#### Exercise 9.4

**Question 1. If 12 ^{th} of an A.P is 82 and 18^{th} term is 124. Then find out the 24^{th} term.**

**Solution: **Given: a_{12} = 82 and a_{18} = 124

General term of an AP : a_{n} = a + (n – 1)d

⇒ a_{12} = a + (12-1)d

⇒ 82 = a + 11d —(1)

⇒ 124 = a + (18-1)d

⇒ 124 = a + 17d —(2)

Subtracting (2) from (1)

⇒ ( a + 17d ) – ( a + 11d ) = 124 – 82

⇒ a +17d – a – 11d = 42

⇒ 6 d = 42

⇒ d = 7, common difference

Putting d = 7 in equation (1), we get

⇒ a + 11 X 7 = 82

⇒ a = 82 – 77

⇒ a = 5, first term

**To find:** 24^{th} term

a_{24} = a + ( 24 -1 )d

= 5 + 23 X 7

= 5 + 161

= 166

**Question 2. In an A.P. the 24 ^{th} term is twice the 10^{th} term. Prove that the 72^{nd} term is twice the 34^{th}**

**term.**

**Solution: **

24^{th} term is twice the 10^{th} term (Given)

Which means, a_{24} = 2 X a_{10 }. . . . . . (1)

Let, “a” be the first term and “d” be the common difference.

And, n^{th} term is a_{n} = a + (n – 1)d

from equation (1), we have

a + ( 24-1 )d = 2( a + (10-1)d )

⇒ a + 23d = 2 ( a + 9d )

⇒ a + 23 d = 2a + 18d

⇒ (23 – 18)d = a

⇒ a = 5d

To Prove: 72^{nd} term is twice the 34^{th} term

⇒ a_{72} = 2 X a_{34}

⇒ a + (72-1)d = 2 [ a + (34-1)d ]

⇒ a + 71d = 2( a + 33d )

⇒ a + 71d = 2a + 66d

putting the value of a = 5 in the above equation,

⇒ 5d + 71d = 2 (5d) + 66d

⇒ 76d = 76d

Hence it is proved.

**Question 3. If the (m+1) ^{th} term of an A.P. is twice the (n+1)^{th} term of the A.P. Then prove that**:

**(3m+1)**

^{th}is twice the (m+n+1)^{th}term.**Solution:**

a_{(m+1)} = 2 a_{(n+1) } (From the question)

Let, First term = a and Common Difference = d

⇒ a + (m + 1 – 1)d = 2[a+(n + 1 -1)d]

⇒ a +md = 2a + 2nd

⇒ a = md – 2nd

⇒ a = (m-2n)d —(2)

**To Prove:** a_{(3m+1)} = 2 a_{(m+n+1)}

⇒ a + (3m + 1 – 1)d = 2 [ a + (m + n + 1 -1)d ]

⇒ a + 3md = 2a + 2(m + n)d

putting the value of a = (m – 2n)d, from equation (1)

⇒ (m – 2n)d + 3md = 2[(m – 2n)d] + 2( m + n)d

⇒ m -2n + 3m = 2m -4n + 2m + 2n

⇒ 4m – 2n = 4m – 2n

Hence proved.

**Question 4. If the n ^{th} term of the A.P. 9,7,5, … is same as the n^{th} term of the A.P. 15, 12 , 9, … find n.**

**Solution: **

First sequence is 9,7,5, …

First term (a) = 9 and Common Difference(d) = 9 – 7 = -2

n^{th} term = a + (n-1)d

⇒ a_{n} = 9 + (n-1)(-2)

= 9 – 2n + 2

= 11 – 2n

Second sequence is 15, 12, 9, …

here, First term (a) = 15

Common Difference (d) = 12 -15 = -3

n^{th} term = a + (n-1)d

⇒ a’_{n} = 15 + (n-1)(-3)

= 15 -3n +3

= 18 -3n

We are given in the question that the n^{th} term of both the A.P.s are same,

So, a^{}_{n }= a’_{n}

⇒ 11 – 2n = 18 – 3n

⇒ n = 7

**Question 5. Find the 13 ^{th} term from the end in the following A.P.**

**(i). 4, 9, 14, … , 254.**

**Solution: **

First term (a) = 4 and common difference (d) = 9 – 4 = 5

last term here (l) = 254

n^{th} term from the end is : l – (n-1)d

we have to find 13^{th} term from end then : l – 12d

= 254 – 12 X 5

= 254 – 60

= 194

**(ii). 3, 5, 7, 9, …, 201.**

**Solution: **we have,

First term (a) = 3 and common difference (d) = 5 – 3 = 2

last term here (l) = 201

n^{th} term from the end is : l – (n-1)d

we have to find 13^{th} term from end then : l – 12d

= 201 – 12 X 2

= 201 – 24

= 177

**(iii). 1, 4, 7, 10, … , 88.**

**Solution: **

First term (a) = 1 and common difference (c.d) = 4 -1 = 3

last term here (l) = 88

n^{th} term from the end is : l – (n-1)d

we have to find 13^{th} term from end then : l – 12d

= 88 – 12 X 3 = 88 – 36 = 52

**Question 6. The 4 ^{th} term of an A.P. is three times the first term and the 7^{th} term exceeds the twice the third term by 1. Find the A.P.**

