Practice Set 3.3

- First term and common difference of an A.P. are 6 and 3 respectively ; find S27. a = 6,…
- Find the sum of first 123 even natural numbers.
- Find the sum of all even numbers from 1 to 350.
- In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.…
- Complete the following activity to find the sum of natural numbers from 1 to 140 which…
- Sum of first 55 terms in an A.P. is 3300, find its 28th term.
- In an A.P. sum of three consecutive terms is 27 and their product is 504 find the…
- Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is…
- If the 9th term of an A.P. is zero then show that the 29th term is twice the 19th term.…

###### Practice Set 3.3

Question 1.

First term and common difference of an A.P. are 6 and 3 respectively ; find S27.

a = 6, d = 3, S27 = ?

Answer:

Given: First term a = 6

Common Difference d = 3

To find: S27 where n = 27

By using sum of nth term of an A.P. is

Where, n = no. of terms

a = first term

d = common difference

Sn = sum of n terms

Thus, Substituting given value in formula we can find the value of S27

Thus, S27 = 1215

Question 2.

Find the sum of first 123 even natural numbers.

Answer:

List of first 123 even natural number is

2,4,6,…….

Where first term a = 2

Second term t1 = 4

Third term t2 = 6

Thus, common difference d = t2 – t1 = 6 – 4 = 2

n = 123

By using sum of nth term of an A.P. is

Where, n = no. of terms

a = first term

d = common difference

Sn = sum of n terms

Thus, Substituting given value in formula we can find the value of Sn

Thus, Sn = 15252

Question 3.

Find the sum of all even numbers from 1 to 350.

Answer:

List of even natural number between 1 to 350 is

2,4,6,…….348

Where first term a = 2

Second term t1 = 4

Third term t2 = 6

Thus, common difference d = t2 – t1 = 6 – 4 = 2

tn = 348 (As we have to find the sum of even numbers between 1 and 350 therefore excluding 350 )

Now, By using nth term of an A.P. formula

tn = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

tn = nth terms

we can find value of “n” by substituting all the value in formula we get,

⇒ 348 = 2 + (n – 1) × 2

⇒ 348 – 2 = 2(n – 1)

⇒ 346 = 2(n – 1)

⇒ n = 173 + 1 = 174

Now, By using sum of nth term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

Sn = sum of n terms

Thus, Substituting given value in formula we can find the value of Sn

Thus, S174 = 30,450

Question 4.

In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.

Answer:

Given: t19 = 52 and t38 = 128

To find: value of “a” and “d”

Using nth term of an A.P. formula

tn = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

tn = nth terms

we will find value of “a” and “d”

Let, t19 = a + (19 – 1) d

⇒ 52 = a + 18 d …..(1)

t38 = a + (38 – 1) d

⇒ 128 = a + 37 d …..(2)

Subtracting eq. (1) from eq. (2), we get,

⇒ 128 – 52 = (a – a) + (37 d – 18 d)

⇒ 76 = 19 d

Substitute value of “d” in eq. (1) to get value of “a”

⇒ 52 = a + 18 ×4

⇒ 52 = a + 72

⇒ a = 52 – 72 = – 20

Now, to find value of S56 we will using formula of sum of n terms

Where, n = no. of terms

a = first term

d = common difference

Sn = sum of n terms

Thus, Substituting given value in formula we can find the value of Sn

⇒S56 = 28 × [ – 40 + 55×4]

⇒S56 = 28 × [ – 40 + 220]

⇒S56 = 28 × 180 = 5040

Thus, S56 = 5040

Question 5.

Complete the following activity to find the sum of natural numbers from 1 to 140 which are divisible by 4.

Sum of numbers from 1 to 140, which are divisible by 4 =

Answer:

List of natural number divisible by 4 between 1 to 140 is

4,8,12,…….136

Where first term a = 4

Second term t1 = 8

Third term t2 = 12

Thus, common difference d = t2 – t1 = 12 – 8 = 4

tn = 136

Now, By using nth term of an A.P. formula

tn = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

tn = nth terms

we can find value of “n” by substituting all the value in formula we get,

⇒ 136 = 4 + (n – 1) × 4

⇒ 136 – 4 = 4(n – 1)

⇒ 132 = 4(n – 1)

⇒ n = 33 + 1 = 34

Now, By using sum of nth term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

Sn = sum of n terms

Thus, Substituting given value in formula we can find the value of S34

⇒S34 = 17 × [8 + 33×4]

⇒S34 = 17 × [8 + 132]

⇒S34 = 17 × 140 = 2380

Thus, S34 = 2380

Question 6.

