$$, $$9 \frac { 1 }{ 2 }$$, $$11 \frac { 1 }{ 4 }$$,…..
Solution:
$$7 \frac { 3 }{ 4 }$$, $$9 \frac { 1 }{ 2 }$$, $$11 \frac { 1 }{ 4 }$$,…..
here a = $$7 \frac { 3 }{ 4 }$$
= $$\\ \frac { 31 }{ 4 }$$

Question 10.
How many terms are there in the series:
(i) 4, 7, 10, 13,…..148 ?
(ii) 0.5, 0.53, 0.56,…..1.1 ?
(iii) $$\\ \frac { 3 }{ 4 }$$, 1, $$1 \frac { 1 }{ 4 }$$,….3 ?
Solution:
(i) 4, 7, 10, 13,…..148 ?
Here a = 4, d = 7 – 4
= 3
Let 148 be the nth term, then

Question 11.
Which term of the A.P. 1 + 4 + 7 + 10 +……is 52 ?
Solution:
A.P. is 1, 4, 7, 10,……is 52
Here a = 1, d = 4 – 1 = 3
Let 52 be the nth term, then

Question 12.
If 5th and 6th terms of an A.P. are respectively 6 and 5, find the 11th term of the A.P.
Solution:
5th term (T5) = 6
=> 6 = a + 4d
and T6 = 5
=> 5 = a + 5d

Question 13.
If tn represents nth term of an A.P., t2 + t5 – t3 = 10 and t2 + t9 = 17, find its first term and its common difference.
Solution:
Let first term of an A.P. be a
and common difference = d
Then Tn is the nth term

Question 14.
Find the 10th term from the end of the A.P. 4, 9, 14,…. 254.
Solution:
We know that rth term from the end
= (n – r + 1)th from the beginning
Here, a = A, d = 9 – 4 = 5,
nth term = 254

Question 15.
Determine the arithmetic progression whose 3rd term is 5 and 7th term is 9.
Solution:
Let a be the first term and d be the common difference, then
T3 = a + 2d – 5
T7 = a + 6d = 9

Question 16.
Find the 31st term of an A.P. whose 10th term is 38 and 16th term is 74.
Solution:
Let a be the first term and d be the common difference, then
T10 = a + 9d =38
and T16 = a + 15d = 74

Question 17.
Which term of the series :
21, 18, 15, is – 81 ?
Can any term of this series be zero ? If yes, find the number of term.
Solution:
21, 18, 15, 1….. – 81
Here a = 21,d = 18 – 21 = – 3,
nth term = – 81

Question 18.
An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find the 31st term.
Solution:
Given : An A.P. consists of 60 terms
a = 7
and a60 = 125
=> a + (60 – 1 )d = 125
=> 7 + 59 d = 125
=> 59d= 125 – 7
=> d = $$\\ \frac { 118 }{ 59 }$$ = 2
now, a31 = a + (n – 1)d
= 7 + (31 – 1) x 2
= 7 + 30 x 2
= 67

Question 19.
The sum of the 4th and the 8th terms of an A.P. is 24 and the sum of the sixth term and the tenth term is 34. Find the first three terms of the A.P.
Solution:
In an A.P.
T4 + T8 = 24
T6 + T10 =34
Let a be the first term and d be the common difference

Question 20.
If the third term of an A.P. is 5 and the seventh terms is 9, find the 17th term
Solution:
Let the first term of an A.P. = a
and the common difference of the given A.P. = d
As, we know that,

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C. https://ncertmcq.com/selina-concise-mathematics-class-10-icse-solutions/

Other Exercises

Question 1.
Find the sum of the first 22 terms of the A.P. : 8, 3, – 2,…..
Solution:
A.P. = 8, 3, – 2,…..
Here, a = 8, d = 3 – 8 = – 5, n = 22

Question 2.
How many terms of the A.P. : 24, 21, 18, must be taken so that their sum is 78 ?
Solution:
Let n term of the given A.P. be taken
and A.P. = 24, 21, 18……
Let n be the number of terms.
Here, a = 24, d = 21 – 24 = – 3, Sn = 78

Question 3.
Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
Solution:
nth term (Tn) = 8n – 5
T1 = 8 – 5 = 3
T2 = 8 x 2 – 5 = 16 – 5 = 11
T3 = 8 x 3 – 5 = 24 – 5 = 19
A.P. is 3, 11, 19,
Here, a = 3,d = 11 – 3 = 8 and n = 28

Question 4.
Find the sum of :
(i) all odd natural numbers less than 50.
(ii) first 12 natural numbers each of which is a multiple of 7.
Solution:
(i) Sum of all odd natural numbers less then 50
1 + 3 + 5 + 7 +…….+ 49
Here a = 1, d = 3 – 1 = 2
Let n be the number of term, then
49 = a + (n – 1)d
=> 49 = 1 + (n – 1) x 2
=> 49 – 1 = (n – 1) x 2

Question 5.
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Solution:
Sum of first 51 terms of an A.P. in which T2 = 14 and T3 = 18
d = T3 – T2 = 18 – 14 = 4
and T2 = a + d
=> 14 = a + 4
=> a = 14 – 4 = 10
A.P. = 10, 14, 18, 22,….

Question 6.
The sum of first 7 terms of an A.P. is 49 and that of first 17 terms of it is 289. Find the sum of first tt terms.
Solution:
S7 = 49,
S17 = 289

Question 7.
The first term of an A.P. is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
Solution:
First term of an AP (a) = 5
Last term = 45
and Sn = 1000

Question 8.
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
Solution:
All natural numbers between 250 and 1000 which are divisible by 9 are
252, 261, 270……999

Question 9.
The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?
Solution:
In an A.P.
T1 = a = 34, l = 700, d = 18
Let n be the number of terms, then
Tn = a + (n – 1 )d
=> 700 = 34 + (n – 1) x 18
=> 700 – 34 = 18(n – 1)
=> 180 – 0 = 666

Question 10.
In an A.P. the first term is 25, nth term is – 17 and the sum of n terms is 132. Find n and the common difference.
Solution:
In an A.P.
First term (a) = 25
nth term = – 17
and Sn = 132
Let d be the common difference

Question 11.
If the 8th term of an A.P. is 37 and the 15th term is 15 more than the 12th term, find the A.P. Also, find the sum of first 20 terms of this A.P.
Solution:
In an A.P.
8th term = 37
15th term = 12th term + 15
Let a be the first term and d be the common difference, then

Question 12.
Find the sum of all multiples of 7 lying between 300 and 700.
Solution:
Multiples of 7 lying between 300 and 700 are 301, 308, 315, 322,…., 693

Question 13.
The sum of n natural numbers is 5n² + 4n. Find its 8th term.
Solution:
Sum of n natural number = 5n² + 4n
Sn = 5n² + 4n
S1(a) = 5 x (1)² + 4 x 1
= 5 + 4 = 9
S2 = 5(2)² + 4 x 2 = 20 + 8 = 28
S2 – S1 = T2 = 28 – 9 = 19
=> a + d = 19 => 9 + d = 19
d = 19 – 9 = 10
a = 9, d = 10
T8 = a + (n – 1 )d = 9 + (8 – 1) x 10
= 9 + 7 x 10 = 9 + 70 = 79

Question 14.
The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.
Solution:
In an A.P.
T= 11
T8 = 2T4 + 5
Now, a + 3d = 11
a + 7d = 2 x 11 + 5 = 22 + 5 = 27
Subtracting, 4d = 16
=> d = $$\\ \frac { 16 }{ 4 }$$ = 4

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.