**Solution: **As per the statement, 4^{th} term of the A.P = thrice the first term

⇒ a_{4} = 3 first term

Using nth term formula, we have, a + ( 4 -1 )d = 3 a

Where, first term to be ‘a’ and the common difference be ‘d’

⇒ a + 3d = 3 a

⇒ a = \(\frac{3}{2}d\) —(1)

and also it is given that,

the 7^{th} term exceeds the twice of the 3^{rd} term by 1

⇒ a_{7} + 1 = 2 X a_{3}

⇒ a + (7-1)d +1 = 2[ a + (3-1)d ]

⇒ a + 6d +1 = 2a + 4d

⇒ a = 2d + 1 —(2)

putting the value of a = \(\frac{3}{2}d\) from equation (1) in equation (2)

\(\frac{3}{2}d\) = 2d + 1

⇒ \(\frac{3}{2}d\) – 2d = 1

⇒ \(\frac{3d – 4d}{2} = 1\)

⇒ -d = 2

⇒ d = -2

put d = -2 in \(a = \frac{3}{2}d\)

⇒ a = \(\frac{3}{2}(-2)\)

⇒ a = -3

Now, we have a = -3 and d = -2, so the A.P. is -2, -5, -8, -11 , …

**Question 7. Calculate the third term and the n ^{th }term of an A.P. whose 8^{th} term and 13^{th} term are 48 and 78 respectively.**

**Solution: **Given: a_{8} = 48 and a_{13} = 78

n^{th} term of an A.P. is: a_{n} = a + (n-1)d

so,

a_{8} = a + (8 – 1)d = a + 7d …………(1)

a_{13} = a + (13 – 1)d = a + 12d ……….(2)

Equating (1) and (2), we get.

⇒ a + 12d – (a + 7d) = 78 – 48

⇒ a +12d – a -7d = 30

⇒ 5d = 30

⇒ **d = 6**

Putting the value of d = 6 in equation (1),

a + 7 X 6 = 48

⇒ a + 42 = 48

⇒ **a = 4**

Now, n^{th} term will be: a_{n }= a + (n-1)d

= 4 + (n-1)6

= 4 + 6n – 6

**a _{n }= 6n – 2**

and the 3^{rd} term will be

a_{3} = 6 X 3 – 2

**a _{n} = 16**

**Question 8. How many three digit numbers are divisible with 3?**

**Solution:** First and last 3 digit numbers which are divisible by 3 are 102 and 999 respectively.

So, here we have

First term (a) = 102

Common Difference (d) = 3

Last term (l) = 999

⇒ a_{n }= 999

⇒ a + (n – 1)d = 999

⇒ 102 + (n – 1)3 = 999

⇒ 102 + 3n – 3 = 999

⇒ 99 + 3n = 999

⇒ 3n = 900

⇒ n = 300

**Therefore, there are 300 terms which are divisible by 3.**

**Question 9. An A.P. has 50 terms and the first term is 8 and the last term is 155. Find the 41 ^{st} term**

**from the A.P.**

**Solution:**

First term (a) = 8

Number of terms (n) = 50

Last term (a_{n}) = 148

⇒ a_{n }= a + (n – 1)d

⇒ 155 = 8 + (50 – 1)d

⇒ 49 d = 147

⇒ d = 3

41^{st} term will be: a + (41-1)d

⇒ 8 + 40 X 3 = 128

41^{st} term = **128**

**Question 10. The sum of 4 ^{th} and 8^{th} term of an A.P. is 24 and the sum of the 6^{th} and 10^{th} terms is 34.**

**Find the first term and the common difference of the A.P.**

**Solution:**

Let’s assume first term be “a” and common difference “d”

Given : 4^{th} term + 8^{th} term = 24

⇒ a_{4} + a_{8} = 24

⇒ ( a + (4 – 1)d ) + ( a + (8 -1)d ) = 24

⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24 …………(1)

Again, 6^{th} term + 10^{th} term = 34

⇒ a_{6} + a_{10} = 34

⇒ (a + 5d) + (a + 9d) = 34

⇒ 2a + 14d = 34 ………… (2)

