Arithmetic & Geometric Progression — Chapter Test

Write the first four terms of the A.P. when its first term is -5 and common difference is -3.

Answer

Given, a = -5 and d = -3.

We know that

   a_n = a + (n - 1)d
∴ a₂ = -5 + (2 - 1) × (-3) = -5 + (-3) = -5 - 3 = -8.
   a₃ = -5 + (3 - 1) × (-3) = -5 + 2 × -3 = -5 - 6 = -11.
   a₄ = -5 + (4 - 1) × (-3) = -5 + 3 × -3 = -5 - 9 = -14.

Hence, the first four terms of the A.P. are -5, -8, -11, -14.

Verify that each of the following lists of numbers is an A.P., and then write its next three terms:

(i) 0, $\dfrac{1}{4}, \dfrac{1}{2}, \dfrac{3}{4}, ...$

(ii) 5, $\dfrac{14}{3}, \dfrac{13}{3}, 4, ...$

Answer

(i) Here,

$a_2 - a_1 = \dfrac{1}{4} - 0 = \dfrac{1}{4},\\[1em] a_3 - a_2 = \dfrac{1}{2} - \dfrac{1}{4} = \dfrac{2 - 1}{4} = \dfrac{1}{4}. \\[1em]$

i.e. any term - preceding term = $\dfrac{1}{4}$ = fixed number.

Hence, the given list of numbers forms an A.P.

For the next terms, we have,

$a_5 = a_4 + d = \dfrac{3}{4} + \dfrac{1}{4} = \dfrac{4}{4} = 1, \\[1em] a_6 = a_5 + d = 1 + \dfrac{1}{4} = \dfrac{4 + 1}{4} = \dfrac{5}{4}, \\[1em] a_7 = a_6 + d = \dfrac{5}{4} + \dfrac{1}{4} = \dfrac{6}{4} = \dfrac{3}{2}.$

So, the next three terms are $1, \dfrac{5}{4}, \dfrac{3}{2}.$

(ii) Here,

$a_2 - a_1 = \dfrac{14}{3} - 5 = \dfrac{14 - 15}{3} = -\dfrac{1}{3},\\[1em] a_3 - a_2 = \dfrac{13}{3} - \dfrac{14}{3} = \dfrac{13 - 14}{3} = -\dfrac{1}{3}. \\[1em]$

i.e. any term - preceding term = $-\dfrac{1}{3}$ = fixed number.
Hence, the given list of numbers forms an A.P.

For the next terms, we have,

$a_5 = a_4 + d = 4 + \Big(-\dfrac{1}{3}\Big) = 4 - \dfrac{1}{3} = \dfrac{12 - 1}{3} = \dfrac{11}{3}, \\[1em] a_6 = a_5 + d = \dfrac{11}{3} + \Big(-\dfrac{1}{3}\Big) = \dfrac{11}{3} - \dfrac{1}{3} = \dfrac{10}{3}, \\[1em] a_7 = a_6 + d = \dfrac{10}{3} + \Big(-\dfrac{1}{3}\Big) = \dfrac{10}{3} - \dfrac{1}{3} = \dfrac{9}{3} = 3.$

So, the next three terms are $\dfrac{11}{3}, \dfrac{10}{3}, 3.$

The nth term of an A.P. is 6n + 2. Find the common difference.

Answer

Given, a_n = 6n + 2.

∴ a₁ = 6(1) + 2 = 6 + 2 = 8,
    a₂ = 6(2) + 2 = 12 + 2 = 14.

We know that, d = any term - preceding term

∴ d = a₂ - a₁ = 14 - 8 = 6.

Hence, the common difference of the A.P. is 6.

Show that the list of numbers 9, 12, 15, 18, ... form an A.P. Find its 16th term and the nth term.

Answer

Here,
a₂ - a₁ = 12 - 9 = 3,
a₃ - a₂ = 15 - 12 = 3.

i.e. any term - preceding term = 3 = fixed number,

Hence, given list of numbers forms an A.P. with first term = a = 9 and common difference = d = 3.

We know that,

a_n = a + (n - 1)d = 9 + (n - 1) × 3 = 9 + 3n - 3 = 6 + 3n.

Since, a_n = 6 + 3n, so

a₁₆ = 6 + 3 × 16 = 6 + 48 = 54.

Hence, the a_n = 3n + 6 and a₁₆ = 54.

Find the 6th term from the end of the A.P. 17, 14, 11, ...., -40.

