Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 3.3 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 3 Arithmetic Progression.

## Practice Set 3.3 Algebra 10th Std Maths Part 1 Answers Chapter 3 Arithmetic Progression

**Arithmetic Progression Practice Set 3.3 Question 1.
**First term and common difference of an A.P. are 6 and 3 respectively; find S

_{27}.

Solution:

**Arithmetic Progression Class 10 Practice Set 3.3 Question 2.
**Find the sum of first 123 even natural numbers.

Solution:

The even natural numbers are 2, 4, 6, 8,…

The above sequence is an A.P.

∴ a = 2, d = 4 – 2 = 2, n = 123

∴ The sum of first 123 even natural numbers is 15252.

**Practice Set 3.3 Question 3.
**Find the sum of all even numbers between 1 and 350.

Solution:

The even numbers between 1 and 350 are 2, 4, 6,…, 348.

The above sequence is an A.P.

∴ a = 2, d = 4 – 2 = 2, t

_{n}= 348

Since, t

_{n}= a + (n – 1)d

∴ 348 = 2 + (n – 1)2

∴ 348 – 2 = (n – 1)2

∴ 346 = (n – 1)2

∴ n – 1 = \(\frac { 346 }{ 2 } \)

∴ n – 1 = 173

∴ n = 173 + 1 = 174

Now, S

_{n}= \(\frac { n }{ 2 } \) [2a + (n – 1)d]

∴ S

_{174}= \(\frac { 174 }{ 2 } \) [2 (2) + (174 – 1)2]

= 87(4 + 173 × 2)

= 87(4 + 346)

= 87 × 350

∴ S

_{174}= 30450

∴ The sum of all even numbers between 1 and 350 is 30450.

**Arithmetic Progression 3.3 Question 4.
**In an A.P. 19

^{th}term is 52 and 38

^{th}term is 128, find sum of first 56 terms.

Solution:

For an A.P., let a be the first term and d be the common difference.

t

_{19}= 52, t

_{38}= 128 …[Given]

Since, t

_{n}= a + (n – 1)d

∴ t

_{19}= a + (19 – 1)d

∴ 52 = a + 18d

i.e. a + 18d = 52 …(i)

Also, t

_{38}= a + (38 – 1)d

∴ 128 = a + 37d

i.e. a + 37d = 128 …(ii)

Adding equations (i) and (ii), we get

**3 Arithmetic Progression Question 5.
**Complete the following activity to find the sum of natural numbers between 1 to 140 which are divisible by 4.

Solution:

**10th Algebra Practice Set 3.3 Question 6.
**Sum of first 55 terms in an A.P. is 3300, find its 28

^{th}term.

Solution:

For an A.P., let a be the first term and d be the common difference.

S

_{55}=3300 …[Given]

Since, Sn = \(\frac { n }{ 2 } \) [2a + (n – 1)d]

∴ S

_{55}= \(\frac { 55 }{ 2 } \) [2a + (55 – 1)d]

**Arithmetic Practice Set Question 7.
**In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms. (Assume that three consecutive terms in A.P. are a – d, a, a + d.)

Solution:

Let the three consecutive terms in an A.P. be

a – d, a and a + d.

According to the first condition,

a – d + a + a + d = 27

∴ 3a = 27

∴ a = \(\frac { 27 }{ 3 } \)

∴ a = 9 ….(i)

According to the second condition,

(a – d) a (a + d) = 504

∴ a(a

^{2}– d

^{2}) = 504

∴ 9(a

^{2}– d

^{2}) = 504 …[From (i)]

∴ 9(81 – d

^{2}) = 504

∴ 81 – d

^{2}= \(\frac { 504 }{ 9 } \)

∴ 81 – d

^{2}= 56

∴ d

^{2}= 81 – 56

∴ d

^{2}= 25

Taking square root of both sides, we get

d = ± 5

When d = 5 and a =9,

a – d 9 – 5 = 4

a = 9

a + d 9 + 5 = 14

When d = -5 and a = 9,

a – d = 9 – (-5) = 9 + 5 = 14

a = 9

a + d = 9 – 5 = 4

∴ The three consecutive terms are 4, 9 and 14 or 14, 9 and 4.

**10th Maths 1 Practice Set 3.3 Question 8.
**Find four consecutive terms in an A.P. whose sum is 12 and sum of 3

^{rd}and 4

^{th}term is 14. (Assume the four consecutive terms in A.P. are a – d, a, a + d, a + 2d.)

Solution:

Let the four consecutive terms in an A.P. be

a – d, a, a + d and a + 2d.

According to the first condition,

a – d + a + a + d + a + 2d = 12

∴ 4a + 2d =12

∴ 2(2a + d) = 12

∴ 2a + d = \(\frac { 12 }{ 2 } \)

∴ 2a + d = 6 …(i)

According to the second condition,

a + d + a + 2d = 14

∴ 2a + 3d = 14 …(ii)

Subtracting equation (i) from (ii), we get

∴ The four consecutive terms are -3,1,5 and 9.

**Math 1 Practice Set 3.3 Question 9.
**If the 9

^{th}term of an A.P. is zero, then show that the 29

^{th}term is twice the 19th term.

To prove: t

_{29}= 2t

_{19}

Proof:

For an A.P., let a be the first term and d be the common difference.

t

_{9}= 0 …[Given]

Since, tn = a + (n – 1)d

∴ t

_{9}= a + (9 – 1)d

∴ 0 = a + 8d

∴ a = -8d …(i)

Also, t

_{19 }= a + (19 – 1)d

= a + 18d

= -8d + 18d … [From (i)]

∴ t

_{19}= 10d …(ii)

and t

_{29}= a + (29 – 1)d

= a + 28d

= -8d + 28d …[From (i)]

∴ t

_{29}= 20d = 2(10d)

∴ t

_{29}= 2(t

_{19}) … [From (ii)]

∴ The 29

^{th}term is twice the 19th term.

**10 Class Math Part 1 Practice Set 3.3 Question 1.
**Find the sum of all odd numbers from 1 to 150. (Textbook pg, no. 71)

Solution:

Odd numbers from 1 to 150 are 1,3, 5, 7,…, 149

Here, difference between any two consecutive terms is 2.

∴ It is an A.P.

∴ a = 1, d = 2

Let us find how many odd numbers are there from 1 to 150, i.e. find the value of n if

t

_{n}= 149

t

_{n}= a + (n – 1)d

∴ 149 = 1 + (n – 1)2

∴ 149 – 1 = (n – 1)2

∴ \(\frac { 148 }{ 2 } \) = n – 1

∴ 74 = n – 1

∴ n = 74 + 1 = 75

ii. Now, let’s find the sum of 75 numbers

i. e. 1 + 3 + 5 + 7 + … + 149