Arithmetic & Geometric Progression — Exercise 9.2

Find the A.P. whose nth term is 7 - 3n. Also find the 20th term.

Answer

Given, a_n = 7 - 3n     (Eq 1)

Putting n = 1, 2, 3, 4, .... in (Eq 1), we get

a₁ = 7 - 3 × 1 = 7 - 3 = 4,   a₂ = 7 - 3 × 2 = 7 - 6 = 1
a₃ = 7 - 3 × 3 = 7 - 9 = -2,   a₄ = 7 - 3 × 4 = 7 - 12 = -5, ...

∴ Series = 4, 1, -2, -5 .....

Putting n = 20, for 20th term we get,

a₂₀ = 7 - 3 × 20 = 7 - 60 = -53.

Hence, the A.P. is 4, 1, -2, -5 ..... and the 20th term is -53.

Find the indicated terms in each of the following A.P.s :

(i) 1, 6, 11, 16, ....; a₂₀

(ii) -4, -7, -10, -13, ...., a₂₅, a_n

Answer

(i) The given list of numbers is 1, 6, 11, 16, ...

Here, a₂ - a₁ = 6 - 1 = 5,     a₃ - a₂ = 11 - 6 = 5,

          a₄ - a₃ = 16 - 11 = 5

i.e. any term - preceding term = 5, a fixed number.
So, the given list of numbers forms an A.P. with a = 1 and d = 5.

∴ nth term = a_n = a + (n - 1)d = 1 + (n - 1)5 = 1 + 5n - 5 = 5n - 4.

Putting n = 20, we get
a₂₀ = 5 × 20 - 4 = 100 - 4 = 96.

Hence, a₂₀ = 96.

(ii) The given list of numbers is -4, -7, -10, -13, ....

Here, a₂ - a₁ = -7 - (-4) = -3,     a₃ - a₂ = -10 - (-7) = -3,

          a₄ - a₃ = -13 - (-10) = -3

i.e. any term - preceding term = -3, a fixed number.
So, the given list of numbers forms an A.P. with a = -4 and d = -3.

∴ nth term = a_n = a + (n - 1)d = -4 + (n - 1)(-3) = -4 - 3n + 3 = -3n - 1.

Putting n = 25, we get
a₂₅ = -3 × 25 - 1 = -75 - 1 = -76.

Hence, a₂₅ = -76 and a_n = -3n - 1.

Find the nth term and the 12th term of the list of numbers : 5, 2, -1, -4, ...

Answer

The given list of numbers is 5, 2, -1, -4, ....

Here, a₂ - a₁ = 2 - 5 = -3,     a₃ - a₂ = -1 - 2 = -3,

          a₄ - a₃ = -4 - (-1) = -3

i.e. any term - preceding term = -3, a fixed number.
So, the given list of numbers forms an A.P. with a = 5 and d = -3.

∴ nth term = a_n = a + (n - 1)d = 5 + (n - 1)(-3) = 5 - 3n + 3 = 8 - 3n.

Putting n = 12, we get
a₁₂ = 8 - 3(12) = 8 - 36 = -28.

Hence, a₁₂ = -28 and a_n = 8 - 3n.

If the common difference of an A.P. is -3 and the 18th term is -5, then find its first term.

Answer

The nth term of an A.P. is given by,

a_n = a + (n - 1)d.    (Eq 1)

Given , d = -3 and a₁₈ = -5.

Putting d = -3, a₁₈ = -5 and n = 18 in Eq 1 we get,

⇒ a_n = a + (n - 1)d
⇒ -5 = a + (18 - 1)(-3)
⇒ -5 = a + 17 × (-3)
⇒ -5 = a - 51
⇒ a = 51 - 5
⇒ a = 46.

Hence, the first term of A.P. is 46.

If the first term of an A.P. is -18 and its 10th term is zero, then find its common difference.

Answer

The nth term of an A.P. is given by,

a_n = a + (n - 1)d.    (Eq 1)

Given , a = -18 and a₁₀ = 0.

Putting a = -18, a₁₀ = 0 and n = 10 in Eq 1 we get,

⇒ a_n = a + (n - 1)d
⇒ 0 = -18 + (10 - 1)d
⇒ 0 = -18 + 9d
⇒ 9d = 18
⇒ d = 2.