Answer

Given, a = 17, l = -40 and d = 14 - 17 = -3.

nth term from the end = l - (n - 1)d

∴ 6th term from the end = -40 - (6 - 1) × (-3) = -40 - 5 × -3 = -40 + 15 = -25.

Hence, 6th term from the end is -25.

If the 8th term of an A.P. is 31 and the 15th term is 16 more than its 11th term, then find the A.P.

Answer

Given, a₈ = 31 and a₁₅ - a₁₁ = 16.

We know that

⇒ a_n = a + (n - 1)d
∴ a₈ = a + 7d      (Eq 1)
   a₁₅ = a + 14d and
   a₁₁ = a + 10d.

Given, a₁₅ - a₁₁ = 16.
∴ a + 14d - (a + 10d) = 16
⇒ a - a + 14d - 10d = 16
⇒ 4d = 16
⇒ d = 4.

Putting value of d in Eq 1
⇒ a₈ = a + 7d
⇒ a + 7 × 4 = 31
⇒ a + 28 = 31
⇒ a = 31 - 28
⇒ a = 3.

Hence, the terms of A.P. are

a₂ = a₁ + d = 3 + 4 = 7,
a₃ = a₂ + d = 7 + 4 = 11,
a₄ = a₃ + d = 11 + 4 = 15.

Hence, the terms of the A.P. are 3, 7, 11, 15, ....

The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, then find the nth term.

Answer

We know that

a_n = a + (n - 1)d

According to question,

⇒ a₁₇ = 2(a₈) + 5
⇒ a + 16d = 2(a + 7d) + 5
⇒ a + 16d = 2a + 14d + 5
⇒ 2a - a + 14d - 16d + 5 = 0
⇒ a - 2d + 5 = 0
⇒ a = 2d - 5.     (Eq 1)

Given, a₁₁ = 43

⇒ a + 10d = 43
⇒ 2d - 5 + 10d = 43
⇒ 12d = 43 + 5
⇒ 12d = 48
⇒ d = 4.

Putting value of d in Eq 1,

⇒ a = 2d - 5
⇒ a = 2 × 4 - 5 = 8 - 5 = 3.

∴ a_n = 3 + (n - 1) × 4 = 3 + 4n - 4 = 4n - 1.

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

The 19th term of an A.P. is equal to three times its 6th term. If its 9th term is 19, find the A.P.

Answer

We know that

    a_n = a + (n - 1)d

According to question,

⇒ a₁₉ = 3(a₆)
⇒ a + 18d = 3(a + 5d)
⇒ a + 18d = 3a + 15d
⇒ 3a - a = 18d - 15d
⇒ 2a = 3d
⇒ a = $\dfrac{3}{2}d$     (Eq 1)

Given, a₉ = 19

$\Rightarrow a + 8d = 19 \\[1em] \Rightarrow \dfrac{3}{2}d + 8d = 19 \\[1em] \Rightarrow \dfrac{3d + 16d}{2} = 19 \\[1em] \Rightarrow \dfrac{19d}{2} = 19 \\[1em] \Rightarrow d = \dfrac{19}{19} \times 2 \\[1em] \Rightarrow d = 2. \\[1em]$

Putting value of d in Eq 1 we get,

$\Rightarrow a = \dfrac{3}{2}d \\[1em] \Rightarrow a = \dfrac{3}{2} \times 2 \\[1em] \Rightarrow a = 3.$

Hence, the terms of A.P. are

a₂ = a₁ + d = 3 + 2 = 5,
a₃ = a₂ + d = 5 + 2 = 7,
a₄ = a₃ + d = 7 + 2 = 9.

Hence, the terms of the A.P. are 3, 5, 7, 9, ....

If the 3rd and the 9th terms of an A.P. are 4 and -8 respectively, then which term of the A.P. is zero ?

Answer

We know that

    a_n = a + (n - 1)d

Given,

    a₃ = 4 and a₉ = -8
∴ a + 2d = 4 and a + 8d = -8

Subtracting the two Equations,

⇒ a + 8d - (a + 2d) = -8 - 4
⇒ a - a + 8d - 2d = -12
⇒ 6d = -12
⇒ d = -2.

Putting value of d in a + 2d = 4

⇒ a + 2 × (-2) = 4
⇒ a - 4 = 4
⇒ a = 4 + 4
⇒ a = 8.

Let nth term of the A.P. be zero so, a_n = 0.

