Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F

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 10F

Other Exercises

Question 1.
The 6th term of an A.P. is 16 and the 14th term is 32. Determine the 36th term.
Solution:
Let the first term and common difference of an A.P. be a and d
As, we know that,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q1.1

Question 2.
If the third and the 9th terms of an A.P. term is 12 and the last term is 106. Find the 29th term of the A.P.
Solution:
Let the first term and common difference of an A.P. be a and d.
As, we know that,
an = a + (n – 1 )d
a3 = a + (3 – 1 )d = a + 2d
Similarly,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q1.3

Question 3.
An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term of the A.P.
Solution:
Number of terms in an A.P. = 50
T3= 12, l = 106
To find T29
Let a be the first term and d be the common difference
=> a + 2d = 12 …(i)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q3.1

Question 4.
Find the arithmetic mean of :
(i) – 5 and 41
(ii) 3x – 2y and 3x + 2y
(iii) (m + n)² and (m – n)²
Solution:
(i) Arithmetic mean between – 5 and 41
= \(\\ \frac { -5+41 }{ 2 } \)
= \(\\ \frac { 36 }{ 2 } \)
= 18
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q4.1

Question 5.
Find the sum of first 10 terms of the A.P. 4 + 6 + 8 +…..
Solution:
A.P. = 4 + 6 + 8 +…….
Here, a = 4, d = 6 – 4 = 2, n = 10
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q5.1

Question 6.
Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 60.
Solution:
Sum of first 20 terms of an A.P. in which
a = 3 and a20 = 60
a20 = a + (20 – 1) x d
60 = 3 + 19 x d
19d = 60 – 3
d = \(\\ \frac { 57 }{ 19 } \)
= 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q6.1

Question 7.
How many terms of the series 18 + 15 + 12 +…..when added together will give 45 ?
Solution:
A.P. is 18 + 15 + 12 +…..
Here, a = 18, d = 15 – 18 = – 3
Given : Sn = 45
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q7.1

Question 8.
The nth term of a sequence is 8 – 5n. Show that the sequence is an A.P.
Solution:
Given, an = 8 – Sn
a1 = 8 – 5 x (1) = 8 – 5 = 3
a2 = 8 – 5 x (2) = 8 – 10 = – 2
a3 = 8 – 5 x (3) = 8 – 15 = – 7
We see that
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q8.1

Question 9.
The the general term (nth term) and 23rd term of the sequence 3, 1, – 1, – 3,……
Solution:
The progression 3, 1, – 1, – 3,…..is A.P.
with first term (a) = 3 and common difference (d) = 1 – 3 = – 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q9.1

Question 10.
Which term of the sequence 3, 8, 13,…..is 78 ?
Solution:
Let 78 be the nth term
a = 3, d = 8 – 3 = 5, an = 78, n = ?
a + (n – 1)d = an
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q10.1

Question 11.
Is – 150 a term of 11, 8, 5, 2,….. ?
Solution:
11, 8, 5, 2,….1st term, a = 11
Common difference, d = 8 – 11 = – 3
an = – 150
=> a + (n – 1 )d = – 150
=> 11 + (n – 1) ( – 3) = – 150
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q11.1

Question 12.
How many two digit numbers are divisible by 3 ?
Solution:
Numbers divisible by 3 are 3, 6, 9, 12,….
Hence, lowest two digit number divisible by 3 = 12
and highest two digit number divisible by 3 = 99
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q12.1

Question 13.
How many multiples of 4 lie between 10 and 250 ?
Solution:
Multiples of 4 between 10 and 250 are
12, 16, 20, 24,……, 248
Here, a = 12, d = 4, l = 248
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q13.1

Question 14.
The sum of the 4th term and the 8th term of an A.P. is 24 and the sum of 6th term and the 10th term is 44. Find the first three terms of the A.P.
Solution:
In an A.P.
T4 + T8 = 24
T6 + T10 = 44
Let a be the first term and d be the common difference
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q14.1

Question 15.
The sum of first 14 terms of an A.P. is 1050 and its 14th term is 140. Find the 20th term.
Solution:
Given a14 = 140
we know, an = a + (n – 1) x d
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression Ex 10F Q15.2

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