Hence, the common difference of A.P. is 2.

Which term of the A.P.

(i) 3, 8, 13, 18, ... is 78 ?

(ii) $18, 15\dfrac{1}{2}, 13, ...$ is -47 ?

Answer

(i) The given A.P. is 3, 8, 13, 18 ....

Here, first term = a = 3 and common difference = d = 8 - 3 = 5.

Let 78 be nth term of the A.P., or a_n = 78

The nth term of A.P. is given by a_n = a + (n - 1)d

Putting value of d and a_n in above equation we get,

⇒ a_n = a + (n - 1)d
⇒ 78 = 3 + (n - 1)5
⇒ 78 = 3 + 5n - 5
⇒ 78 = 5n - 2
⇒ 5n = 78 + 2
⇒ 5n = 80
⇒ n = 16.

Hence, 78 is 16th term of the A.P.

(ii) The given A.P. is $18, 15\dfrac{1}{2}, 13, ...$

Here, first term = a = 18 and common difference = d = $\dfrac{31}{2} - 18 = \dfrac{31 - 36}{2} = -\dfrac{5}{2}$.

Let -47 be nth term of the A.P., or a_n = -47

The nth term of A.P. is given by a_n = a + (n - 1)d

Putting value of a, d and a_n in above equation we get,

$a_n = a + (n - 1)d \\[1em] \Rightarrow -47 = 18 + (n - 1)\Big(-\dfrac{5}{2}\Big) \\[1em] \Rightarrow -47 = 18 -\dfrac{5n}{2} + \dfrac{5}{2} \\[1em] \Rightarrow -47 - 18 = \dfrac{5 - 5n}{2} \\[1em] \Rightarrow -65 \times 2 = 5 - 5n \\[1em] \Rightarrow -130 = 5 - 5n \\[1em] \Rightarrow 5n = 130 + 5 \\[1em] \Rightarrow 5n = 135 \\[1em] \Rightarrow n = 27.$

Hence, -47 is 27th term of the A.P.

Check whether -150 is a term of the A.P. 11, 8, 5, 2, ...

Answer

Here, a₂ - a₁ = 8 - 11 = -3,     a₃ - a₂ = 5 - 8 = -3,

          a₄ - a₃ = 2 - 5 = -3

i.e. any term - preceding term = -3, a fixed number.
So, the given list of numbers forms an A.P. with a = 11 and d = -3.

Now we want to find whether there exists a natural number n for which a_n = -150

⇒ a + (n - 1)d = -150
⇒ 11 + (n - 1)(-3) = -150
⇒ 11 - 3n + 3 = -150
⇒ 14 - 3n = -150
⇒ 3n = 14 + 150
⇒ 3n = 164
⇒ n = $54\dfrac{2}{3}$, which is not a natural number.

Hence, there is no term in the given list of numbers which is -150.

Find whether 55 is a term of the A.P. 7, 10, 13, .... or not. If yes find which term is it.

Answer

Here, a₂ - a₁ = 10 - 7 = 3,     a₃ - a₂ = 13 - 10 = 3,

i.e. any term - preceding term = 3, a fixed number.
So, the given list of numbers forms an A.P. with a = 7 and d = 3.

Now we want to find whether there exists a natural number n for which a_n = 55

⇒ a + (n - 1)d = 55
⇒ 7 + (n - 1)3 = 55
⇒ 7 + 3n - 3 = 55
⇒ 4 + 3n = 55
⇒ 3n = 51
⇒ n = 17.

Hence, 55 is the 17th term of the A.P. 7, 10, 13 .....

Find the 20th term from the last term of the A.P. 3, 8, 13, ..., 253.

Answer

Here, common difference = d = 8 - 3 = 5 and last term = l = 253.

We know that the nth term from last is given by the formula:
nth term from end = l - (n - 1)d

∴ 20th term from end = 253 - (20 - 1)5 = 253 - 19 × 5 = 253 - 95 = 158.

Hence, the 20th term from last term of the A.P. 3, 8, 13, ...., 253 is 158.

Find the 12th term from the end of A.P. -2, -4, -6, ...., -100.

Answer

Here, common difference = d = -4 - (-2) = -2 and last term = l = -100.

