Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.2 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.2

Question 1.

Fill in the blanks in the following table, given that a is the first term, d the common difference and a_{n} the n^{th} term of the A.P :

Solution:

Question 2.

Find the

i) 30^{th} term of the A.P. : 10, 7, 4, …… (A.P. June 15)

ii) 11^{th} term of the A.P. : -3, \(\frac{-1}{2}\) ,2, …….

Solution:

i) Given A.P. = 10, 7, 4, ….

a_{1} = 10; d = a_{2} – a_{1} = 7 – 10 = – 3

a_{n} = a + (n – 1) d

a_{30} = 10 + (30 – 1)(-3)

= 10 + 29 × (- 3)

= 10 – 87

= -77

ii) Given A.P. = -3, \(\frac{-1}{2}\), 2,……..

a_{1} = -3; d = a_{2} – a_{1} = \(\frac{-1}{2}\) – (-3)

= \(\frac{-1}{2}\) + 3 = = \(\frac{-1+6}{2}\) = \(\frac{5}{2}\)

a_{n} = a + (n – 1)d

a_{11} = -3 + (11 – 1) × (\(\frac{5}{2}\))

= -3 + 10 × \(\frac{5}{2}\)

= -3 + 5 × 5

= -3 + 25 = 22

Question 3.

Find the respective terms for the following APs.

i) a_{1} = 2; a_{3} = 26, find a_{2}.

Solution:

Given : a_{1} = a = 2 —— (1)

a_{3} = a + 2d = 26 —— (2)

Equation (2) – equation (1)

⇒ (a + 2d) – a = 26 – 2

⇒ 2d = 24

d = \(\frac{24}{2}\) = 12

Now a_{2} = a + d = 2 + 12 = 14

ii) a_{2} = 13; a_{4} = 3, find a_{1}, a_{3}.

Solution:

Given : a_{2} = a + d = 13 —— (1)

a_{4} = a + 3d = 3 —- (2)

Solving equations (1) and (2);

Substituting, d = – 5 in equation (1) we get

a + (-5) = 13

∴ a = 13 + 5 = 18

i.e., a_{1} = 18

a_{3} = a + 2d = 18 + 2(-5)

= 18 – 10 = 8

iii) a_{1} = 5; a_{4} = 9\(\frac{1}{2}\), a_{2}, a_{3}.

Solution:

Given : a_{1} = a = 5 —– (1)

a_{4} = a + 3d = 9\(\frac{1}{2}\) —– (2)

Solving equations (1) and (2);

iv) a_{1} = – 4; a_{6} = 6, find a_{2}, a_{3}, a_{4}, a_{5}.

Solution:

Given : a_{1} = a = – 4 —– (1)

a_{6} = a + 5d = 6 —– (2)

Solving equations (1) and (2);

(-4) + 5d = 6 ⇒ 5d = 6 + 4

⇒ 5d = 10 ⇒ d = \(\frac{10}{5}\) = 2

Now, a_{2} = a + d = -4 + 2 = -2;

a_{3} = a + 2d = -4 + 2 × 2

= -4 + 4 = 0;

a_{4} = a + 3d = (-4) + 3 × 2

= -4 + 6 = 2;

a_{5} = a + 4d = -4 + 4 × 2

= -4 + 8 = 4

v) a_{2} = 38; a_{6} = -22, find a_{1}, a_{3}, a_{4}, a_{5}.

Solution:

Given : a_{2} = a + d = 38 —– (1)

a_{6} = a + 5d = -22 —– (2)

Subtracting (2) from (1) we get

Now substituting, d = -15 in equation (1), we get

a + (-15) = 38 ⇒ a = 38 + 15 = 53 Thus,

a_{1} = a = 53;

a_{3} = a + 2d = 53 + 2 × (-15) = 53 – 30 = 23;

a_{4} = a + 3d = 53 + 3 × (-15) = 53 – 45 = 8;

a_{5} = a + 4d = 53 + 4 × (-15) = 53 – 60 = -7

Question 4.

Which term of the AP :

3, 8, 13, 18, ……. is 78 ?

Solution:

Given : 3, 8, 13, 18,

Here a = 3; d = a_{2} – a_{1} = 8 – 3 = 5

Let ’78’ be the nth term of the given A.P.

∴ a_{n} = a + (n – 1) d

78 = 3 + (n – 1)5

78 = 3 + 5n – 5

⇒ 5n = 78 + 2 ⇒ n = \(\frac{80}{5}\) = 16

∴ 78 is the 16th term of the given A.P.

Question 5.