⇒ a + (n - 1)d = 0
⇒ 8 + (n - 1) × (-2) = 0
⇒ 8 - 2n + 2 = 0
⇒ 10 - 2n = 0
⇒ 2n = 10
⇒ n = 5.

Hence, the 5th term of the A.P. is zero.

Which term of the list of the numbers 5, 2, -1, -4, .... is -55?

Answer

The above list is an A.P. with a = 5 and d = 2 - 5 = -3.

Let the nth term of the A.P. be -55 or, a_n = -55.

We know that

a_n = a + (n - 1)d

⇒ -55 = 5 + (n - 1) × (-3)
⇒ -55 = 5 - 3n + 3
⇒ -55 = 8 - 3n
⇒ 3n = 8 + 55
⇒ 3n = 63
⇒ n = 21.

Hence, the 21st term of the A.P. is -55..

The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is four times its 15th term.

Answer

We know that

a_n = a + (n - 1)d

Given,
a₂₄ = 2(a₁₀)
∴ a + 23d = 2(a + 9d)
⇒ a + 23d = 2a + 18d
⇒ 2a - a = 23d - 18d
⇒ a = 5d.

By using formula for nth term and a = 5d

a₇₂ = a + 71d = 5d + 71d = 76d
a₁₅ = a + 14d = 5d + 14d = 19d

∴ 4a₁₅ = 4 × 19d = 76d = a₇₂.

Hence, proved that 72nd term is four times the 15th term of the A.P.

Which term of the list of the numbers $20, 19\dfrac{1}{4}, 18\dfrac{1}{2}, 17\dfrac{3}{4}, ...$ is the first negative term?

Answer

The above series is an A.P. with a = 20 and

$d = 19\dfrac{1}{4} - 20 \\[1em] = \dfrac{77}{4} - 20 \\[1em] = \dfrac{77 - 80}{4} \\[1em] = -\dfrac{3}{4}.$

Let nth term be the first negative number.

We know that

a_n = a + (n - 1)d

$\therefore a_n = 20 + (n - 1) \times -\dfrac{3}{4} \\[1em] = 20 - \dfrac{3}{4}n + \dfrac{3}{4} \\[1em] = \dfrac{80 + 3}{4} - \dfrac{3}{4}n \\[1em] = \dfrac{83}{4} - \dfrac{3}{4}n \\[1em]$

Since a_n is a negative number

$\therefore a_n \lt 0 \\[1em] \Rightarrow \dfrac{83}{4} - \dfrac{3}{4}n \lt 0 \\[1em] \Rightarrow \dfrac{3}{4}n \gt \dfrac{83}{4} \\[1em] \Rightarrow n \gt \dfrac{83}{4} \times \dfrac{4}{3} \\[1em] \Rightarrow n \gt \dfrac{83}{3} \\[1em] \Rightarrow n \gt 27\dfrac{2}{3} \\[1em] \therefore n = 28.$

Hence, 28th term is the first negative number in the A.P.

How many three digit numbers are divisible by 9 ?

Answer

The three digit numbers that are divisible by 9 are 108, 117, 126, ....., 999.

The above series is in A.P. with a = 108, d = 117 - 108 = 9 and l = 999.

Let the nth term be last term

∴ a_n = 999.

We know that

a_n = a + (n - 1)d

⇒ 999 = 108 + (n - 1) × 9
⇒ 108 + 9n - 9 = 999
⇒ 9n + 99 = 999
⇒ 9n = 900
⇒ n = 100.

Hence, 100 three digit numbers are divisible by 9.

The sum of three numbers in A.P. is -3 and the product is 8. Find the numbers.

Answer

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

Given, S_n = -3

⇒ a - d + a + a + d = -3
⇒ 3a = -3
⇒ a = -1.

Given, product of numbers = 8

⇒ (a - d)a(a + d) = 8

Putting value of a = -1

⇒ (-1 - d)(-1)(-1 + d) = 8
⇒ (d + 1)(d - 1) = 8
⇒ d² - 1 = 8
⇒ d² = 9
⇒ d = -3, 3.

Putting d = -3
Numbers ⇒ a - d = -1 - (-3) = -1 + 3 = 2, a = -1, a + d = -1 + (-3) = -4.

Putting d = 3
Numbers ⇒ a - d = -1 - 3 = -4, a = -1, a + d = -1 + 3 = 2.

Hence, the three numbers in A.P. are -4, -1, 2 or 2, -1, -4.