We know that the nth term from last is given by the formula:
nth term from end = l - (n - 1)d

∴ 20th term from end = -100 - (12 - 1)(-2) = -100 - 11 × (-2) = -100 + 22 = -78.

Hence, the 12th term from last term of the A.P. -2, -4, -6, ...., -100 is -78.

Find the sum of the two middle most terms of the A.P.

$-\dfrac{4}{3}, -1, -\dfrac{2}{3}, ...., 4\dfrac{1}{3}.$

Answer

Here, a = $-\dfrac{4}{3} \text{ and d } = -1- \big(-\dfrac{4}{3}\big) = \dfrac{-3 + 4}{3} = \dfrac{1}{3}$

$\text{ and l } = 4\dfrac{1}{3} = \dfrac{13}{3}.$

We need to find the number of terms in the A.P.

$\Rightarrow a_n = a + (n - 1)d \\[1em] \Rightarrow \dfrac{13}{3} = -\dfrac{4}{3} + (n - 1)\dfrac{1}{3} \\[1em] \Rightarrow \dfrac{13}{3} + \dfrac{4}{3} = \dfrac{(n - 1)}{3} \\[1em] \Rightarrow \dfrac{17}{3} = \dfrac{(n - 1)}{3}$

On multiplying both sides by 3,

$\Rightarrow 17 = n - 1 \\[1em] \Rightarrow 17 + 1 = n \\[1em] \Rightarrow n = 18.$

Since, A.P. has 18 terms, therefore 9th and 10th terms are two middle most terms.

a₉ = a + (9 - 1)d = a + 8d        (Eq 1)

a₁₀ = a + (10 - 1)d = a + 9d     (Eq 2)

Adding Eq 1 and Eq 2,

a₉ + a₁₀ = a + 8d + a + 9d = 2a + 17d.

Hence, the sum of middle most terms

$= 2a + 17d \\[0.5em] = 2 \times -\dfrac{4}{3} + 17 \times \dfrac{1}{3} \\[0.5em] = -\dfrac{8}{3} + \dfrac{17}{3} \\[0.5em] = \dfrac{-8 + 17}{3} \\[0.5em] = \dfrac{9}{3} \\[0.5em] = 3.$

Hence, the sum of the two middle most terms of the A.P. is equal to 3.

Which term of the A.P. 53, 48, 43, .... is the first negative term?

Answer

Let us first evaluate which term is 0 because the next term will be the term containing first negative number.

a = 53, d = 48 - 53 = -5, a_n = 0.

By formula, a_n = a + (n - 1)d
⇒ a_n = 53 + (n - 1)(-5)
⇒ 0 = 53 - 5n + 5
⇒ 5n = 58
⇒ n = $11\dfrac{3}{5}$

The next term will be containing the first negative number.

Hence, the 12th term will contain the first negative number.

Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.

Answer

Given, a₃ = 16 (Eq 1)     and    a₇ - a₅ = 12 (Eq 2)

By using formula a_n = a + (n - 1)d on Eq 1 we get,

⇒ a₃ = a + (3 - 1)d = 16
⇒ a + 2d = 16
⇒ a = 16 - 2d.             (Eq 3)

By using formula a_n = a + (n - 1)d on Eq 2 we get,

⇒ a₇ - a₅ = 12
⇒ a + (7 - 1)d - [a + (5 - 1)d] = 12
⇒ a + 6d - a - 4d = 12
⇒ 2d = 12 ⇒ d = 6.

∴ a = 16 - 2d = 16 - 2(6) = 16 - 12 = 4.

    a₂ = a + d = 4 + 6 = 10,    a₃ = a₂ + d = 10 + 6 = 16,
    a₄ = a₃ + d = 16 + 6 = 22.

Hence, the A.P. is 4, 10, 16, 22 ....

Find the 20th term of the A.P. whose 7th term is 24 less than the 11th term, first term being 12.

Answer

Given, a = 12 and a₁₁ - a₇ = 24 (Eq 1)

By using formula a_n = a + (n - 1)d for Eq 1 we get,

⇒ a₁₁ - a₇ = 24
⇒ a + (11 - 1)d - [a + (7 - 1)d] = 24
⇒ a + 10d - a - 6d = 24
⇒ 4d = 24
⇒ d = 6.