Subtracting equation (1) from (2),we get

⇒ (2a + 14d) – (2a + 10d) = 34 -24

⇒ 2a + 14d – 2a – 10d = 10

⇒ 4d = 10

⇒ d = \(\frac{5}{2}\)

Put d = \(\frac{5}{2}\) in equation (1)

⇒ 2a + 10 X \(\frac{5}{2}\) = 24

⇒ 2a + 25 = 24

⇒ 2a = -1

⇒ a = \(-\frac{1}{2}\)

Therefore, we have a = \(-\frac{1}{2}\) and d = \(\frac{5}{2}\)

**Question 11. The first term of an A.P. is 7 and its 100 ^{th} term is -488, Find the 50^{th} term of the same A.P.**

**Solution: **Given,

First term (a) = 7

100^{th} term (a_{100}) = -488

we know, a_{n} = a + (n – 1)d

⇒ (a_{100}) = a + ( 100 -1 )d

⇒ 7 + 99d = -488

⇒ 99d = -495

⇒ **d = -5**

Now, we have the common difference (d) = 5

We have to find out the 50^{th} term of the A.P.

Then, a_{50} = a + 49 d

= 7 + 49 X (-5)

= 7 – 245

= **– 238**

So, the 50^{th} term of the A.P. is -238

**Question 12. Find a _{40 }– a_{30} of the following A.P.**

**(i). 3, 5, 7, 9, . . .**

**Solution: ** A.P. is 3, 5, 7, 9, . . .

So, we have first term ( a ) = 3 and the common difference ( d ) is 5 – 3 = 2

we have to find a_{40 }– a_{30 }= (a + 39d) – (a + 29d)

= a + 39 d – a -29 d

= 10 d

= 10 X 2

=** 20**

**(ii). 4, 9, 14, 19, . . .**

**Solution: **Given A.P. is 4, 9, 14, 19, . . .

Common difference ( d ) = 9 – 4 = 5

we have to find a_{40 }– a_{30 }= 10 d

= 10 X 5 = 50

**Question 13. Write the expression a _{m} – a_{n} for the A.P. a, a + d, a + 2d, . . . Find the common differences for given expressions.**

**Solution: **General Arithmetic Progression

a, a + d, a + 2d, . . .

a_{m} – a_{n} = ( a + (m – 1)d) – ( a + ( n – 1 )d )

⇒ a + md – d – a – nd +d

⇒ md – nd

a_{m} – a_{n} = (m – n) d ………….(1)

**Let’s find the common difference of the A.P. for which**

**(i). 11 ^{th} term is 5 and 13^{th} term is 79**

11^{th} term ( a_{11} ) = 5

and 13^{th} term (a_{13}) = 79

from equation (1),

taking m = 11 and n = 13

⇒ a_{m} – a_{n }= ( 13 -11 ) d

⇒ 79 – 5 = 2d

⇒ 74 = 2d

⇒ **d = 37**

**(ii). a _{10} – a_{5} = 200**

**Solution: **The difference between the 10^{th} term and 5^{th} term

Putting the value of m and n in equation (1) as 10 and 5, we have

⇒ a_{10} – a_{5} = ( 10 – 5 )d

⇒ 200 = 5 d

⇒ **d = 40**

**(iii). 20 ^{th} term is 10 more than the 18^{th} term**

**Solution: **Given,

a_{20} + 10 = a_{18}

⇒ a_{20} – a_{18} = 10

from equation (1), we have

a_{m} – a_{n }= ( m – n ) d

⇒ a_{20} – a_{18 }= ( 20 – 18 ) d

⇒ 10 = 2d

⇒ **d = 5**

**Question 15. Find n if the given value of x is the n term. The given A.P.**

**(i) \(1,\frac{21}{11}, \frac{31}{11}, \frac{41}{11}, . . . . : x = \frac{141}{11}\)**

**(ii) \(5\frac{1}{2}, 11, 16\frac{1}{2}, 22, . . . : x = 550\)**

**(iii) -1 , -3, -5, -7, . . . : x = -151**

**(iv) 25, 50, 70, 100, . . . : x = 1000**

**Solution:**

**(i) **Given sequence is

\(1, \frac{21}{11}, \frac{31}{11}, \frac{41}{11}, . . . . : x = \frac{141}{11}\)

first term (a) = 1

Common difference ( d) = \(\frac{21}{11} – 1\)

= \( \frac{21 – 11}{11}\)

= \(\frac{10}{11}\)

n^{th} term a_{n} = a + (n-1) X d

⇒ \(\frac{171}{11} = 1 + (n – 1).\frac{10}{11}\)

⇒ \(\frac{171}{11} – 1 = (n – 1).\frac{10}{11}\)

⇒ \(\frac{171 – 11}{11} = (n – 1).\frac{10}{11}\)

