Arithmetic Progression — Test Yourself

If A, B and C are three arithmetic progressions (APs) as given below :

A = 2, 4, 6, 8, ........ upto n terms

B = 3, 6, 9, 12, ...... upto n terms

C = 0, 4, 8, 12, ...... upto n terms, then

out of A + B, A - C, C - B and B - A which is/are A.P. ?

1. A + B

2. A - C

3. C - B

4. All are A.P.

Answer

Given,

A = 2, 4, 6, 8, ........ upto n terms

B = 3, 6, 9, 12, ...... upto n terms

C = 0, 4, 8, 12, ...... upto n terms, then

A + B = 2 + 3, 4 + 6, 6 + 9, 8 + 12, ........

= 5, 10, 15, 20, ..........

The above list is an A.P. with common difference = 5.

A - C = 2 - 0, 4 - 4, 6 - 8, 8 - 12, ......

= 2, 0, -2, -4, .........

The above list is an A.P. with common difference = -2.

C - B = 0 - 3, 4 - 6, 8 - 9, 12 - 12,.........

= -3, -2, -1, 0, .........

The above list is an A.P. with common difference = 1.

Hence, Option 4 is the correct option.

In an A.P. a = -36, d = 18 and l = 36, then n is :

1. 10

2. 5

3. 15

4. 20

Answer

Let nth be the last term.

By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ 36 = -36 + (n - 1)18

⇒ 36 = -36 + 18n - 18

⇒ 36 + 36 + 18 = 18n

⇒ 18n = 90

⇒ n = $\dfrac{90}{18}$ = 5.

Hence, Option 2 is the correct option.

Do the numbers 1², 5², 7², 73 ..... form an A.P. ? If yes, its next term will be :

1. Yes, 11²

2. No

3. Yes, 97

4. Yes, 24

Answer

Given,

List = 1², 5², 7², 73 .....

= 1, 25, 49, 73, ........

Here,

a₂ - a₁ = 25 - 1 = 24

a₃ - a₂ = 49 - 25 = 24.

Since, common difference between consecutive terms are equal.

Hence, it forms an A.P.

Next term = 73 + d = 73 + 24 = 97.

Hence, Option 3 is the correct option.

The sum of first 10 even natural numbers is :

1. 120

2. 110

3. 65

4. 120

Answer

First 10 even natural numbers are : 2, 4, 6, 8, 10, .... upto 10 terms.

The above list is an A.P. with,

First term (a) = 2

Common difference (d) = 2

By formula,

Sum of n terms = $\dfrac{n}{2}$[2a + (n - 1)d]

Substituting values we get :

$= \dfrac{10}{2}[2 \times 2 + (10 - 1) \times 2] \\[1em] = 5 \times [4 + 9 \times 2] \\[1em] = 5 \times [4 + 18] \\[1em] = 5 \times 22 \\[1em] = 110.$

Hence, Option 2 is the correct option.

For the given numbers $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}............$

Assertion (A):

To find whether these terms form an A.P. or not. Express each term as the product of a natural number and $\sqrt{3 }$ i.e.,

$\sqrt{3} = 1 \times \sqrt{3}, \sqrt{12} = 2\sqrt{3}, \sqrt{27} = 3\sqrt{3}, \sqrt{48} = 4\sqrt{3}$, etc.

Reason (R): Since, for the given number difference between the consecutive term is same. It is an A.P.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Given, sequence

$\Rightarrow \sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}............\\[1em] \Rightarrow \sqrt{3}, \sqrt{4 \times 3}, \sqrt{9 \times 3}, \sqrt{16 \times 3}............\\[1em] \Rightarrow \sqrt{3}, \sqrt{4} \times \sqrt{3}, \sqrt{9} \times \sqrt{3}, \sqrt{16} \times \sqrt{3}............\\[1em] \Rightarrow \sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3}............$

Difference between first term and second term = $2\sqrt{3} - \sqrt{3} = \sqrt{3}$

Difference between second term and third term = $3\sqrt{3} - 2\sqrt{3} = \sqrt{3}$

So, the first term = $\sqrt{3}$, common difference = $\sqrt{3}$

So, assertion is true.

Since, for the given number difference between the consecutive term is same. Its an A.P., means reason is true and it clearly explain assertion.