Find the number of terms in each of the following APs :

i) 7, 13, 19, ……., 205 (A.P. Mar. ’16)

Solution:

Given : A.P : 7, 13, 19, ………

Here a_{1} = a = 7; d = a_{2} – a_{1} = 13 – 7 = 6

Let 205 be the n^{th} term of the given A.P.

Then, a_{n}= a + (n – 1) d

205 = 7 + (n – 1) 6

⇒ 205 = 7 + 6n – 6

⇒ 205 = 6n + 1

⇒ 6n = 205 – 1 = 204

∴ n = \(\frac{204}{6}\) = 34

∴ 34 terms are there.

ii) 18, 15\(\frac{1}{2}\), 13, ………, -47.

Solution:

Given A.P = 18, 15\(\frac{1}{2}\), 13, ….

Here a_{1} = a = 18

d = a_{2} – a_{1} = 15\(\frac{1}{2}\) – 18 = -2\(\frac{1}{2}\) = –\(\frac{5}{2}\)

Let ‘-47’ be the n^{th} term of the given A.P.

a_{n} = a + (n – 1) × d

⇒ -47 = \(\frac{18 \times 2+(n-1)(-5)}{2}\)

⇒ – 94 = 36 – 5n + 5

⇒ 5n = 94 + 41 ⇒ n = \(\frac{135}{5}\) = 27

i.e., 27 terms are there.

Question 6.

Check whether, -150 is a term of the AP : 11, 8, 5, 2, ….

Solution:

Given A.P. = 11, 8, 5, 2,

Here a = a_{2} – a_{1} = 8 – 11 = -3

If possible, take -150 as the n,sup>th term of the given A.P.

a_{n} = a + (n – 1) d

-150 = 11 + (n – 1) × (- 3)

⇒ -150 = 11 – 3n + 3

⇒ 14 – 3n = -150

⇒ 3n = 14 + 150 = 164

∴ n = \(\frac{164}{3}\) = 54\(\frac{2}{3}\)

Here n is not an integer.

∴ -150 is not a term of the given A.P.

Question 7.

Find the 31^{st} term of an A.P. whose 11^{th} term is 38 and the 16th term is 73.

Solution:

Given : An A.P. whose

Substituting d = 7 in the equation (1) we get,

a + 10 × 7 = 38

⇒ a + 70 = 38

⇒ a = 38 – 70 = – 32

Now, the 31^{st} term = a + 30d

= (- 32) + 30 × 7

= – 32 + 210 = 178

Question 8.

If the 3^{rd} and the 9^{th} terms of an A.P are 4 and – 8 respectively, which term of this A.P is zero ?

Solution:

Given : An A.P whose

Substituting d = – 2 in equation (1) we get a + 2 × (- 2) = 4

⇒ a – 4 = 4

⇒ a = 4 + 4 = 8

Let n^{th} term of the given A.P be equal to zero.

a_{n} = a + (n – 1) d

⇒ 0 = 8 + (n – 1) × (-2)

⇒ 0 = 8 – 2n + 2

⇒ 10 – 2n = 0

⇒ 2n = 10 and n = \(\frac{10}{2}\) = 5

∴ The 5^{th} term of the given A.P is zero.

Question 9.

The 17^{th} term of an A.P exceeds its 10^{th} term by 7. Find the common difference.

Solution:

Given an A.P. in which a_{17} = a_{10} + 7

⇒ a_{17} – a_{10} = 7

We know that a_{n} = a + (n – 1) d

Question 10.

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms ?

Solution:

Let the first A.P be : a, a + d, a + 2d,

Second A.P be : b, b + d, b + 2d, b + 3d,….

Also, general term, a_{n} = a + (n – 1) d

Given that, a_{100} – b_{100} = 100

⇒ a + 99d – (b + 99d) = 100

⇒ a – b = 100

Now the difference between their 1000 terms,

∴ The difference between their 1000th terms is (a – b) = 100.

Note : If the common difference for any two A.Ps are equal then difference n^{th} terms of two A.Ps is same for all natural values of n.

Question 11.

How many three-digit numbers are divisible by 7 ?

Solution:

The list of three digit numbers divisible by 7 , is,

105, 112, 119, ………, 994

Here a = 105; d = 7; a_{n} = 994

a_{n} = a + (n – 1) d

994 = 105 + (n – 1) 7

(n – 1)7 = 889

n -1 = \(\frac{889}{7}\)

n – 1 = 127

∴ n = 127 + 1 = 128

There are 128 three digit numbers which are divisible by 7.

Question 12.

How many multiples of 4 lie between 10 and 250 ?