⇒ \(\frac{160}{11} = (n – 1).\frac{10}{11}\)

⇒ \((n – 1) = \frac{160}{11} X \frac{11}{10}\)

⇒ n = 17

**(ii) **Given sequence is

\(5\frac{1}{2}, 11, 16\frac{1}{2}, 22, . . . : x = 550\)

first term (a) = \(5\frac{1}{2} = \frac{11}{2}\)

Common Difference ( d ) = \(11 – \frac{11}{2} = \frac{11}{2}\)

n^{th} term a_{n} = a + (n-1) X d

⇒ \(550 = \frac{11}{2} + ( n – 1 ). \frac{11}{2}\)

⇒ \(550 = \frac{11}{2} [ 1 + n – 1 ]\)

⇒ n = 550 X \(\frac{2}{11}\)

⇒ 100

**(iii) **Given sequence is,

-1 , -3, -5, -7, . . . : x = -151

first term ( a ) = -1

Common Difference ( d ) = -3 – (-1)

= -3 + 1

= -2

n^{th} term a_{n} = a + (n-1) X d

⇒ -151 = -1 + ( n – 1 ) X -2

⇒ -151 = -1 – 2n + 2

⇒ – 151 = 1 – 2n

⇒ 2n = 152

⇒ n = 76

**(iv)** Given sequence is,

25, 50, 70, 100, . . . : x = 1000

First term ( a ) = 25

Common Difference ( d ) = 50 – 25 = 25

n^{th} term a_{n} = a + (n-1) X d

we have a_{n} = 1000

⇒ 1000 = 25 + ( n – 1 ) 25

⇒ 975 = ( n – 1)25

⇒ n – 1 = 39

⇒ n = 40

**Question 16. If an A.P. consists of n terms with the first term a and n ^{th} term 1. Show that the sum of the m^{th} term from the beginning and the m^{th} term from the end is (a + 1).**

**Solution: **First term = a

Common Difference = d

Last term ( l ) = a + ( n – 1 ) d

Total no. of terms = n

m^{th} term from the beginning a_{m} = a + ( n – 1 )d

m^{th} term from the end = l + ( n – 1 )( -d )

⇒ a_{(n – m + 1)} = l – (n – 1)d

⇒ a_{m} + a_{(n – m + 1)} = a + (n – 1)d + ( l – ( n – 1)d )

= a + (n-1)d + l – (n-1)d

= a + l

**Question 17. Find the A.P. whose third term is 16 and seventh term exceeds its fifth term by 12.**

**Solution:** Given: a_{3} = 16 and a_{7} – 12 = a_{5}

a_{3} = a + ( 3 – 1 )d = 16

⇒ a + 2d = 16 … (i)

and a_{7} – 12 = a_{5}

⇒ a + 6d -12 = a + 4d

⇒ 2d = 12

⇒ d = 6

Put d = 6 in equation (1)

a + 2 X 6 = 16

⇒ a + 12 = 16

⇒ a = 4.

So, the sequence is 4, 10, 16, . . .

**Question 18. The 7 ^{th} term of an A.P is 32 and its 13^{th} term is 62. Find the A.P.**

**Solution:** Given

a _{7} = 32

⇒ a + ( 7 – 1 )d = 32

⇒ a + 6d = 32 … (i)

and a_{13} = 62

⇒ a + ( 13 – 1 )d = 62

⇒ a + 12d = 62 …(ii)

equation (ii) – (i), we have

(a + 12d) – (a + 6d) = 62 – 32

⇒ 6d = 30

⇒ d = 5

Putting d = 5 in equation (i)

a + 6 X 5 = 32

⇒ a = 32 – 30

⇒ a = 2

Answer: A.P. = 2, 7, 12, 17, . . .

**Question 19. Which term of the A.P. 3, 10, 17, . . . will be 84 more than its 13 ^{th} term?**

**Solution:**

Given A.P. is 3, 10 , 17, . . .

First term ( a ) = 3

Common Difference (d) = 10 -3 = 7

Let n^{th} term of the A.P. will be 84 more than its 13^{th} term, then

a_{n} = 84 + a_{13}

⇒ a + (n – 1)d = a + ( 13 – 1 )d + 84

⇒ ( n – 1 ) X 7 = 12 X 7 + 84

⇒ n – 1 = 24

⇒ n = 25

Hence, 25^{th} term, of the given A.P. is 84 more than the 13^{th} term.

**Question 20. Two arithmetic progressions have the same common difference. The difference between their 100 ^{th} terms is 100. What is the difference between their 1000^{th} terms?**

**Solution:**

Let us consider,

First A.P. = a_{1}, a_{2} , a_{3} , . . . and

Second A.P. = b_{1}, b_{2} , b_{3}, . . .

a_{n} = a_{1 }+ ( n – 1 )d and b_{n} = a_{1} + ( n – 1 )d

Since, the difference between their 100^{th} terms is 100.