Hence, option 3 is the correct option.

5, 8, 11, 14, ............... are in AP.

Assertion (A): $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7,$ ............... are also in AP.

Reason (R): If each term of a given A.P. is divided by the same non zero number, the resulting sequence is an A.P..

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Given, 5, 8, 11, 14, ............... are in AP.

Here, first term = 5, common difference = 8 - 5 = 11 - 8 = 3

Now, new sequence : $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7,.................$

The above sequence is found by dividing 5, 8, 11, 14, ............... the sequence by 2.

In new sequence,

Here,

Difference between second and first term = $4 - \dfrac{5}{2} = \dfrac{8 - 5}{2} = \dfrac{3}{2}$

Difference between third and second term = $\dfrac{11}{2} - 4 = \dfrac{11 - 8}{2} = \dfrac{3}{2}$

Difference between fourth and third term = $7 - \dfrac{11}{2} = \dfrac{14 - 11}{2} = \dfrac{3}{2}$

So, the common difference is same, means the given sequence is also in A.P..

So, Assertion is true.

The sequence $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7,.................$ this sequence is found by each term of the A.P. 5, 8, 11, 14, ............... is divided by 2.

If each term of a given A.P. is divided by the same non-zero number, the resulting sequence is an A.P.

So, Reason is true.

Hence, option 3 is the correct option.

An A.P. with 3^rd term = -8 and 9^th term = 4.

Assertion (A): Common difference = -2.

Reason (R): If first term of the A.P. is a, then (a + 8d) - (a + 2d) = -8 - 4.

1. A is true, R is false.

2. Both A and R are false.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Given, in an A.P. 3^rd term = -8 and 9^th term = 4.

Let a be the first term of the A.P. and d be the common difference.

By formula :

⇒ a_n = a + (n - 1)d

⇒ a₃ = a + (3 - 1)d

⇒ -8 = a + 2d .................(1)

⇒ a₉ = a + (9 - 1)d

⇒ 4 = a + 8d .................(2)

Subtracting equation (1) from (2), we get :

⇒ (a + 8d) - (a + 2d) = 4 - (-8)

⇒ a + 8d - a - 2d = 4 + 8

⇒ 6d = 12

⇒ d = $\dfrac{12}{6}$

⇒ d = 2.

According to assertion d = -2, which is incorrect.

So, assertion (A) is false.

From above calculation, we get :

⇒ (a + 8d) - (a + 2d) = 4 - (-8)

⇒ (a + 8d) - (a + 2d) = 12

According to reason,

⇒ (a + 8d) - (a + 2d) = -8 - 4

⇒ (a + 8d) - (a + 2d) = -12, which is incorrect.

So, reason (R) is false.

Hence, option 2 is the correct option.

The n^th term of a sequence = 5n² - 3.

Statement (1): The sequence is an A.P.

Statement (2): If the n^th term of a sequence is not linear, the sequence does not form an A.P.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, a_n = 5n² - 3

⇒ a₁ = 5(1)² - 3 = 5 - 3 = 2

⇒ a₂ = 5(2)² - 3 = 5 × 4 - 3 = 20 - 3 = 17

⇒ a₃ = 5(3)² - 3 = 5 × 9 - 3 = 45 - 3 = 42.

Difference between terms :

⇒ a₃ - a₂ = 42 - 17 = 25

⇒ a₂ - a₁ = 17 - 2 = 15

Since, the difference between consecutive terms is not equal. Thus, the sequence is not in an A.P.

So, statement 1 is false.

The general form of an A.P. is a_n = a + (n - 1)d, which is a linear expression in n.

So, if T_n is not linear in n, then the sequence cannot be an A.P.

So, statement 2 is true.

Hence, option 4 is the correct option.

The sum of first 10 term of an A.P. = 3 and the sum of its first 15 term = 16.

Statement (1): The sum of first five terms of the given AP equals to 16 - 3 = 13.