20th term of A.P. is,

a₂₀ = 12 + (20 - 1)6 = 12 + 19 × 6 = 12 + 114 = 126.

Hence, the 20th term of the A.P. is 126.

Find the 31st term of an A.P. whose 11th term is 38 and 6th term is 73.

Answer

Given,
a₁₁ = 38     (Eq 1)
a₆ = 73       (Eq 2)

By using formula a_n = a + (n - 1)d for Eq 1 we get,

⇒ a₁₁ = a + (11 - 1)d = 38
⇒ a + 10d = 38
⇒ a = 38 - 10d              (Eq 3)

By using formula a_n = a + (n - 1)d for Eq 2 we get,

⇒ a₆ = a + (6 - 1)d = 73
⇒ a + 5d = 73                (Eq 4)

Putting value of a from Eq 3 in Eq 4 we get,

⇒ a + 5d = 73
⇒ 38 - 10d + 5d = 73
⇒ 38 - 5d = 73
⇒ 5d = 38 - 73
⇒ 5d = -35
⇒ d = -7.

∴ a = 38 - 10d = 38 - 10(-7) = 38 + 70 = 108.

and 31st term = a₃₁ = a + (31 - 1)d = 108 + 30(-7) = 108 - 210 = -102.

Hence, the 31st term of the A.P. is -102.

If the seventh term of an A.P. is $\dfrac{1}{9}$ and its ninth term is $\dfrac{1}{7},$ find its 63rd term.

Answer

Given,

a₇ = $\dfrac{1}{9}$     (Eq 1)

a₉ = $\dfrac{1}{7}$     (Eq 2)

By using formula a_n = a + (n - 1)d for Eq 1 we get,

⇒ a₇ = a + (7 - 1)d = $\dfrac{1}{9}$

⇒ a + 6d = $\dfrac{1}{9}$

⇒ 9(a + 6d) = 1

⇒ 9a + 54d = 1

⇒ 9a = 1 - 54d

⇒ a = $\dfrac{1 - 54d}{9}$     (Eq 3)

By using formula a_n = a + (n - 1)d for Eq 2 we get,

⇒ a₉ = a + (9 - 1)d = $\dfrac{1}{7}$

⇒ a + 8d = $\dfrac{1}{7}$     (Eq 4)

Putting value of a from Eq 3 in Eq 4 above,

$\Rightarrow \dfrac{1 - 54d}{9} + 8d = \dfrac{1}{7} \\[1em] \Rightarrow \dfrac{1 - 54d + 72d}{9} = \dfrac{1}{7} \\[1em] \Rightarrow \dfrac{1 + 18d}{9} = \dfrac{1}{7} \\[1em] \Rightarrow 7 + 126d = 9 \\[1em] \Rightarrow 126d = 2 \\[1em] \Rightarrow d = \dfrac{2}{126} \\[1em] \Rightarrow d = \dfrac{1}{63} \\[1.5em] \therefore a = \dfrac{1 - 54d}{9} \\[1em] = \dfrac{1 - 54 \times \dfrac{1}{63}}{9} \\[1em] = \dfrac{63 - 54}{9 \times 63} \\[1em] = \dfrac{9}{9 \times 63} \\[1em] = \dfrac{1}{63}. \\[1em]$

63rd term = a₆₃ = $a + (63 - 1)d$

$= a + 62d \\[1em] = \dfrac{1}{63} + 62 \times \dfrac{1}{63} \\[1em] = \dfrac{1}{63} + \dfrac{62}{63} \\[1em] = \dfrac{63}{63} \\[1em] = 1.$

Hence, 63rd term of the A.P. is 1.

The 15th term of an A.P. is 3 more than twice its 7th term. If the 10th term of the A.P. is 41, find its nth term.

Answer

Given,

a₁₀ = 41,          (Eq 1)
a₁₅ = 2a₇ + 3.  (Eq 2)

By using formula a_n = a + (n - 1)d for Eq 1 we get,
⇒ a₁₀ = a + (10 - 1)d = 41
⇒ a + 9d = 41
⇒ a = 41 - 9d     (Eq 3)

By using formula a_n = a + (n - 1)d for Eq 2 we get,
⇒ a₁₅ = 2a₇ + 3
⇒ a + (15 - 1)d = 2(a + (7 - 1)d) + 3
⇒ a + 14d = 2(a + 6d) + 3
⇒ a + 14d = 2a + 12d + 3

Putting value of a from Eq 3 in above equation
⇒ 41 - 9d + 14d = 2(41- 9d) + 12d + 3
⇒ 41 + 5d = 82 - 18d + 12d + 3
⇒ 41 + 5d = 82 - 6d + 3
⇒ 5d + 6d = 85 - 41
⇒ 11d = 44
⇒ d = 4.