⇒ a_{100} – b_{100} = 100

⇒ a + ( 100 – 1 )d – [ b + ( 100 – 1)d ] = 100

⇒ a + 99d – b – 99d = 100

⇒ a + b = 100 … (1)

Difference between 100^{th} term is

⇒ a_{1000} – b_{1000}

= a + (1000 – 1)d – [ b + (1000 – 1)d ]

= a + 999d – b – 999d = a – b = 100 (from equation 1)

Therefore, Difference between 1000^{th} terms of two A.P. is 100.

**Question 21. For what value of n, the n ^{th} terms of the Arithmetic Progression 63, 65, 67, . . . and , 3, 10, 17, . . . are equal?**

**Solution:**

Given two A.P.s are:

63, 65, 67, . . . and 3, 10, 17, . . .

First term for first A.P. is (a) = 63

Common difference ( d ) is 65 – 63 = 2

n^{th} term (a_{n}) = a + ( n – 1)d

= 63 + ( n – 1 ) 2

First term for second A.P. is (a’) = 3

Common Difference ( d’ ) = 10 – 3 = 7

n^{th} term ( a’_{n }) = a’ + ( n – 1 )d

= 3 + ( n – 1) 7

Let n^{th }term of the two sequence be equal then,

⇒ 63 + (n – 1)2 = 3 + (n – 1)7

⇒ 60 = ( n – 1 ).7 – ( n – 1 ).2

⇒ 60 = 5( n – 1 )

⇒ n – 1 = 12

⇒ n = 13

The value of n is 13.

**Question 22. How many multiple of 4 lie between 10 and 250?**

**Solution:** Multiple of 4 after 10 is 12 and multiple of 4 before 250 is 120/4, remainder is 2, so,

250 – 2 = 248

248 is the last multiple of 4 before 250

the sequence is

12,. . . . . . . . , 248

with first term ( a ) = 12

Last term (l) = 258

Common Difference ( d ) = 4

n^{th} term ( a_{n} ) = a + ( n – 1)d

Here n^{th} term a _{n} = 248

⇒ 248 = a + ( n – 1 )d

⇒ 12 + (n – 1)4 = 248

⇒ (n – 1)4 = 236

⇒ n – 1 = 59

⇒ n = 59 + 1 = 60

Therefore, there are 60 terms between 10 and 250 which are multiples of 4.

**Question 23. How many three digit numbers are divisible by 7?**

**Solution: **All three digit numbers lies in the range: 100, . . . . . . , 999

Here 105 is the first 3 digit number which is divisible by 7.

994 Is the last three digit number which is divisible by 7 .

The required sequence is

105, .. . . . . . . . , 994

Here, First term (a) = 105

Last term (l ) = 994

Common Difference (d) = 7

Let there are n numbers in the sequence then,

⇒ a_{n} = 994

⇒ a + ( n – 1 )d = 994

⇒ 105 + ( n – 1 )7 = 994

⇒ (n – 1) X 7 = 889

⇒ n – 1 = 127

⇒ n = 128

Therefore, there are 128 three digit numbers which are divisible by 7.

**Question 24. Which term of the A.P. 8, 14, 20, 26, . . . will be 72 more than its 41 ^{st} term?**

**Solution: **Given sequence

8, 14, 20, 26, . . .

Let its n term be 72 more than its 41^{st} term

⇒ a_{n} = a_{41} + 72 … (1)

For the given sequence,

first term (a) = 8,

Common Difference (d) = 14 – 8 = 6

from equation (1), we have

a_{n} = a_{41} + 72

⇒ a + ( n – 1 )d = a + ( 41 – 1 )d + 72

⇒ 8 + ( n – 1 )6 = 8 + 40 X 6 + 72

⇒ ( n – 1 )6 = 312

⇒ n – 1 = 52

⇒ n = 53

Therefore, 53^{rd} term is 72 more than its 41^{st} term.

**Question 25. Find the term of the Arithmetic Progression 9, 12, 15, 18, . . . which is 39 more than its 36 ^{th} term.**

**Solution:** Given A.P. is

9, 12, 15, 18 , . . .

Here we have,

First term ( a ) = 9

Common Difference ( d ) = 12 – 9 = 3

Let its nth term is 39 more than its 36th term

So, a_{n} = 39 + a_{36}

⇒ a + ( n – 1 )d = 39 + a + ( 36 – 1 )d

⇒ ( n -1 )3 = 39 + 35 X 3

⇒ ( n -1 )3 = ( 13 + 35 ) 3

⇒ n – 1 = 48

⇒ n = 49

Therefore, 49^{th} term of the A.P. 39 more than its 36^{th} term.