Statement (2): The sum of last 5 terms of the given AP equals to sum of first 15 term minus sum of first 10 terms.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

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

Using the formula; S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Given, the sum of first 10 term of an A.P. = 3

$\Rightarrow S_{10} = 3 \\[1em] \Rightarrow \dfrac{10}{2}[2a + (10 - 1)d] = 3\\[1em] \Rightarrow 5[2a + 9d] = 3\\[1em] \Rightarrow 10a + 45d = 3 .............................(1)$

Given, the sum of its first 15 term = 16

$\Rightarrow \dfrac{15}{2}[2a + (15 - 1)d] = 16\\[1em] \Rightarrow \dfrac{15}{2}[2a + 14d] = 16\\[1em] \Rightarrow \dfrac{15}{2} \times 2[a + 7d] = 16 \\[1em] \Rightarrow 15[a + 7d] = 16\\[1em] \Rightarrow 15a + 105d = 16 .............................(2)$

Subtract equation (1) from (2),

⇒ (15a + 105d) - (10a + 45d) = 16 - 3

⇒ 15a + 105d - 10a - 45d = 13

⇒ 5a + 60d = 13.

Sum of first 5 terms = S₅

$S_5 = \dfrac{5}{2}[2a + (5 - 1)d]\\[1em] = \dfrac{5}{2}[2a + 4d]\\[1em] = \dfrac{5}{2}[2(a + 2d)]\\[1em] = 5[a + 2d]\\[1em] = 5a + 10d.$

Since, value of 5a + 10d cannot be equal to 13.

So, statement 1 is false.

The sum of last 5 terms of the given AP = a₁₁ + a₁₂ + a₁₃ + a₁₄ + a₁₅

= [a + (11 - 1)d] + [a + (12 - 1)d] + [a + (13 - 1)d] + [a + (14 - 1)d] + [a + (15 - 1)d]

= (a + 10d) + (a + 11d) + (a + 12d) + (a + 13d) + (a + 14d)

= 5a + 60d.

We have calculated earlier that,

Sum of first 15 term - Sum of first 10 terms = 5a + 60d

Thus, we can say that the sum of last 5 terms of the given AP equals to sum of first 15 term - sum of first 10 terms.

So, statement 2 is true.

Hence, option 4 is the correct option.

The 6^th term of an A.P. is 16 and the 14^th term is 32. Determine the 36^th term.

Answer

Let the first term be a and common difference be d of the A.P.

Given,

⇒ a₆ = a + (6 - 1)d

⇒ a + 5d = 16 ........(i)

⇒ a₁₄ = a + (14 - 1)d

⇒ a + 13d = 32 ........(ii)

Subtracting (i) from (ii) we get,

⇒ a + 13d - (a + 5d) = 32 - 16

⇒ 8d = 16

⇒ d = 2.

Substituting value of d in (i) we get,

⇒ a + 5(2) = 16

⇒ a + 10 = 16

⇒ a = 6.

a₃₆ = a + (36 - 1)d = 6 + (35)(2) = 76.

Hence, 36^th term = 76.

If the third and the 9^th terms of an A.P. be 4 and -8 respectively, find which term is zero?

Answer

Let the first term be a and common difference be d of the A.P.

Given,

⇒ a₃ = a + (3 - 1)d

⇒ a + 2d = 4 ........(i)

⇒ a₉ = a + (9 - 1)d

⇒ a + 8d = -8 ........(ii)

Subtracting (i) from (ii) we get,

⇒ a + 8d - (a + 2d) = -8 - 4

⇒ 8d - 2d = -12

⇒ 6d = -12

⇒ d = -2.

Substituting value of d in (i) we get,

⇒ a + 2(-2) = 4

⇒ a - 4 = 4

⇒ a = 8.

Let n^th term be zero.

⇒ a_n = 0

⇒ a + (n - 1)d = 0

⇒ 8 + (n - 1)(-2) = 0

⇒ 8 - 2n + 2 = 0

⇒ 2n = 10

⇒ n = 5.

Hence, 5^th term = 0.

An A.P. consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th term of the A.P.

Answer

Let the first term be a and common difference be d of the A.P.

Given,

⇒ a₃ = a + (3 - 1)d

⇒ a + 2d = 12 ........(i)

⇒ a₅₀ = a + (50 - 1)d

⇒ a + 49d = 106 ........(ii)

Subtracting (i) from (ii) we get,

⇒ a + 49d - (a + 2d) = 106 - 12

⇒ 49d - 2d = 94

⇒ 47d = 94

⇒ d = 2.