∴ a = 41 - 9d = 41 - 9(4) = 41 - 36 = 5.

nth term = a_n = a + (n - 1)d = 5 + 4(n - 1) = 5 + 4n - 4 = 4n + 1.

Hence, the nth term of the A.P. is 4n + 1.

The sum of 5th and 7th terms of an A.P. is 52 and the 10th term is 46. Find the A.P.

Answer

Given,

a₅ + a₇ = 52   (Eq 1)

a₁₀ = 46         (Eq 2)

By using formula a_n = a + (n - 1)d for Eq 1 we get,
⇒ a₅ + a₇ = 52
⇒ a + (5 - 1)d + a + (7 - 1)d = 52
⇒ a + 4d + a + 6d = 52
⇒ 2a + 10d = 52
⇒ 2(a + 5d) = 52
⇒ a + 5d = 26
⇒ a = 26 - 5d    (Eq 3)

By using formula a_n = a + (n - 1)d for Eq 2 we get,
⇒ a₁₀ = 46
⇒ a + (10 - 1)d = 46
⇒ a + 9d = 46

Putting value of a from Eq 3 in above equation
⇒ 26 - 5d + 9d = 46
⇒ 26 + 4d = 46
⇒ 4d = 46 - 26
⇒ 4d = 20
⇒ d = 5.

∴ a = 26 - 5d = 26 - 5 × 5 = 26 - 25 = 1.

We know a₁ = a so,

a₂ = a₁ + d = 1 + 5 = 6,
a₃ = a₂ + d = 6 + 5 = 11,
a₄ = a₃ + d = 11 + 5 = 16.

Hence, the required A.P. is 1, 6, 11, 16, 21, ...

If 8th term of an A.P. is zero, prove that its 38th term is triple of its 18th term.

Answer

Given,
a₈ = 0

By using formula a_n = a + (n - 1)d for a₈ we get,

⇒ a + (8 - 1)d = 0
⇒ a + 7d = 0
⇒ a = -7d.

So,
a₃₈ = a + (38 - 1)d
      = a + 37d
      = -7d + 37d
      = 30d.

a₁₈ = a + (18 - 1)d
      = a + 17d
      = -7d + 17d
      = 10d.

∴ a₃₈ = 30d = 3 × 10d = 3 × a₁₈.

Hence, proved that 38th term of the A.P. is triple of the 18th term.

Which term of the A.P. 3, 10, 17, ... will be 84 more than its 13th term?

Answer

Here, common difference = d = 10 - 3 = 7 and a = 3.

Let the nth term of A.P. be 84 more than its 13th term so,

⇒ a_n - a₁₃ = 84
⇒ a + (n - 1)d - (a + (13 - 1)d) = 84
⇒ 3 + 7(n - 1) - (3 + 12 × 7) = 84
⇒ 3 + 7n - 7 - (3 + 84) = 84
⇒ 7n - 4 - 87 = 84
⇒ 7n - 91 = 84
⇒ 7n = 84 + 91
⇒ 7n = 175
⇒ n = 25.

Hence, the 25th term of A.P. will be 84 more than 13th term.

How many two digits numbers are divisible by 3?

Answer

Two digit numbers which are divisible by 3 are :

12, 15, 18, 21, 24, ......, 99.

First number = a = 12 and common difference = d = 15 - 12 = 3.

Last number = 99.

By formula, a_n = a + (n - 1)d

⇒ 99 = 12 + 3(n - 1)
⇒ 99 = 12 + 3n - 3
⇒ 99 = 9 + 3n
⇒ 99 - 9 = 3n
⇒ 3n = 90
⇒ n = 30.

Hence, there are 30 two digit numbers that are divisible by 3.

Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.

Answer

The numbers which are divisible by both 2 and 5 are
110, 120, 130, 140 .....990.