The angles of a quadrilateral are in A.P. If the greatest angle is double of the smallest angle, find all the four angles.

Answer

Let the angles of quadrilateral be a, (a + d), (a + 2d), (a + 3d).

According to question greatest angle is twice the smallest angle,

⇒ (a + 3d) = 2a
⇒ 2a - a = 3d
⇒ a = 3d.

By putting values of a = 3d , the angles become,

a = 3d

a + d = 3d + d = 4d

a + 2d = 3d + 2d = 5d

a + 3d = 3d + 3d = 6d

Sum of angles of quadrilateral = 360°.

∴ 3d + 4d + 5d + 6d = 360°
⇒ 18d = 360°
⇒ d = 20°.

Hence, the angles are

⇒ 3d = 3 × 20° = 60°

⇒ 4d = 4 × 20° = 80°

⇒ 5d = 5 × 20° = 100°

⇒ 6d = 6 × 20° = 120°.

Hence, the angles of the quadrilateral are 60°, 80°, 100°, 120°.

Find the sum of first 20 terms of an A.P. whose nth term is 15 - 4n.

Answer

Since, a_n = 15 - 4n

∴ a₁ = 15 - 4(1) = 15 - 4 = 11 and a₂₀ = 15 - 4(20) = 15 - 80 = -65.

We know that

$S_n = \dfrac{n}{2}[a + l] \\[1em] \therefore S_{20} = \dfrac{20}{2}[11 + (-65)] \\[1em] = 10[-54] \\[1em] = -540.$

Hence, the sum of first 20 terms of an A.P. is -540.

Ten students of a class are lined up according to their heights. Height of first student is 150 cm, and every next student is 2 cm more in height. What is the total sum of heights of these ten students ?

Answer

Total number of students (n) = 10

Height of first student (a) = 150 cm

Difference between height of every next student (d) = 2 cm

Since, the height difference between every next student is equal, so it can be considered an A.P.

By formula,

Sum of A.P. (S) = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$S = \dfrac{10}{2} \times [2 \times 150 + (10 - 1) \times 2] \\[1em] = 5 \times [300 + 9 \times 2] \\[1em] = 5 \times [300 + 18] \\[1em] = 5 \times 318 \\[1em] = 1590 \text{ cm}.$

Hence, the total sum of heights of ten students = 1590 cm.

Find the geometric progression whose 4th term is 54 and 7th term is 1458.

Answer

Given, a₄ = 54 and a₇ = 1458.

We know that in G.P.

   a_n = ar^{n - 1}
∴ a₄ = ar³ and a₇ = ar⁶.

Dividing a₇ by a₄,

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{1458}{54} \\[1em] \Rightarrow r^3 = 27 \\[1em] \Rightarrow r = \sqrt[3]{27} \\[1em] \therefore r = 3. \\[1em]$

Putting value of r in ar³ = 54,

$\Rightarrow a(3)^3 = 54 \\[1em] \Rightarrow 27a = 54 \\[1em] \Rightarrow a = 2. \\[1em]$

Terms of G.P. are

⇒ a₂ = ar = 2 × 3 = 6,
⇒ a₃ = ar² = 2 × 3² = 18,
⇒ a₄ = ar³ = 2 × 3³ = 54.

Hence, the G.P. is 2, 6, 18, 54, ....

The fourth term of a G.P. is the square of its second term and the first term is -3. Find its 7th term.

Answer

Given, a₄ = (a₂)² and a = -3.

   a_n = ar^{n - 1}
∴ a₄ = ar³ and a₂ = ar.

Since, a₄ = (a₂)²
∴ ar³ = (ar)²
⇒ (-3)r³ = (-3r)²
⇒ -3r³ = 9r²
⇒ 9r² + 3r³ = 0
⇒ r²(9 + 3r) = 0
⇒ r² = 0 or 9 + 3r = 0
⇒ r = 0 or r = -3.

As common ratio cannot be equal to zero so, r ≠ 0.

a₇ = ar⁶ = (-3)(-3)⁶ = -3 × 729 = -2187.

Hence, the 7th term of the G.P. is -2187.

If the 4th, 10th and 16th terms of a G.P. are x, y and z respectively, prove that x, y, z are in G.P.

Answer

Given, a₄ = x, a₁₀ = y and a₁₆ = z.

We know that
   a_n = ar^{n - 1}
∴ a₄ = ar³ = x, a₁₀ = ar⁹ = y and a₁₆ = ar¹⁵ = z.