**Question 26. Find the 8 ^{th} term from the end of the A.P. 7, 10, 13, . . . , 184.**

**Solution:**

Given A.P. is 7, 10, 13, . . . , 184

First term (a) = 7

Common Difference (d) = 10 – 7 = 3

last term (l) = 184

n^{th} term from end = l – ( n – 1 )d

8^{th} term from end = 184 – ( 8 – 1 )3

= 184 – 7 X 3

= 184 – 21

= 183

Therefore, 8^{th} term from the end is 183.

**Question 27. Find the 10 ^{th} term from the end of the A.P. 8, 10, 12, . . . , 126**

**Solution:** Given A.P. is 8, 10, 12, . . . , 126

First term ( a ) = 8

Common Difference ( d ) = 10 – 8 = 2

Last term ( l ) = 126

n^{th} term from end is : l – ( n -1 )d

So, 10^{th} term from end is : l – ( 10 – 1 )d

= 126 – 9 X 2

= 126 – 18

= 108

Therefore, 108 is the 10^{th} term from the last in the gievn A.P.

**Question 28. The sum of 4 ^{th} and 8^{th} term of an A.P. is 24 and the sum of 6^{th} and 10^{th} term is 44. Find the Arithmetic Progression.**

**Solution:** Given: a_{4} + a_{8} = 24

⇒ a + ( 4 – 1 )d + a + ( 8 – 1 )d = 24

⇒ 2a + 3d + 7d = 24

⇒ 2a + 10d = 24 …(1)

and a_{6} + a_{10} = 44

⇒ a + ( 6 – 1 )d + a + ( 10 – 1 )d = 44

⇒ 2a + 5d + 9 d = 44

⇒ 2a + 14d = 44 …(2)

Subtract equation (1) from equation (2), we get

2a + 14d – ( 2a + 10d ) = 44 -24

⇒ 4d = 20

⇒ d = 5

Put d = 5 in equation (1), we get

2a + 10X5 = 24

⇒ 2a = 24 – 50

⇒ 2a = -26

⇒ a = -13

The A.P is -13, – 7, -2, . . .

**Question 29: Which term of the A.P. is 3, 15, 27, 39, . . . will be 120 more than its 21 ^{st} term?**

**Solution:** Given A.P. is 3, 15, 27, 39, . . .

First term ( a ) = 3

Common Difference ( d ) = 15 – 3 = 12

Let n^{th} term is 120 more than 21^{st} term

⇒ a_{n} = 120 + a_{21}

⇒ a + ( n – 1 )d = 120 + a + ( 21 – 1 )d

⇒ ( n -1 )d = 120 + 20d

⇒ ( n – 1 )12 = 120 + 20 X 12

⇒ n – 1 = 10 + 20

⇒ n = 31

Therefore, 31^{st} term of the A.P. is 120 more than the 21^{st} term.

**Question 30. The 17 ^{th} term of an A.P. is 5 more than twice its 8^{th} term. If the 11^{th} term of the A.P. is 43. Find the n^{th} term.**

**Solution:**

17^{th} term of an A.P is 5 more than twice its 8^{th} term

⇒ a_{17} = 5 + 2a_{8}

⇒ a + ( 17 – 1 )d = 5 + 2 [ a + ( 8 – 1 )d ]

⇒ a + 16d = 5 + 2a + 14d

⇒ a + 5 = 2d … (1)

and 11^{th} term of the A>P. is 43

a_{11} = 43

⇒ a +( 11 – 1 )d = 43

⇒ a + 10d = 43

⇒ a + 5 X 2d = 43

from equation (1)

⇒ a + 5 X ( a + 5 ) = 43

⇒ a + 5a + 25 = 43

⇒ 6a = 18

⇒ a = 3

Putting the value of a = 3, in equation (1), we get

3 + 5 = 2d

⇒ 2d = 8

⇒ d = 4

We have to find the n^{th} term (a_{n}) = a + ( n – 1 )d

= 3 + ( n – 1 )4

= 3 + 4n – 4

= 4n – 1

Therefore, n^{th} term is 4n – 1

**Question 31. Find the number of all three digit natural numbers which are divisible by 9.**

**Solution:** First 3-digit number that is divisible by 9 is 108.

The second number is 108 + 9 = 117.

And the last three-digit number that is divisible by 9 is 999.

Thus, the A.P. will be 108, 117, …. , 999.