Substituting value of d in (i) we get,

⇒ a + 2(2) = 12

⇒ a + 4 = 12

⇒ a = 8.

a₂₉ = a + (29 - 1)d

a + 28d = 8 + 28(2) = 8 + 56 = 64.

Hence, 29^th term = 64.

Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 57.

Answer

$S = \dfrac{n}{2}(a + l) \\[1em] = \dfrac{20}{2}(3 + 57) \\[1em] = 10 \times 60 \\[1em] = 600.$

Hence, sum of first 20 terms = 600.

How many terms of the series 18 + 15 + 12 + ....... when added together will give 45?

Answer

Let n terms be added to get sum of 45.

In above A.P., a = 18 and d = 15 - 18 = -3.

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 45 = \dfrac{n}{2}(2 \times 18 + (n - 1) \times (-3)) \\[1em] \Rightarrow 45 = \dfrac{n}{2}(36 - 3n + 3) \\[1em] \Rightarrow 45 = \dfrac{n}{2}(39 - 3n)\\[1em] \Rightarrow 90 = n(39 - 3n) \\[1em] \Rightarrow 3n(13 - n) = 90 \\[1em] \Rightarrow n(13 - n) = 30 \\[1em] \Rightarrow 13n - n^2 = 30 \\[1em] \Rightarrow n^2 - 13n + 30 = 0 \\[1em] \Rightarrow n^2 - 10n - 3n + 30 = 0 \\[1em] \Rightarrow n(n - 10) - 3(n - 10) = 0 \\[1em] \Rightarrow (n - 3)(n - 10) = 0 \\[1em] \Rightarrow n = 3, 10.$

Hence, n = 3 or 10.

Find the general term (n^th term) and 23^rd term of the sequence, 3, 1, -1, -3, .........

Answer

The above A.P. has common difference = 1 - 3 = -2 and a = 3.

a_n = a + (n - 1)d

= 3 + (n - 1)(-2)

= 3 - 2n + 2

= 5 - 2n.

a₂₃ = 5 - 2(23) = 5 - 46 = -41.

Hence, a_n = 5 - 2n and a₂₃ = -41.

Is -150 a term of 11, 8, 5, 2, .........?

Answer

In above sequence,

8 - 11 = -3 and 5 - 8 = -3.

Hence, above sequence is an A.P. with common difference = -3.

Let -150 be n^th term of the A.P.

∴ a_n = -150

⇒ a + (n - 1)d = -150

⇒ 11 + (-3)(n - 1) = -150

⇒ 11 - 3n + 3 = -150

⇒ 14 - 3n = -150

⇒ -3n = -164

⇒ n = $\dfrac{164}{3} = 54\dfrac{2}{3}$.

Since, no. of terms cannot be in fraction.

Hence, -150 is not a term of 11, 8, 5, 2, .........

How many multiples of 4 lie between 10 and 250?

Answer

Multiples of 4 lying between 10 and 250 are,

12, 16, 20, 24, .........., 248.

The above sequence is an A.P. with d = 4.

Let 248 be n^th term.

∴ a_n = 248

⇒ 12 + 4(n - 1) = 248

⇒ 12 + 4n - 4 = 248

⇒ 4n + 8 = 248

⇒ 4n = 240

⇒ n = 60.

Hence 60, multiples of 4 lie between 10 and 250.

The 25^th term of an A.P. exceeds its 9^th term by 16. Find its common difference.

Answer

Given,

⇒ a₂₅ - a₉ = 16

⇒ a + (25 - 1)d - (a + (9 - 1)d) = 16

⇒ a - a + 24d - 8d = 16

⇒ 16d = 16

⇒ d = 1.

Hence, common difference = 1.

If the n^th term of the A.P. 58, 60, 62, ...... is equal to the n^th term of the A.P. -2, 5, 12, ......., find the value of n.

Answer

n^th term of the A.P. 58, 60, 62, ......

In above sequence,

60 - 58 = 2 and 62 - 60 = 2.

Hence, above sequence is an A.P. with common difference = 2.

a_n = a + (n - 1)d

= 58 + (n - 1)2

= 58 + 2n - 2

= 2n + 56.

n^th term of the A.P. -2, 5, 12, .......