Solution:

Given numbers : 10 to 250

∴ Multiples of 4 between 10 and 250 are

First term : 10 + (4 – 2) = 12

Last term : 250 – 2 = 248

∴ 12, 16, 20, 24, ………, 248

a = a_{1} = 12; d = 4: a_{n} = 248

a_{n} = a + (n – 1) d

248 = 12 + (n – 1) × 4

⇒ (n – 1) 4 = 248 – 12

⇒ n – 1 = \(\frac{236}{4}\) = 59

∴ n = 59 + 1 = 60

There are 60 numbers between 10 and 250 which are divisible by 4.

Question 13.

For what value of n, are the n^{th} terms of two APs : 63, 65, 67, … and 3, 10, 17, …. equal ?

Solution:

Given : The first A.P is 63, 65, 67, ….

where a = 63, d = a_{2} – a_{1}

⇒ d = 65 – 63 = 2

and the second A.P is 3, 10, 17, ….

where a = 3, d = a_{2} – a_{1} = 10 – 3 = 7

Suppose the n^{th} terms of the two A.Ps are equal, where a_{n} = a + (n – 1) d

⇒ 63 + (n -1) 2 = 3 + (n – 1) 7

⇒ 63 + 2n – 2 = 3 + 7n – 7

⇒ 61 + 2n = 7n – 4

⇒ 7n – 2n = 61 +4

⇒ 5n = 65

⇒ n = \(\frac{65}{5}\) = 13

∴ 13^{th} terms of the two A.Ps are equal.

Question 14.

Determine the AP whose third term is 16 and the 7^{th} term exceeds the 5^{th} term by 12.

Solution:

Given : An A.P in which

a_{3} = a + 2d = 16 —– (1)

and a_{7} = a_{5} + 12

i.e., a + 6d = a + 4d + 12

⇒ 6d – 4d = 12

⇒ 2d = 12

⇒ d = \(\frac{12}{2}\) = 6

Substituting d = 6 in equation (1) we get

a + 2 × 6 = 16

⇒ a = 16 – 12 = 4

∴ The series/A. P is a, a + d, a + 2d, a + 3d, …

⇒ 4, 4 + 6, 4 + 12, 4 + 18, …

⇒ A.P: 4, 10, 16, 22,….

Question 15.

Find the 20^{th} term from the end of the AP : 3, 8, 13,…….., 253.

Solution:

Given : An A.P : 3, 8, 13, …., 253

Here a = a_{1} = 3

d = a_{2} – a_{1} = 8 – 3 = 5

a_{n} = 253, where 253 is the last term

a_{n} = a + (n – 1) d

∴ 253 = 3 + (n – 1) 5

⇒ 253 = 3 + 5n – 5

⇒ 5n = 253 + 2

⇒ n = \(\frac{255}{5}\) = 51

∴ The 20^{th} term from the other end would be

1 + (51 – 20) = 31 + 1 = 32

∴ a_{32} = 3 + (32 – 1) × 5

= 3 + 31 × 5

= 3 + 155

= 158

Question 16.

The sum of the 4^{th} and 8^{th} terms of an AP is 24 and the sum of the 6^{th} and 10^{th} terms is 44. Find the first three terms of the (A.P. Mar. ’15)

Solution:

Given an A.P in which

a_{4} + a_{8} = 24 ⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24

⇒ a + 5d = 12 —– (1)

and a_{6} + a_{10} = 44

⇒ a + 5d + a + 9d = 44

⇒ 2a + 14d = 44

⇒ a + 7d = 22 —– (2)

Also a + 5d = 12

⇒ a + 5(5) = 12

⇒ a + 25 = 12

⇒ a = 12 – 25 = -13

∴ The A.P is a , a + d, a + 2d, ……..

i.e., – 13, (- 13 + 5), (- 13 + 2 × 5)

⇒ -13, -8, -3, ………

Question 17.

Subba Rao started work in 1995 at an annual salary of ₹ 5,000 and received an increment of ₹ 200 each year. In which year did his income reach Rs. 7000?

Solution:

Given : Salary of Subba Rao in 1995 = ₹ 5000

Annual increment = ₹ 200

i.e., His salary increases by Rs. 200 every year.

Clearly 5000, 5200, 5400, ……… forms an A.P in which a = 5000 and d = 200.

Now suppose that his salary reached ₹ 7000

after x – years,

i.e., a_{n} = 7000

But, a_{n} = a + (n – 1) d

7000 = 5000 + (n – 1) 200

⇒ 7000 – 5000 = (n – 1) 200

⇒ n – 1 = \(\frac{2000}{200}\)

= 10

⇒ n = 10 + 1 = 11

∴ In 11^{th} year his salary reached ₹ 7000.