CBSE 10th Maths Term 2 2022: Important 2 Marks Questions With Solution

Chapter5arithmeticProgression

Chapter – 5: Arithmetic Progressions

Que 1. How many two digits numbers are divisible by 3?

Solution: First two digit number divisible by 3 = 12

Last two digit number divisible by 3 = 99

An arithmetic progression (A.P) is a progression in which the difference between two consecutive terms is constant.

A.P = 12,15,18,…,99

here

First term (a) = 12

Common difference (d) = 3

Let us consider there are n numbers then

an = 99

a + (n – 1)d = 99

12 + (n – 1)3 = 99

12 + 3n – 3 = 99

n = 29+1

n = 30

∴ Two digit numbers divisible by 3 = 30.

Que 2. In an AP,  if the common difference (d) = -4 and the seventh term (a7) is 4, then find the the first term.

Solution: we know,a= a + (n−1)d

so, a​= a + (7−1)d

or, 4 = a + 6 (−4)

or, 24 + 4 = a

∴ a = 28

Que 3. Which term of AP 8,14,20,26…will be 72 more than its 41st term?

Solution: 

Given :  

8, 14, 20, 26, ..

first term , a = 8 , common difference , d = 14 – 8 = 6

nth term = 72 + a41

a + (n -1)d = 72 + a41

[nth term = a + (n -1)d]

8 + (n – 1) 6 = 72 + a + (41 – 1) d

8 + 6n – 6 = 72 + 8 + 40 ×  6

2 + 6n  = 72 + 8 + 240

2 + 6n   = 80 + 240

2 + 6n  = 320

6n = 320 – 2

6n = 318

n = 318/6

n = 53

Hence, 53rd term of the A.P   8, 14, 20, 26, … will be 72 more than its 41st term.

Que 4. Write the nth term of A.P. 1/m, (1+m)/m, (1+2m)/m, ……..?

Solution: We know that,

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Que 5. Find the middle term of 6, 13, 20… 216?

Solution: The given AP is 6,13,20,……..,216.

First term, a = 6

Common difference d = 13-6 = 7

Suppose these are n terms in the given AP. Then,

79872a96c9

Hence, the middle term of the given AP is 111.

Que 6. The 8th term of AP is zero. Prove that 38th term is triple of 18th term.

Solution: T= a + 7d = 0

⇒ a = −7d

T38​ = a + 37d

= −7d + 37d

= 30d

T18​ = a + 17d

= −7d + 17d

= 10d.

Clearly T38​=3(T18​).

Que 7. Find the sum of all two digit positive odd numbers?

Solution: All two-digit odd positive numbers are 11, 13, 15, 17, …., 99. which are in AP with a = 11, d = 2, l = 99

Let the number of terms be n.

an​ = 99

a + (n − 1)d = 99

11 + (n − 1) × 2 = 99

n = 45

Sum of n terms is given by

Sn​ = n/2​(a + l)

Sn​ = 45/2​(11 + 99) = 2475

Therefore, the sum of all two-digit odd positive numbers is 2475.

Que 8. Three numbers are in AP and their sum is 21, find the middle number ?

Solution: Let the numbers be, 

a − d, a, a + d

Given,

a − d + a + a + d = 21

3a = 21

∴ a = 7 

Hence the middle number.

Que 9. If 7 times the 7th term of an AP is equal to 11 times its 11th term. Then find its 18th term.

Solution: Let a and d be first term and common difference respectively

According to the question,

7 times the 7th term of an AP is equal to 11 times its 11th term

⇒ 7a7 = 11a11

As we know, nth term of an AP is

an = a + (n – 1)d

where a = first term

an is nth term

d is the common difference

we have,

7(a + 6d) = 11(a + 10d)

⇒ 7a + 42d = 11a + 110d

⇒ 4a + 68d = 0

⇒ a + 17d = 0

⇒ a18 = 0

18th term of AP is 0

Que 10. The consecutive terms of an AP are 2, x, 26, find the value of x?

Solution: x = 14

Step-by-step explanation:

2, x, 26 are in AP.

Hence,

x – 2 = 26 – x

2x = 28

x = 14

Que 11. For what value of p is 2p+1, 13, 5p-3, are 3 consecutive terms of an AP?

Solution: Given 2p+1, 13 , 5p-3 are in AP.

To find: value of P

If a,b,c are in AP then we know that

2b = ( a+c)

Here a =  2p + 1

b = 13

c = 5p – 3

2 (13) = (2p+1+5p-3)

26 = 7p-2

26 + 2 = 7p

28 = 7p

p = 28/7 = 4

Therefore the value of P is 4.

Que 12. Which term of the AP: 3, 8, 13, 18… is 78?

Solution: The formula for nth term of an AP is aₙ = a + (n – 1) d

The given arithmetic progression is 3, 8, 13, 18, …

First term: a = 3

Second term: a + d = 8

Common difference: d = 8 – 3 = 5

an = a + (n – 1) d 

an = 78, n = ?

3 + (n – 1) 5 = 78

5(n – 1) = 78 – 3

n – 1 = 15

n = 16

78 is the 16th term of the given AP.

Que 13. Write the 5th term from the end of the AP 3,5,7,9 .. 201?

Solution:

Given :  

A.P. 3, 5, 7, 9, ….,201.

Let ‘l’ be the last term of an A.P

Here,  l = 201, a = 3, a2 = 5

d = a2 – a

d = 5 – 3 =  2

nth term from the end = l – (n – 1)d

5th term from the end = l – (n – 1)d

5th term from the end = 201 – (5 – 1) × 2

5th term from the end = 201 – 4 × 2

5th term from the end = 201 – 8

5th term from the end = 193

Hence, the 5th term from the end of the A.P is 193.

Que 14. From the given AP: 8, 10, 12… Find the sum of its last 10 terms if it has 60 terms?

Solution: A.P is 8,10,12………………….

Then first term is 8 and common difference is 2.

a​= a + (n−1)d.

Then a60​ = 8 + (60−1)2 = 8 + 59 × 2 = 126

Last 10 term =a51​, a52​………………a60

Then a51​ = 8 + (51 − 1)2 = 108

So total of last 10 terms = 10/2​ [108 × 2 + (10 − 1)2] = 216 + 81 = 234 × 5 = 1170

Que 15. Find the number of terms of an AP 5, 9, 13 …185.

Solution:

Given:

l = 185

a = 5

d = 4

To find:

n = ?

Solution:

l = a + (n – 1)d

185 = 5 + (n – 1)4

185 = 5 + 4n -4

185 = 1 + 4n

185 – 1 = 4n

184 = 4n

184/4 = n

46 = n

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