In above sequence,

5 - (-2) = 7 and 12 - 5 = 7.

Hence, above sequence is an A.P. with common difference = 7.

a_n = a + (n - 1)d

= -2 + (n - 1)7

= -2 + 7n - 7

= 7n - 9.

Since, n^th term of the A.P. 58, 60, 62, ...... is equal to the n^th term of the A.P. -2, 5, 12, .......

∴ 2n + 56 = 7n - 9

⇒ 7n - 2n = 56 + 9

⇒ 5n = 65

⇒ n = 13.

Hence, n = 13.

Which term of the A.P. 105, 101, 97, ........., is the first negative term?

Answer

Let the first negative term be n.

⇒ a_n < 0

⇒ a + (n - 1)d < 0

⇒ 105 + (n - 1)(-4) < 0

⇒ 105 - 4n + 4 < 0

⇒ 109 - 4n < 0

⇒ 4n > 109

⇒ n > $\dfrac{109}{4} = 27\dfrac{1}{4}.$

Given, n > $27\dfrac{1}{4}$.

So after 27^th term next terms will be negative so the first term to be negative will be 28^th.

Hence, 28^th term is the first negative term.

Divide 216 into three parts which are in A.P. and the product of two smaller parts is 5040.

Answer

Let numbers be a - d, a, a + d

Sum = 216

⇒ a - d + a + a + d = 216

⇒ 3a = 216

⇒ a = 72.

According to question,

⇒ (a - d)(a) = 5040

⇒ (72 - d)72 = 5040

⇒ 5184 - 72d = 5040

⇒ 72d = 5184 - 5040

⇒ 72d = 144

⇒ d = 2.

Numbers = (72 - 2), 72, (72 + 2)

Hence, numbers are 70, 72, 74.

Can 2n² - 7 be the n^th term of an A.P. Explain.

Answer

Given,

a_n = 2n² - 7

a₁ = 2(1)² - 7 = 2 - 7 = -5

a₂ = 2(2)² - 7 = 8 - 7 = 1

a³ = 2(3)² - 7 = 18 - 7 = 11

Here, a₂ - a₁ = 1 - (-5) = 6 and a₃ - a₂ = 11 - 1 = 10.

Since, difference is not common.

Hence, 2n² - 7 cannot be the n^th term of an A.P.

The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31^st term of this A.P.

Answer

$\Rightarrow S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 2100 = \dfrac{7}{2}[2 \times 20 + (7 - 1)d] \\[1em] \Rightarrow 2100 = \dfrac{7}{2}[40 + 6d] \\[1em] \Rightarrow 2100 = 7(20 + 3d) \\[1em] \Rightarrow 300 = 20 + 3d \\[1em] \Rightarrow 3d = 280 \\[1em] \Rightarrow d = \dfrac{280}{3}.$

We know that,

a_n = a + (n - 1)d

⇒ 31^st term = a₃₁

= a + (31 - 1)d

= $20 + 30 \times \dfrac{280}{3}$

= 20 + 2800

= 2820.

Hence, 31^st term of A.P. = 2820.

Find the sum of last 8 terms of the A.P.

-12, -10, -8, ........, 58.

Answer

Sum of last 8 terms of A.P. -12, -10, -8, ........, 58 = Sum of first 8 terms of A.P. 58, 56, 54, ........., -10, -12.

In A.P. 58, 56, 54, ........., -10, -12,

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{8}{2}[2 \times 58 + (8 - 1) \times (-2)] \\[1em] = 4[116 - 14] \\[1em] = 4 \times 102 \\[1em] = 408.$

Hence, sum = 408.

An A.P. consists of 57 terms of which 7^th term is 13 and the last term is 138. Find the 45^th term of this A.P.

