## RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions Ex 11A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11A

**Other Exercises**

- RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11A
- RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11B
- RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11C
- RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D
- RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions MCQS

**Question 1.**

**Solution:**

(i) 9, 15, 21, 27, …

Here, 15 – 9 = 6,

21 – 15 = 6,

27 – 21 = 6

d = 6 and a = 9

Next term = 27 + 6 = 33

(ii) 11, 6, 1, -4, …

Here, 6 – 11 = -5,

1 – 6 = -5,

-4 – 1 = -5

d = -5 and a = 11

Next term = -4 – 5 = -9

**Question 2.**

**Solution:**

(i) AP is 9, 13, 17, 21, ……

Here, a = 9, d = 13 – 9 = 4

**Question 3.**

**Solution:**

**Question 4.**

**Solution:**

If the terms are in AP, then

a_{2} – a_{1} = a_{3} – a_{2} = …….

a_{2} = 3p + 1

a_{1} = 2p – 1

a_{3} = 11

⇒ (3p + 1) – (2p – 1) = 11 – (3p + 1)

⇒ 3p + 1 – 2p + 1 = 11 – 3p – 1

⇒ p + 2 = 10 – 3p

⇒ 4p = 8

⇒ p = 2

Then for p = 2, these terms are in AP.

**Question 5.**

**Solution:**

(i) AP is 5, 11, 17, 23, ……

Here, a = 5, d = 11 – 5 = 6

T_{n} = a (n – 1)d = 5 + (n – 1) x 6 = 5 + 6n – 6 = (6n – 1)

(ii) AP is 16, 9, 2, -5, ……

Here, a = 16 d = 9 – 16 = -7

T_{n} = a + (n – 1)d = 16 + (n – 1) (-7)

= 16 – 7n + 7 = (23 – 7n)

**Question 6.**

**Solution:**

nth term = 4n – 10

Substituting the value of 1, 2, 3, 4, …, we get

4n – 10

= 4 x 1 – 10 = 4 – 10 = -6

= 4 x 2 – 10 = 8 – 10 = -2

= 4 x 3 – 10 = 12 – 10 = 2

= 4 x 4 – 10 = 16 – 10 = 6

We see that -6, -2, 2, 6,… are in AP

(i) Whose first term = -6

(ii) Common difference = -2 – (-6) = -2 + 6 = 4

(iii) 16th term = 4 x 16 – 10 = 64 – 10 = 54

The common difference calculator takes the input values of sequence and difference and shows you the actual results.

**Question 7.**

**Solution:**

In AP 6, 10, 14, 18,…, 174

Here, a = 6, d= 10 – 6 = 4

nth or l = 174

T_{n} = a + (n – 1)d

⇒ 174 = 6 + (n – 1) x 4

⇒ 174 – 6 = (n – 1) x 4

⇒ n – 1 = \(\frac { 168 }{ 4 }\) = 42

n = 42 + 1 = 43

Hence, there are 43 terms in the given AP.

**Question 8.**

**Solution:**

In AP 41, 38, 35,…, 8

a = 41, d = 38 – 41 = -3, l = 8

Let l be the nth term

l = T_{n} = a + (n – 1) d

⇒ 8 = 41 + (n – 1)(-3)

⇒ 8 – 41 = (n – 1)(-3)

⇒ n – 1 = 11

⇒ n = 11 + 1 = 12

There are 12 terms in the given AP.

**Question 9.**

**Solution:**

the AP is 8, 15\(\frac { 1 }{ 2 }\) , 13, …, -47

There are 27 terms in the given AP.

**Question 10.**

**Solution:**

Let 88 be the nth term

Now, in AP 3, 8, 13, 18, …

a = 3, d = 8 – 3 = 5

T_{n} = a + (n – 1) d

88 = 3 + (n – 1)(5)

⇒ 88 – 3 = (n – 1) x 5

⇒ \(\frac { 88 }{ 5 }\) = n – 1

⇒ 17 = n – 1

n= 17 + 1 = 18

88 is the 18th term.

**Question 11.**

**Solution:**

In the AP 72, 68, 64, 60, …..

Let 0 be the nth term

Here, a = 72, d = 68 – 72 = -4

T_{n} = a + (n – 1)d

0 = 72 + (n – 1)(-4)

⇒ -72 = -4(n – 1)

⇒ n – 1 = 18

⇒ n = 18 + 1 = 19

0 is the 19th term.

**Question 12.**

**Solution:**

n = 13 + 1 = 14

3 is the 14th term.

