## 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.**

**Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each.**

**(i) 9, 15, 21, 27, …..**

**(ii) 11, 6, 1, -4, ……**

**(iii) -1, , , , …………**

**(iv) √2, √8, √18, √32, ……..**

**(v) √20, √45, √80, √125, ……**

**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.**

**Find:**

**(i) the 20th term of the AP 9, 13, 17, 21, ……**

**(ii) the 35th term of the AP 20, 17, 14, 11, ……**

**(iii) the 18th term of the AP √2, √18 , √50, √98, ………**

**(iv) the 9th term of the AP , , , , …….**

**(v) the 15th term of the AP -40, -15, 10, 35, ……**

**Solution:**

(i) AP is 9, 13, 17, 21, ……

Here, a = 9, d = 13 – 9 = 4

**Question 3.**

**(i) Find the 37th term of the AP 6, 7, 9, 11 , ………..**

**(ii) Find the 25th term of the AP 5, 4, 4, 3, 3, ……….**

**Solution:**

**Question 4.**

**Find the value of p for which the numbers 2p – 1, 3p + 1, 11 are in AP. Hence, find the numbers. [CBSE 2017]**

**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.**

**Find the nth term of each of the following APs:**

**(i) 5, 11, 17, 23, ……**

**(ii) 16, 9, 2, -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.**

**If the nth term of a progression is (4n – 10) show that it is an AP. Find its**

**(i) first term,**

**(ii) common difference, and**

**(iii) 16th term.**

**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

**Question 7.**

**How many terms are there in the AP 6, 10, 14, 18, …. 174.?**

**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 = = 42

n = 42 + 1 = 43

Hence, there are 43 terms in the given AP.

**Question 8.**

**How many terms are there in the AP 41, 38, 35, …, 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.**

**How many terms are there in the AP is 8, 15 , 13, …, -47 ?**

**Solution:**

the AP is 8, 15 , 13, …, -47

There are 27 terms in the given AP.

**Question 10.**

**Which term of the AP 3, 8, 13, 18,… is 88?**

**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

⇒ = n – 1

⇒ 17 = n – 1

n= 17 + 1 = 18

88 is the 18th term.

**Question 11.**

**Which term of the AP 72, 68, 64, 60, ….. is 0?**

**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.**

**Which term of the AP , 1, 1, 1, ….. is 3?**

**Solution:**

n = 13 + 1 = 14

3 is the 14th term.

**Question 13.**

**Which term of the AP 21, 18, 15,… is -81?**

**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.**

**Which term of the AP 3, 8, 13, 18, …will be 55 more than its 20th term? [CBSE 2007C]**

**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.**

**Which term of the AP 5, 15, 25,… will be 130 more than its 31st term? [CBSE2006C]**

**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.**

**If the 10th term of an AP is 52 and 17th term is 20 more than its 13 th term, find the AP. [CBSE2009C]**

**Solution:**

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

**Question 17.**

**Find the middle term of the AP 6, 13, 20,…, 216. [CBSE 2015]**

**Solution:**

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

**Question 18.**

**Find the middle term of the AP 10, 7, 4, ………. (-62). [CBSE 2009C]**

**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 = th = 13th term

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

**Question 19.**

**Find the sum of two middle most terms of the AP , -1, , …, 4. [CBSE 2013C]**

**Solution:**

**Question 20.**

**Find the 8th term from the end of the AP 7, 10, 13, ……, 184. [CBSE 2005]**

**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.**

**Find the 6th term from the end of the AP 17, 14, 11, …, (-40). [CBSE 2005]**

**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.**

**Is 184 a term of the AP 3, 7, 11, 15, …..?**

**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

⇒ = n – 1

⇒ n = + 1 = = 46

Which is in fraction.

184 is not a term of the given AP.

**Question 23.**

**Is -150 a term of the AP 11, 8, 5, 2,…?**

**Solution:**

Let -150 be the nth term of the AP

11, 8, 5, 2,…

**Question 24.**

**Which term of the AP 121, 117, 113,… is its first negative term?**

**Solution:**

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

**Question 25.**

**Which term of the AP 20, 19, 18 , 17 , …………. is its first negative term?**

**Solution:**

**Question 26.**

**The 7th term of an AP is -4 and its 13th term is -16. Find the AP. [CBSE 2014]**

**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.**

**The 4th term of an AP is zero, Prove that its 25th term is triple its 11th term. [CBSE 2005]**

**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.**

**If the sixth term of an AP is zero then show that its 33rd term is three times its 15th term. [CBSE2017]**

**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.**

**The 4th term of an AP is 11. The sum of the 5th and 7th terms of this AP is 34. Find its common difference. [CBSE2015]**

**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.**

**The 9th term of an AP is -32 and the sum of its 11th and 13th terms is -94. Find the common difference of the AP.[CBSE2015]**

**Solution:**

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

**Question 31.**

**Determine the wth term of the AP whose 7th term is -1 and 16th term is 17. [CBSE2014]**

**Solution:**

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

**Question 32.**

**If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term. [CBSE 2012]**

**Solution:**

In an AP,

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

**Question 33.**

**If 10 times the 10th term of an AP is equal to 15 times the 15th term, show that its 25th term is zero.**

**Solution:**

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

**Question 34.**

**Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms. [CBSE 2012]**

**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.**

**The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th term, find the AP. [CBSE 2014]**

**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.**

**For what value of n, the nth terms of the arithmetic progressions 63, 65, 67, … and 3, 10, 17, … are equal? [CBSE 2008]**

**Solution:**

In AP 63, 65, 67, …..

**Question 37.**

**The 17th term of AP is 5 more than twice its 8th term. If the 11th term of the AP is 43, find its nth term. [CBSE 2012]**

**Solution:**

Let first term of AP = a

and common difference = d

**Question 38.**

**The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term. [CBSE 2013]**

**Solution:**

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

**Question 39.**

**The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP. [CBSE 2013]**

**Solution:**

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

**Question 40.**

**If the pth term of an AP is q and its qth term is p then show that its (p + q)th term is zero.**

**Solution:**

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

**Question 41.**

**The first and last terms of an AP are a and l respectively. Show that the sum of the nth term from the beginning and the nth term from the end is (a + l).**

**Solution:**

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

**Question 42.**

**How many two-digit numbers are divisible by 6? [CBSE 2011]**

**Solution:**

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

12, 18, 24, 30, …, 96

**Question 43.**

**How many two-digit numbers are divisible by 3? [CBSE 2012]**

**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.**

**How many three-digit numbers are divisible by 9? [CBSE 2013]**

**Solution:**

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

108, 117, 126, 999

**Question 45.**

**How many numbers are there between 101 and 999, which are divisible by both 2 and 5? [CBSE 2014]**

**Solution:**

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

110, 120, 130,…, 990

**Question 46.**

**In a flower bed, there are 43 rose plants in the first row, 41 in the second, 39 in the third, and so on. There are 11 rose plants in the last row. How many rows are there in the flower bed?**

**Solution:**

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

**Question 47.**

**A sum of ₹ 2800 is to be used to award four prizes. If each prize after the first is ₹ 200 less than the preceding prize, find the value of each of the prizes.**

**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.**

**Find how many integers between 200 and 500 are divisible by 8. [CBSE 2017]**

**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) =

⇒ 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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