Answer

Number of terms in A.P. (n) = 57

Given,

We know that,

a_n = a + (n - 1)d

⇒ a₇ = a + 6d

⇒ a + 6d = 13 ....(1)

⇒ a₅₇ = a + 56d

⇒ a + 56d = 138 ....(2)

Subtracting equation (1) from equation (2), we get:

⇒ (a + 56d) - (a + 6d) = 138 - 13

⇒ 50d = 125

⇒ d = $\dfrac{125}{50}$

⇒ d = 2.5

Substituting value of d in equation (1), we get :

⇒ a + 6(2.5) = 13

⇒ a + 15 = 13

⇒ a = 13 - 15

⇒ a = -2

45th term:

⇒ a₄₅ = a + (45 - 1)d

⇒ a₄₅ = -2 + (45 - 1)(2.5)

⇒ a₄₅ = -2 + 110

⇒ a₄₅ = 108.

Hence, the 45^th term of this A.P. = 108.

Ten times the tenth term of an A.P. is equal to fifteen times its fifteenth term. Find the twenty-fifth term of this A.P.

Answer

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

We know that,

a_n = a + (n - 1)d

Given,

Ten times the tenth term of an A.P. is equal to fifteen times its fifteenth term.

⇒ 10a₁₀ = 15a₁₅

⇒ 10(a + 9d) = 15(a + 14d)

⇒ 10a + 90d = 15a + 210d

⇒ 10a - 15a + 90d - 210d = 0

⇒ -5a - 120d = 0

⇒ 5a + 120d = 0

⇒ 5(a + 24d) = 0

⇒ a + 24d = 0

⇒ a = -24d .....(1)

25th term :

⇒ a₂₅ = a + (25 - 1)d

⇒ a₂₅ = -24d + (24)d [From equation 1]

⇒ a₂₅ = 0

Hence, the twenty-fifth term of this A.P. = 0

The sum of the 2^nd term and the 7^th term of an A.P. is 30. If its 15^th term is 1 less than twice of its 8^th term, find this A.P.

Answer

Let first term be a and common difference be d.

According to question,

⇒ a₂ + a₇ = 30

⇒ a + (2 - 1)d + a + (7 - 1)d = 30

⇒ a + d + a + 6d = 30

⇒ 2a + 7d = 30 .........(i)

Also,

⇒ 2a₈ - 1 = a₁₅

⇒ 2[a + (8 - 1)d] - 1 = a + (15 - 1)d

⇒ 2[a + 7d] - 1 = a + 14d

⇒ 2a + 14d - 1 = a + 14d

⇒ 2a - a - 1 = 14d - 14d

⇒ a - 1 = 0

⇒ a = 1.

Substituting value of a in (i) we get,

⇒ 2(1) + 7d = 30

⇒ 2 + 7d = 30

⇒ 7d = 28

⇒ d = 4.

A.P. = a, (a + d), (a + 2d),..........

= 1, (1 + 4), (1 + 2.4),.........

= 1, 5, 9,.......

Hence, A.P. = 1, 5, 9,........

Refer the given sequence 23, $21\dfrac{1}{2}$, 20, ....

(a) Find the general term of the given sequence.

(b) Which term is the last positive term in the sequence.

Answer

(a) Given,

23, $21\dfrac{1}{2}$, 20, .... is A.P.

a = 23

d = $21\dfrac{1}{2} - 23 = -1\dfrac{1}{2} = -\dfrac{3}{2}$.

We know that,

a_n = a + (n - 1)d

$\Rightarrow a_n = 23 - \Big(\dfrac{3}{2}\Big)(n - 1) \\[1em] \Rightarrow a_n = 23 - \Big(\dfrac{3n - 3}{2}\Big) \\[1em] \Rightarrow a_n = \Big(\dfrac{46 - 3n + 3}{2}\Big) \\[1em] \Rightarrow a_n = \Big(\dfrac{49 - 3n}{2}\Big)$

Hence, a_n = $\Big(\dfrac{49 - 3n}{2}\Big)$ .

(b) The last positive term occurs when :

⇒ a_n > 0

⇒ $\dfrac{49 - 3n}{2} \gt 0$

⇒ 49 - 3n > 0

⇒ 3n < 49

⇒ n < $\dfrac{49}{3}$

⇒ n < $16\dfrac{1}{3}$

Thus, 16th term will be the last positive term of the sequence.

⇒ a₁₆ = $\Big(\dfrac{49 - 3 \times 16}{2}\Big) = \dfrac{49 - 48}{2} = \dfrac{1}{2}$ = 0.5

Hence, the 16th term is the last positive term.