**Question 13.**

**Solution:**

In the AP 21, 18, 15, ……

Let -81 is the nth term

a = 21, d = 18 – 21 = -3

T_{n} = a + (n – 1)d

⇒ -81 = 21 + (n – 1)(-3)

⇒ -81 – 21 = (n – 1)(-3)

⇒ -102 = (n – 1)(-3)

⇒ n – 1 = 34

n = 34 + 1 = 35

-81 is the 35th term

**Question 14.**

**Solution:**

In the given AP 3, 8, 13, 18,…

a = 3, d = 8 – 3 = 5

T_{20} = a + (n – 1)d = 3 + (20 – 1) x 5 = 3 + 19 x 5 = 3 + 95 = 98

The required term = 98 + 55 = 153

Let 153 be the nth term, then

T_{n} = a + (n – 1)d

⇒ 153 = 3 + (n – 1) x 5

⇒ 153 – 3 = 5(n – 1)

⇒ 150 = 5(n – 1)

⇒ n – 1 = 30

⇒ n = 30 + 1 = 31

Required term will be 31st term.

**Question 15.**

**Solution:**

AP is 5, 15, 25,…

a = 5, d = 15 – 5 = 10

T_{31} = a + (n – 1)d = 5 + (31 – 1) x 10 = 5 + 30 x 10 = 5 + 300 = 305

Now the required term = 305 + 130 = 435

Let 435 be the nth term, then

T_{n} = a + (n – 1)d

⇒ 435 = 5 + (n – 1)10

⇒ 435 – 5 = (n – 1)10

⇒ n – 1 = 43

⇒ n = 43 + 1 = 44

The required term will be 44th term.

**Question 16.**

**Solution:**

Let a be the first term and d be the common difference, then

**Question 17.**

**Solution:**

T_{16} = 6 + (16 – 1)7 = 6+ 15 x 7 = 6 + 105 = 111

**Question 18.**

**Solution:**

AP is 10, 7, 4, …, (-62)

a = 10, d = 7 – 10 = -3, l = -62

l = T_{n} = a + (n – 1)d

⇒ -62 = 10 + (n – 1) x (-3)

⇒ -62 – 10 = -3(n- 1)

-72 = -3(n – 1)

n = 24 + 1 = 25

Middle term = \(\frac { 25 + 1 }{ 2 }\) th = 13th term

T_{13} = 10 + (13 – 1)(-3) = 10+ 12 x (-3)= 10 – 36 = -26

**Question 19.**

**Solution:**

**Question 20.**

**Solution:**

AP is 7, 10, 13,…, 184

a = 7, d = 10 – 7 = 3, l = 184

nth term from the end = l – (n – 1)d

8th term from the end = 184 – (8 – 1) x 3 = 184 – 7 x 3 = 184 – 21 = 163

**Question 21.**

**Solution:**

AP is 17, 14, 11, …,(-40)

Here, a = 17, d = 14 – 17 = -3, l = -40

6th term from the end = l – (n – 1)d

= -40 – (6 – 1) x (-3)

= -40 – [5 x (-3)]

= -40 + 15

= -25

**Question 22.**

**Solution:**

Let 184 be the nth term of the AP

3, 7, 11, 15, …

Here, a = 3, d = 7 – 3 = 4

T_{n} = a + (n – 1)d

⇒ 184 = 3 + (n – 1) x 4

⇒ 184 – 3 = (n – 1) x 4

⇒ \(\frac { 181 }{ 4 }\) = n – 1

⇒ n = \(\frac { 181 }{ 4 }\) + 1 = \(\frac { 185 }{ 4 }\) = 46\(\frac { 1 }{ 4 }\)

Which is in fraction.

184 is not a term of the given AP.

**Question 23.**

**Solution:**

Let -150 be the nth term of the AP

11, 8, 5, 2,…

**Question 24.**

**Solution:**

Let nth of the AP 121, 117, 113,… is negative

**Question 25.**

**Solution:**

**Question 26.**

**Solution:**

Let a be the first term and d be the common difference of an AP

T_{n} = a + (n – 1)d

T_{7} = a + (7 – 1)d = a + 6d = -4 …(i)

T_{13} = a + 12d = -16 …..(ii)

Subtracting (i) from (ii),

6d = -16 – (-4) = -16 + 4 = -12

From (i), a + 6d = -4

a + (-12) = -4

⇒ a = -4 + 12 = 8

a = 8, d = -2

AP will be 8, 6, 4, 2, 0, ……

**Question 27.**

**Solution:**

Let a be the first term and d be the common difference of an AP.

