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

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*Chapter 5, ARITHMETIC PROGRESSION (AP) *

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__NCERT Solutions
for Mathematics Textbook Exercise 5.2__

__NCERT Solutions for Mathematics Textbook Exercise 5.2__

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*(nth Term of an arithmetic progression
/ AP)*

*(Page 105, 106, 107)*

**2: Choose the correct choice in the following and justify:**

**(i) 30**

^{th}term of the AP: 10, 7, 4, . . . . . is**(A) 97 (B) 77 (C) –77 (D) –87**

**(ii) 11**

^{th}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 = a

_{2}- a

_{1 }= 7 - 10 = –3 and n = 30

We have,

a

_{n}= a + (n - 1)d
=> a

_{30}= 10 + (30 - 1) (–3) = 10 - 87 = –77.__Solution.2(ii)__: Here, a = –3, d = a

_{2}- a

_{1 }= 5/2 and n = 11

We have,

a

_{n}= a + (n - 1)d
=> a

_{11}= –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, a

_{2}= 13, a_{4}= 3
Using formula, a

_{n}= a_{1}+ (n - 1)d
or, 13 = a

_{1}+ (2 - 1)d
or, 13 = a

_{1}+ d .............Eq(1)
or, 3 = a

_{1}+ (4 - 1)d
or, a

_{1}= 3 - 3d ..............Eq(2)
Substituting the value of a

_{1}in Eq(1), we get,
13 = 3 - 3d + d = 3 - 2d

or, d = –5.

So, a

_{1}= a_{2}- d = 13 - (-5) = 18 and similarly
a

_{3}= a_{2}+ 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 a

_{1}= –4, a_{6}= 6. We know,
a

_{n}= a_{1}+ (n - 1)d
=> a

_{6}= 6 = (–4) + 5d = 5d - 4
=> d = 2

Now, a

_{2}= a_{1}+ d = -4 + 2 = -2
a

_{3}= a_{2}+ d = -2 + 2 = 0
a

_{4}= a_{3}+ d = 0 + 2 = 2
a

_{5}= a_{4}+ 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, a

_{1}= 3, d = 8 - 3 = 5 and a

_{n}= 78.

Using formula: a

_{n}= a_{1}+ (n - 1)d
We can write,

78 = 3 + (n - 1)5

or, 5n =78 - 3 + 5 = 80

or, n = 16.

Hence, 78 is the 16

^{th}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)__:

a

_{1}= 7, d = a_{2 }- a_{1}= 13 - 7 = 6, a_{n }= 205
Using formula, a

_{n}= a_{1}+ (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__:

a

_{1}= 11, a_{2 }= 8, a_{3 }= 5, a_{4 }= 2
So, a

_{2 }- a_{1}= 8 - 11 = –3
a

_{3 }- a_{2}= 5 - 8 = –3
a

_{4 }- a_{3}= 2 - 5 = –3
Clearly the sequence forms an AP with
common difference (d) = –3. Let 150 be the n

^{th}term of this AP.
Since, a

_{n}= a_{1}+ (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 31**

^{st}term of an AP whose 11^{th}term is 38 and the 16^{th}term is 73.__Solution.7__:

a

_{11}= 38, a_{16}= 73
By the formula, a

_{n}= a_{1}+ (n - 1)d we can write,
38 = a

_{1}+ (11 - 1)d
38 = a

_{1}+ 10d ..............Eq(1) and
73 = a

_{1}+ 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 = a

_{1}+ 70
or, a

_{1}= –32
a

_{31}= a_{1}+ (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 29**

^{th}term.__Solution.8__: Try to solve it yourself using the same method as in Solution.7.

**9: If the 3**

^{rd}and the 9^{th}terms of an AP are 4 and –8 respectively, which term of this AP is zero ?__Solution.9__:

Given that a

_{3}= 4, a_{9}= –8, a_{n}= 0, n = ?
a

_{n}= a_{1}+ (n - 1)d
a

_{1}+ 2d = 4 ..........Eq(i) and
a

_{1}+ 8d = –8 ..........Eq(ii)
From Eq(i) we get

a

_{1}= 4 - 2d
Substituting the value of ‘a

_{1}’ 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,

a

_{1}+ 8d = –8
or, a

_{1}+ 8 (–2) = –8
or, a

_{1}= 8
Once again using the formula: a

_{n}= a_{1}+ (n - 1)d {where a_{n}= 0, a_{1}= 8, d = –2}
0 = 8 + (n - 1)2

or, n = 5

Hence, 5

^{th}term of this Arithmetic Progression is zero.**10: The 17**

^{th}term of an AP exceeds its 10^{th}term by 7. Find the common difference.__Solution.9__

**:**

Let the first term and common
difference be ‘a

_{1}’ and ‘d’. Given that
a

_{17}- a_{10}= 7
=> (a

_{1}+ 16d) - (a_{1}+ 9d) = 7
=> a

_{1}+ 16d - a_{1}- 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 54**

^{th}term ?**12: Two APs have the same common difference. The difference between their 100**

^{th}terms is 100, what is the difference between their 1000^{th}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 3**

^{rd}term is 16 and 7^{th}term exceeds the 5^{th}term by 12.**17: Find the 20**

^{th}term from the last term of the AP: 3, 8, 13, . . . . , 253.**18: The sum of the 4**

^{th}and 8^{th}terms of an AP is 24 and the sum of 6^{th}and 10^{th}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**

*n*th week, her weekly savings become Rs 20.75, find*n.*
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