Dividing y by x we get,

$\dfrac{y}{x} = \dfrac{ar^9}{ar^3} = r^6$

Dividing z by y we get,

$\dfrac{z}{y} = \dfrac{ar^{15}}{ar^9} = r^6 \\[1em] \therefore \dfrac{y}{z} = \dfrac{z}{y}.$

Hence, proved that x, y, z are in G.P.

How many terms of the G.P. $3, \dfrac{3}{2}, \dfrac{3}{4}, ....$ are needed to give the sum $\dfrac{3069}{512}?$

Answer

Here, a = 3 and r = $\dfrac{\dfrac{3}{2}}{3} = \dfrac{3}{2 \times 3} = \dfrac{1}{2}.$

Let n terms are needed for the sum.

$S_n = \dfrac{a(r^n - 1)}{r - 1} \\[1em] \Rightarrow \dfrac{3069}{512} = \dfrac{3\Big[\Big(\dfrac{1}{2}\Big)^n - 1\Big]}{\dfrac{1}{2} - 1} \\[1em] \Rightarrow \dfrac{3069}{512} = \dfrac{3\Big[\Big(\dfrac{1}{2}\Big)^n - 1\Big]}{\dfrac{1 - 2}{2}} \\[1em] \Rightarrow \dfrac{3069}{512} = -6\Big[\Big(\dfrac{1}{2}\Big)^n - 1\Big] \\[1em] \Rightarrow \dfrac{3069}{512 \times 6} = 1 - \Big(\dfrac{1}{2}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = 1 - \dfrac{3069}{3072} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = \dfrac{3072 - 3069}{3072} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = \dfrac{3}{3072} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = \dfrac{1}{1024} \\[1em] \Rightarrow \Big(\dfrac{1}{2}\Big)^n = \Big(\dfrac{1}{2}\Big)^{10} \\[1em] \therefore n = 10.$

Hence, 10 terms of the G.P. are required for the sum of $\dfrac{3069}{512}$.

15, 30, 60, 120 ...... are in G.P. (Geometric Progression).

(a) Find the nth term of this G.P. in terms of n.

(b) How many terms of the above G.P. will give the sum 945 ?

Answer

Given,

G.P. : 15, 30, 60, 120 ......

First term (a) = 15

Common ratio (r) = $\dfrac{30}{15}$ = 2

(a) nth term of G.P. = ar^{n - 1} = 15 x 2^{n - 1}

= $\dfrac{15}{2}$ x 2ⁿ

= 7.5 x 2ⁿ

Hence, nth term of the given G.P. is 7.5 x 2ⁿ

(b) Let sum of n terms of G.P. is 945.

By formula,

Sum of n terms of G.P. = $\dfrac{a(r^n - 1)}{(r - 1)}$

Substituting values we get :

$\Rightarrow 945 = \dfrac{15.(2^n - 1)}{2 - 1} \\[1em] \Rightarrow 945 = \dfrac{15.(2^n - 1)}{1} \\[1em] \Rightarrow 2^n - 1 = \dfrac{945}{15} \\[1em] \Rightarrow 2^n - 1 = 63 \\[1em] \Rightarrow 2^n = 63 + 1 \\[1em] \Rightarrow 2^n = 64 \\[1em] \Rightarrow 2^n = 2^6 \\[1em] \Rightarrow n = 6.$

Hence, sum of 6 terms of G.P. = 945.

The roots of the equation (q - r)x² + (r - p)x + (p - q) = 0 are equal.

Prove that : 2q = p + r, that is, p, q and r are in A.P.

Answer

Given,

The roots of the equation (q - r)x² + (r - p)x + (p - q) = 0 are equal.

∴ Discriminant (D) = 0

⇒ b² - 4ac = 0

⇒ (r - p)² - 4 × (q - r) × (p - q) = 0

⇒ r² + p² - 2pr - 4(qp - q² - rp + qr) = 0

⇒ r² + p² - 2pr - 4qp + 4q² + 4rp - 4qr = 0

⇒ r² + p² + 2pr - 4qp - 4qr + 4q² = 0

⇒ (p + r)² - 4q(p + r) + 4q² = 0

Let p + r = y

⇒ y² - 4qy + 4q² = 0

⇒ (y - 2q)² = 0

⇒ y - 2q = 0

⇒ y = 2q

⇒ p + r = 2q.

Hence, proved that p + r = 2q.