Sum of first 55 terms in an A.P. is 3300, find its 28th term.

Answer:

Given: S55 = 3300 where n = 55

Now, By using sum of nth term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

Sn = sum of n terms

Thus, on substituting the given value in formula we get,

⇒ 3300 = 55 × [ a + 27d]

⇒ a + 27d = 60 ……(1)

We need to find value of 28th term i.e t28

Now, By using nth term of an A.P. formula

tn = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

tn = nth terms

we can find value of t28 by substituting all the value in formula we get,

⇒ t28 = a + (28 – 1) d

⇒ t28 = a + 27 d

From eq. (1) we get,

⇒ t28 = a + 27 d = 60

⇒ t28 = 60

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.)

Answer:

Let the first term be a – d

the second term be a

the third term be a + d

Given: sum of consecutive three term is 27

⇒ (a – d) + a + (a + d) = 27

⇒ 3 a = 27

Also, Given product of three consecutive term is 504

⇒ (a – d)× a × (a + d) = 504

⇒ (9 – d) × 9 × (9 + d) = 504 (since, a = 9)

⇒ 92 – d2 = 56 (since, (a – b)(a + b) = a2 – b2)

⇒ 81 – d2 = 56

⇒ d2 = 81 – 56 = 25

⇒ d = √25 = ± 5

Case 1:

Thus, if a = 9 and d = 5

Then the three terms are,

First term a – d = 9 – 5 = 4

Second term a = 9

Third term a + d = 9 + 5 = 14

Thus, the A.P. is 4, 9, 14

Case 2:

Thus, if a = 9 and d = – 5

Then the three terms are,

First term a – d = 9 – ( – 5) = 9 + 5 = 14

Second term a = 9

Third term a + d = 9 + ( – 5) = 9 – 5 = 4

Thus, the A.P. is 14, 9, 4

Question 8.

Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14.

(Assume the four consecutive terms in A.P. are a – d, a, a + d, a + 2d.)

Answer:

Let the first term be a – d

the second term be a

the third term be a + d

the fourth term be a + 2 d

Given: sum of consecutive four term is 12

⇒ (a – d) + a + (a + d) + (a + 2d) = 12

⇒ 4 a + 2d = 12

⇒ 2(2 a + d) = 12

⇒ 2a + d = 6 …..(1)

Also, sum of third and fourth term is 14

⇒ (a + d) + (a + 2d) = 14

⇒ 2a + 3d = 14 ……(2)

Subtracting eq. (1) from eq. (2) we get,

⇒(2a + 3d) – (2a + d) = 14 – 6

⇒2a + 3d – 2a – d = 8

⇒ 2d = 8

⇒ d = 4

Substituting value of “d” in eq. (1) we get,

⇒ 2a + 4 = 6

⇒ 2a = 6 – 4 = 2

⇒ a = 1

Thus, a = 1 and d = 4

Hence, first term a – d = 1 – 4 = – 3

the second term a = 1

the third term a + d = 1 + 4 = 5

the fourth term a + 2 d = 1 + 2×4 = 1 + 8 = 9

Thus, the A.P. is – 3, 1, 5, 9

Question 9.

If the 9th term of an A.P. is zero then show that the 29th term is twice the 19th term.

Answer:

Now, By using nth term of an A.P. formula

tn = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

tn = nth terms

Given: t9 = 0

⇒ t9 = a + (9 – 1)d

⇒ 0 = a + 8d

⇒ a = – 8d

To Show: t29 = 2× t19

Now,

⇒ t29 = a + (29 – 1)d

⇒ t29 = a + 28d

⇒ t29 = – 8d + 28d = 20 d (since, a = – 8d )

⇒ t29 = 20 d

⇒ t29 = 2 × 10 d ….(1)

Also,

⇒ t19 = a + (19 – 1)d

⇒ t19 = a + 18d

⇒ t19 = – 8d + 18d = 10 d (since, a = – 8d )

⇒ t19 = 10 d …..(2)

From eq. (1) and eq. (2) we get,

t29 = 2× t19