The above numbers are in A.P. with common difference = d = 120 - 110 = 10 and first term = a = 110.

Last term = 990.

By formula, a_n = a + (n - 1)d

⇒ 990 = 110 + 10(n - 1)
⇒ 990 = 110 + 10n - 10
⇒ 990 = 100 + 10n
⇒ 990 - 100 = 10n
⇒ 10n = 890
⇒ n = 89.

Hence, there are 89 natural numbers between 101 and 999 that are divisible by 2 and 5.

If the numbers n - 2, 4n - 1 and 5n + 2 are in A.P., find the value of n.

Answer

Here, a = a₁ = n - 2, a₂ = 4n - 1 and a₃ = 5n + 2

Since, the numbers are in A.P. so,
a₂ - a₁ = d = a₃ - a₂

⇒ a₂ - a₁ = a₃ - a₂
⇒ 4n - 1 - (n - 2) = 5n + 2 - (4n - 1)
⇒ 4n - n - 1 + 2 = 5n - 4n + 2 + 1
⇒ 3n + 1 = n + 3
⇒ 3n - n = 3 - 1
⇒ 2n = 2
⇒ n = 1.

Hence, the value of n = 1.

The sum of three numbers in A.P. is 3 and their product is -35. Find the numbers.

Answer

Let the three numbers that are in A.P. be a - d, a, a + d.

Given, sum of three numbers = 3
⇒ a - d + a + a + d = 3
⇒ 3a = 3
⇒ a = 1.

Given, product of the numbers = -35
⇒ (a - d)(a)(a + d) = -35

Putting the value of a = 1 in the above eq we get,

⇒ (1 - d)(1)(1 + d) = -35
⇒ (1 - d²) = -35
⇒ d² = 1 + 35
⇒ d² = 36
⇒ d = $\sqrt{36}$
⇒ d = +6, -6.

∴ a - d = 1 - 6 = -5 or 1 - (-6) = 7 and a + d = 1 + 6 = 7 or 1 + (-6) = -5

Hence, the numbers are -5, 1, 7.

The sum of three numbers in A.P. is 30 and the ratio of the first number to the third number is 3 : 7. Find the numbers.

Answer

Let the three numbers that are in A.P. be a - d, a, a + d.

Given, sum of three numbers = 30
⇒ a - d + a + a + d = 30
⇒ 3a = 30
⇒ a = 10.

Given, the ratio of the first number to the third number = 3 : 7

$\Rightarrow \dfrac{a - d}{a + d} = \dfrac{3}{7} \\[0.5em] \Rightarrow 7(a - d) = 3(a + d) \\[0.5em] \Rightarrow 7a - 7d = 3a + 3d \\[0.5em] \Rightarrow 7a - 3a = 3d + 7d \\[0.5em] \Rightarrow 4a = 10d \\[0.5em] \Rightarrow d = \dfrac{4a}{10} \\[0.5em]$

Putting the value of a = 10 we get,

$\Rightarrow d = \dfrac{4 \times 10}{10} \\[0.5em] \Rightarrow d = 4.$

∴ a - d = 10 - 4 = 6 and a + d = 10 + 4 = 14.

Hence, the numbers are 6, 10, 14.

The sum of the first three terms of an A.P. is 33. If the product of the first and the third terms exceeds the second term by 29, find the A.P.

Answer

Let the three numbers be a - d, a, a + d.

Given , sum of these terms = 33
⇒ a - d + a + a + d = 33
⇒ 3a = 33
⇒ a = 11.

Given, the product of the first and the third terms exceeds the second term by 29
⇒ (a + d)(a - d) - a = 29

Putting the value of a = 11 we get,

⇒ (11 + d)(11 - d) - 11 = 29
⇒ (11)² - d² - 11 = 29
⇒ 121 - d² - 11 = 29
⇒ 110 - d² = 29
⇒ 110 - 29 = d²
⇒ d² = 81
⇒ d = $\sqrt{81}$
⇒ d = +9, -9.

∴ a - d = 11 - 9 = 2 or 11 - (-9) = 20 and a + d = 11 + 9 = 20 or 11 + (-9) = 2.

Hence, the A.P. is 2, 11, 20, ... or 20, 11, 2, ....