T_{4} = a + (n- 1)d = a + (4 – 1)d = a + 3d

a + 3d = 0 ⇒ a = -3d

Similarly,

T_{25} = a + 24d and T_{11} = a + 10d = -3d + 24d = 21d

It is clear that T_{25} = 3 x T_{11}

**Question 28.**

**Solution:**

Given, a_{6} = 0

⇒ a + 5d = 0

⇒ a = -5 d

Now, a_{15} = a + (n – 1 )d

= a + (15 – 1)d = -5d + 14d = 9d

and a_{33} = a + (n – 1 )d = a + (33 – 1)d = -5d + 32d = 27d

Now, a_{33} : a_{12}

⇒ 27d : 9d

⇒ 3 : 1

a_{33} = 3 x a_{15}

**Question 29.**

**Solution:**

Let a be the first term and d be the common difference of an AP.

T_{n} = a + (n – 1)d

T_{4} = a + (4 – 1)d = a + 3d

a + 3d = 11 …(i)

Now, T_{5} = a + 4d

and T_{7} = a + 6d

Adding, we get T_{5} + T_{7} = a + 4d + a + 6d = 2a + 10d

2a + 10d = 34

⇒ a + 5d = 17 …(ii)

Subtracting (i) from (ii),

2d = 17 – 11 = 6

⇒ d = 3

Hence, common difference = 3

**Question 30.**

**Solution:**

Let a be the first term and d be the common difference of an AP.

**Question 31.**

**Solution:**

Let a be the first term and d be the common difference, then

**Question 32.**

**Solution:**

In an AP,

Let a be the first term and d be the common difference, then

**Question 33.**

**Solution:**

Let a be the first term and d be the common difference in an AP, then

**Question 34.**

**Solution:**

In an AP,

Let d be the common difference,

First term (a) = 5

Sum of first 4 terms

= a + a + d + a + 2d + a + 3d = 4a + 6d

Sum of next 4 terms

= a + 4d + a + 5d + a + 6d + a + 7d = 4a + 22d

According to the condition,

**Question 35.**

**Solution:**

Let a be the first term and d be the common difference in an AP, then

a = 1, d = 4

AP = 1, 5, 9, 13, 17, …

**Question 36.**

**Solution:**

In AP 63, 65, 67, …..

**Question 37.**

**Solution:**

Let first term of AP = a

and common difference = d

**Question 38.**

**Solution:**

Let a be the first term and d be the common difference, then

**Question 39.**

**Solution:**

Let a be the first term and d be the common difference, then

**Question 40.**

**Solution:**

Let a be the first term and d be the common difference, then

**Question 41.**

**Solution:**

Let a be the first term, d be the common difference, then

**Question 42.**

**Solution:**

Two-digit numbers are 10 to 99 and two digit numbers divisible by 6 will be

12, 18, 24, 30, …, 96

**Question 43.**

**Solution:**

Two digit numbers are 10 to 99 and

Two digit numbers which are divisible by 3 are

12, 15, 18, 21, 24, … 99

**Question 44.**

**Solution:**

Three digit numbers are 100 to 999 and numbers divisible by 9 will be

108, 117, 126, 999

**Question 45.**

**Solution:**

Numbers between 101 and 999 which are divisible both by 2 and 5 will be

110, 120, 130,…, 990

**Question 46.**

**Solution:**

Let number of from a rows are in the flower bed, then

**Question 47.**

**Solution:**

Total amount = ₹ 2800

and number of prizes = 4

Let first prize = ₹ a

Then second prize = ₹ a – 200

Third prize = a – 200 – 200 = a – 400

and fourth prize = a – 400 – 200 = a – 600

But sum of there 4 prizes are ₹ 2800

a + a – 200 + a – 400 + a – 600 = ₹ 2800

⇒ 4a – 1200 = 2800

⇒ 4a = 2800 + 1200 = 4000

⇒ a = 1000

First prize = ₹ 1000

Second prize = ₹ 1000 – 200 = ₹ 800

Third prize = ₹ 800 – 200 = ₹ 600

and fourth prize = ₹ 600 – 200 = ₹ 400

**Question 48.**

**Solution:**

The first term between 200 and 500 divisible by 8 is 208, and last term is 496.

So, first term (a) = 208

Common difference (d) = 8

Now, an = a + (n – 1 )d

⇒ 496 = 208 + (n – 1) x 8

⇒ (n – 1) = \(\frac { 288 }{ 8 }\)

⇒ n – 1 = 36

⇒ n = 36 + 1 = 37

Hence, there are 37 integers between 200 and 500 which are divisible by 8.

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11A are helpful to complete your math homework.

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