Here, first term ( a ) = 108

last term (l) = 999

and the common difference ( d ) as 9

We know that, n^{th} term ( a_{n }) = a + (n – 1)d

According to the question,

999 = 108 + (n – 1)9

⇒ 999 = 108 + 9n – 9

⇒ 999 = 99 + 9n

⇒ 999 = 9n

⇒ 999 – 99

⇒ 9n = 900

⇒ n = 100

Therefore, There are 100 three digit terms which are divisible by 9.

**Question 32. The 19 ^{th} term of an A.P. is equal to three times its 6^{th} term. if its 9^{th} term is 19, find the A.P.**

**Solution: **nth term formula for an A.P. is, a_{n}= a + (n – 1)d

Where “a” be the first term and “d” be the common difference.

According to the question,

a_{19} = 3a_{6}

⇒ a + (19 – 1)d= 3(a + (6 – 1)d)

⇒ a + 18d= 3a + 15d

⇒ 18d- 15d= 3a – a

⇒3d= 2a

⇒ a = 32d …. (1)

Also, a9 = 19

⇒ a+(9 – 1)d= 19

⇒ a+ 8d= 19 ….(2)

On substituting the values of (1) in (2), we get

⇒ 32d + 8d= 19

⇒ 3d+ 16d= 19 x 2

⇒ 19d= 38

⇒d = 2

Now, a = 32×2 [From (1)]

a= 3

Therefore, The A.P. is : 3, 5, 7, 9, . . .

**Question 33. The 9 ^{th} term of an A.P. is equal to 6 times its second term. If its 5^{th} term is 22, find the A.P.**

**Solution:** Let a be the first term and d be the common difference.

We know that, nth term ( a_{n }) = a + (n – 1)d

According to the question,

a_{9 }= 6a_{2}

⇒ a + ( 9 – 1)d = 6(a + (2 – 1)d)

⇒ a+8d= 6a+6d

⇒ 8d-6d= 6a-a

⇒ 2d = 5a

⇒ a = \(\frac{2}{5}\) …. (1)

Also, a_{5} = 22

⇒ a+(6 – 1)d= 22

⇒ a + 4d = 22 ….(2)

On substituting the values of (1) in (2), we get

\(\frac{2}{5}\) d + 4d = 22

⇒ 2d + 20d = 22 X 5

⇒ 22d = 110

⇒ d = 5

Now, a = \(\frac{2}{5}\) X 5 [From (1)]

⇒ a = 2

Thus, the A.P. is : 2, 7, 12, 17, . . .

**Question 34. The 24 ^{th} term of an A.P. is twice its 10^{th} term. Show that its 72^{nd} term is 4 times its 15^{th} term.**

**Solution:** We know that,

n^{th} term ( a_{n} ) = a + (n – 1)d

According to the question,

a_{24} = 2 a_{10}

⇒ a + (24 – 1)d = 2(a + (10 – 1)d)

⇒ a + 23d = 2a + 18d

⇒23d – 18d = 2a – a

⇒ 5d = a

⇒ a = 5d …. (1)

Also,

a_{72} = a + (72 – 1) d

= 5d + 71d [From (1)]

= 76d …. (2)

and

a_{15} = a + (15 – 1) d

= 5d + 14d [From (1)]

= 19d …. (3)

On comparing (2) and (3), we get

⇒ 76 d = 4 X 19 d

⇒ a_{72} = 4 X a_{15}

Thus, 72^{nd} term of the given A.P. is 4 times its 15^{th} term.

**Question 35. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5**.

**Solution:** The number is divisible by both 2 and 5, must be divisible by 10.

In the given numbers, first number that is divisible by 10 is 110.

Next number is 110 + 10 = 120.

The last number that is divisible by 10 is 990.

Thus, the **progression will be 110, 120, …, 990.**

first term = 110

and the common difference = 10.

We know that,

nth term = a_{n}= a + (n – 1)d

According to the question,

990 = 110 + ( n – 1 )10

⇒ 990 = 110 + 10n – 10

⇒ 10n = 990 – 100

⇒ n = 89

Thus, total number of natural numbers between 101 and 999 which are divisible by both 2 and 5 are 89.

**Question 36. If the seventh term of an A.P. is 1/9 and its 9 ^{th} term is 1/7, find the 63^{rd} term.**

**Solution: **We know that, nth term, a_{n }= a + (n – 1)d

where “a” be the first term and “d” be the common difference.

