Class X Arithmetic Progressions - Solutions of Cbse Ncert Maths Textbook Exercise 5.2 - nth term of an AP

 



CBSE Class 10, Ncert Solutions for Mathematics (SA-II / Term II)  

Chapter 5, ARITHMETIC PROGRESSION (AP)

NCERT Solutions for Mathematics Textbook Exercise 5.2

(nth Term of an arithmetic progression / AP)

(Page 105, 106, 107)
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
(ii) 11th term of the AP (Arithmetic Progression): –3, –1/2, 2, . . . is
(A) 28  (B) 22  (C) –38  (D) –48    
Solution.2(i): Here, a = 10, d = a2 - a1 = 7 - 10 = –3 and n = 30
We have,
an = a + (n - 1)d
=> a30 = 10 + (30 - 1) (–3) = 10 - 87 = –77.

Solution.2(ii): Here, a = –3, d = a2 - a1 = 5/2 and n = 11
We have,
an = a + (n - 1)d
=> a11 = –3 + (11 - 1) 5/2 = 22.

3: In the following APs, find the missing terms in the boxes:








Solution.3(i): Do it yourself.
Solution.3(ii):
Here, a2 = 13, a4 = 3
Using formula, an = a1 + (n - 1)d
or, 13 = a1 + (2 - 1)d
or, 13 = a1 + d .............Eq(1)
or, 3 = a1 + (4 - 1)d
or, a1 = 3 - 3d ..............Eq(2)
Substituting the value of a1 in Eq(1), we get,   
13 = 3 - 3d + d = 3 - 2d
or, d = –5.
So, a1 = a2 - d = 13 - (-5) = 18 and similarly
a3 = a2 + d = 13 + (-5) = 8
Hence, the missing terms are 18 and 8.
Solution.3(iii): Taking hint from the above try to solve it.
Solution.3(iv):
Here a1 = –4, a6 = 6. We know,
an = a1 + (n - 1)d
=> a6 = 6 = (–4) + 5d = 5d - 4
=> d = 2
Now, a2 = a1 + d = -4 + 2 = -2
a3 = a2 + d = -2 + 2 = 0
a4 = a3 + d = 0 + 2 = 2
a5 = a4 + d = 2 + 2 = 4
Hence, the missing terms are –2, 0, 2, and 4.  
Solution.3(v): Try to solve it yourself.

4: Which term of the AP: 3, 8, 13, 18, . . . . , is 78 ?
Solution.4: Here, a1 = 3, d = 8 - 3 = 5 and an = 78.
Using formula: an = a1 + (n - 1)d
We can write,
78 = 3 + (n - 1)5
or, 5n =78 - 3 + 5 = 80
or, n = 16.
Hence, 78 is the 16th term of this AP.  

5: Find the number of terms in each of the following APs:
(i) 7, 13, 19, . . . . , 205   
(ii) 18, 15 , 13, . . . . . , –47  
Solution.5(i):
a1 = 7, d = a2 - a1 = 13 - 7 = 6, an = 205
Using formula, an = a1 + (n - 1)d
=> 205 = 7 + (n - 1)6
=> (n - 1)6 = 198
=> n = 33 + 1 = 34.
Hence, number of terms in this AP = 34.
Solution.5(ii): Try to solve it yourself using the same method as in Solution.5(i).

6: Check whether –150 is a term of the Arithmetic Progression (AP): 11, 8, 5, 2, . . .
Solution.6:
a1 = 11, a2 = 8, a3 = 5, a4 = 2
So, a2 - a1 = 8 - 11 = –3
a3 - a2 = 5 - 8 = –3
a4 - a3 = 2 - 5 = –3
Clearly the sequence forms an AP with common difference (d) = –3. Let 150 be the nth term of this AP.
Since, an = a1 + (n - 1)d
=> –150 = 11 + (n - 1) (–3)
Now solving this equation we get,
n = 164/3
Since 164/3 is not an integer so –150 is not a term of the given sequence of numbers. .

7: Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.  
Solution.7:
a11 = 38, a16 = 73
By the formula, an = a1 + (n - 1)d we can write,
38 = a1 + (11 - 1)d
38 = a1 + 10d ..............Eq(1) and
73 = a1 + 15d ..............Eq(2)
Subtracting Eq(1) from Eq(2) we get,
35 = 5d
or, d = 7. Putting the value of ‘d’ in Eq(1) we get,
38 = a1 + 70
or, a1 = –32
a31 = a1 + (31 - 1)d = –32 + (31 - 1)7 = 178.

8: An AP consists of 50 terms of which third term is 12 and the last term is 106. Find the 29th term.
Solution.8: Try to solve it yourself using the same method as in Solution.7.  

9: If the 3rd and the 9th terms of an AP are 4 and –8 respectively, which term of this AP is zero ?
Solution.9:
Given that a3 = 4, a9 = –8, an = 0, n = ?
an = a1 + (n - 1)d
a1 + 2d = 4 ..........Eq(i) and
a1 + 8d = –8 ..........Eq(ii)
From Eq(i) we get
a1 = 4 - 2d
Substituting the value of ‘a1’ in Eq(ii) we have,
a + 8d = –8
or, 4 - 2d + 8d = –8
or, d = –2
Now putting the value of ‘d’ in Eq(ii) we have,
a1 + 8d = –8
or, a1 + 8 (–2) = –8
or, a1 = 8
Once again using the formula: an = a1 + (n - 1)d {where an = 0, a1 = 8, d = –2}
0 = 8 + (n - 1)2
or, n = 5
Hence, 5th term of this Arithmetic Progression is zero.

10: The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution.9:
Let the first term and common difference be ‘a1’ and ‘d’. Given that
a17 - a10 = 7
=> (a1 + 16d) - (a1 + 9d) = 7
=> a1 + 16d - a1 - 9d = 7
=> 16d - 9d = 7
=> d = 1

So, the common difference is 1.
 
11: Which term of the AP (Arithmetic Progression): 3, 15, 27, 39, . . . . will be 132 more than its 54th term ?  
 
12: Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms.
 
13: How many three digit numbers are divisible by 7 ?
 
14: How many multiples of 4 lie between 10 & 250 ?  
 
15: For what value of n, are the nth terms of two APs: 63, 65, 67, . . . . and 3, 10, 17, . . . . equal ?
 
16: Determine the AP whose 3rd term is 16 and 7th term exceeds the 5th term by 12.  
 
17: Find the 20th term from the last term of the AP: 3, 8, 13, . . . . , 253.  
 
18: The sum of the 4th and 8th terms of an AP is 24 and the sum of 6th and 10th terms is 44. Find the first three terms of the AP.   

19: Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000 ?

20: Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.

Click NEXT to view the complete solutions of Question No.11 to 20 of NCERT Mathematics Textbook Lesson 5 (Arithmetic Progression) - Exercise 5.2 

Taking hints from these solutions try to solve remaining problems. At the end of completing all Exercises from each Chapter given in NCERT Math textbook, we shall provide CBSE Notes (Hints) / Solutions of CBSE Questions on Mathematics and many more…    
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