According to the question,

a_{7} = \(\frac{1}{9}\)

⇒ a+(7 – 1)d = \(\frac{1}{9}\)

⇒ a+ 6d = \(\frac{1}{9}\) ….(1)

Also, a_{9} = \(\frac{1}{7}\)

⇒ a + (9 – 1)d = \(\frac{1}{7}\)

⇒ a + 8d = \(\frac{1}{7}\) ….(2)

On Subtracting (1) from (2), we get

⇒ 8d – 6d = \(\frac{1}{7}\) – \(\frac{1}{9}\)

⇒ 2d = \(\frac{9 – 7}{ 63 }\)

⇒ 2d = \(\frac{2}{63}\)

⇒ d= \(\frac{ 1 }{ 63 }\)

Put value of d = \(\frac{ 1 }{ 63 }\) in equation (1), we get

⇒ a + 6 X \(\frac{ 1 }{ 63 }\) = \(\frac{ 1 }{ 9}\)

⇒ a = \(\frac{ 1 }{ 9 }\) – \(\frac{ 6 }{ 63 }\)

⇒ a = \(\frac{ 7 – 6}{ 63 }\)

⇒ a = \(\frac{ 1 }{ 63 }\)

Therefore, a_{63} = a + ( 63 – 1 )d

= \(\frac{ 1 }{ 63 }\) + \(\frac{ 62 }{ 63 }\) = \(\frac{ 63 }{ 63 }\) = 1

Thus, 63^{rd} term of the given A.P. is 1.

**Question 37. The sum of 5 ^{th} and 9^{th} terms of an A.P. is 30. If its 25^{th} term is three times its 8^{th} term, Find the A.P.**

**Solution: **We know that, nth term ( a_{n} ) = a + (n – 1)d

where “a” be the first term and “d” be the common difference.

According to the question,

a_{5} + a_{9} = 30

⇒ a + (5 – 1)d + a + (9 – 1)d = 30

⇒ a + 4d + a + 8d = 30

⇒ 2a + 12d = 30

⇒ a + 6d = 15 …. (1)

Also, a_{25} = 3(a_{8})

⇒ a + (25 – 1)d = 3[a + (8 – 1)d]

⇒ a + 24d = 3a + 21d

⇒ 3a – a = 24d – 21d

⇒ 2a = 3d

⇒ a = \(\frac{ 3 }{ 2 } d\) ….(2)

Substituting the value of (2) in (1), we get 32d+6d= 15

⇒ \(\frac{ 3 }{ 2 } d\) + 6d = 15

⇒ 3d + 12d = 15 x 2

⇒ 15d = 30

⇒ d = 2

now, a = \(\frac{ 3 }{ 2 } d\) X 2 [From (1)]

⇒ a = 3

Therefore, the A.P. is 3, 5, 7, 9, . . .

**Question 38. Find whether 0 (zero) is a term of the A.P. 40, 37, 34, 31, . . . or not.**

**Solution: **We know that, nth term = an = a + (n – 1)d

where “a” be the first term and “d” be the common difference.

Given: a = 40, d = -3

Suppose there is a term whose value is zero, a_{n}= 0

According to the question,

⇒ 0 = 40 + (n – 1)(-3)

⇒ 0 = 40 – 3n + 3

⇒ 3n = 43

⇒ n = \(\frac{ 43 }{ 3 }\) …. (1)

Here, n is the number of terms, so must be an integer.

Thus, there is no term where 0 (zero) is a term of the A.P. 40, 37, 34, 31,. . .

**Question 39. Find the middle term of the A.P. 213, 205, 197, . . . 37.**

**Solution:**

We know that, nth term ( a_{n} ) = a + (n – 1)d

where “a” be the first term and “d” be the common difference.

Given: a = 213,

d = -8

and a_{n} = 37

According to the question,

⇒ 37 = 213 + (n – 1)(-8)

⇒ 37 =213 – 8n+ 8

⇒ 8n = 221 – 37

⇒ an = 184

⇒ n=23 ….(1)

Therefore, the total number of terms are 23.

Let us say, there is an odd number of terms.

So, Middle term will be the 12th term.

a_{12 }=213 + (12 – 1)(-8)

a_{12 }= 213 – 88

= 125

Thus, the middle term of the A.P. 213, 205, 197, . . . , 37 is 125.

**Question 40. If the 5 ^{th} term of the A.P. is 31 and 25^{th} term is 140 more than the 5^{th} term, find the A.P.**

**Solution: **Let a be the first term and d be the common difference.

We know that, nth term ( a_{n} ) = a + (n – 1)d

According to question,

a_{6} = 31

⇒ a + ( 5 – 1 ) = 31

⇒ a + 4d = 31

⇒ a = 31 – 4d . . . .(1)Also, a_{25 }= 140 + a_{5}

⇒ a + ( 25 – 1 ) = 140 + 31

⇒ a + 24d = 171 . . . . (3)

On substituting the values of (1) in (2), we get

31 – 4d + 24d = 171

⇒ 20d = 171 – 31

⇒ 20d = 140

⇒ d = 7

⇒ a = 31 – 4 X 7 [From (1)]

⇒ a = 3

Thus, the A.P. obtained is 3, 10 , 17, 24, . .