Arithmetic and Geometric Progressions

For the following A.P.s, write the first term a and the common difference d.

(i) 3, 1, -1, -3, ....

(ii) $\dfrac{1}{3}, \dfrac{5}{3}, \dfrac{9}{3}, \dfrac{13}{3}, ...$

(iii) -3.2, -3, -2.8, -2.6, ...

Answer

(i) First term = a = 3 and common difference = d = 1 - 3 = -2.

(ii) First term = a = $\dfrac{1}{3}$ and common difference = d = $\dfrac{5}{3} - \dfrac{1}{3} = \dfrac{4}{3}$.

(iii) First term = a = -3.2 and common difference = d = -3 - (-3.2) = 0.2.

Write first four of the terms of the A.P., when the first term a and the common difference a are given as follows :

(i) a = 10, d = 10

(ii) a = -2, d = 0

(iii) a = 4, d = -3

(iv) a = $\dfrac{1}{2}$, d = $-\dfrac{1}{6}$

Answer

(i) Here, a₁ = a = 10,   a₂ = a₁ + d = 10 + 10 = 20,

a₃ = a₂ + d = 20 + 10 = 30,    a₄ = a₃ + d = 30 + 10 = 40.

Hence, the first four terms of A.P. are 10, 20, 30, 40.

(ii) Here, a₁ = a = -2,   a₂ = a₁ + d = -2 + 0 = -2,

a₃ = a₂ + d = -2 + 0 = -2,    a₄ = a₃ + d = -2 + 0 = -2.

Hence, the first four terms of A.P. are -2, -2, -2, -2.

(iii) Here, a₁ = a = 4,   a₂ = a₁ + d = 4 + (-3) = 1,

a₃ = a₂ + d = 1 + (-3) = -2,    a₄ = a₃ + d = -2 + (-3) = -5.

Hence, the first four terms of A.P. are 4, 1, -2, -5.

(iv) Here, a₁ = a = $\dfrac{1}{2}$,   a₂ = a₁ + d = $\dfrac{1}{2} + \big(-\dfrac{1}{6}\big) = \dfrac{2}{6} = \dfrac{1}{3}$,

a₃ = a₂ + d = $\dfrac{2}{6} + \big(-\dfrac{1}{6}\big) = \dfrac{1}{6}$,    a₄ = a₃ + d = $\dfrac{1}{6} + \big(-\dfrac{1}{6}\big)$ = 0.

Hence, the first four terms of A.P. are $\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{6}, 0$.

Which of the following lists of numbers form an A.P. ? If they form an A.P., find the common difference d and write the next three terms:

(i) 4, 10, 16, 22, ....

(ii) -2, 2, -2, 2, ....

(iii) 2, 4, 8, 16, ....

(iv) $2, \dfrac{5}{2}, 3, \dfrac{7}{2}, ....$

(v) -10, -6, -2, 2, ...

(vi) 1², 3², 5², 7², ....

Answer

(i) Given, 4, 10, 16, 22, ....

Here, a₂ - a₁ = 10 - 4 = 6,     a₃ - a₂ = 16 - 10 = 6,

          a₄ - a₃ = 22 - 16 = 6

i.e. any term - preceding term = 6, a fixed number.
Hence, the given list of numbers forms an A.P. with common difference = d = 6.

For the next three terms, we have:
    a₅ = a₄ + d = 22 + 6 = 28,
    a₆ = a₅ + d = 28 + 6 = 34,
    a₇ = a₆ + d = 34 + 6 = 40.

Hence, the given series is in A.P. with common difference d = 6 and the next three terms : 28, 34, 40.

(ii) Given, -2, 2, -2, 2, ....

Here, a₂ - a₁ = 2 - (-2) = 4,     a₃ - a₂ = -2 - 2 = -4,

          a₄ - a₃ = 2 - (-2) = 4

⇒ a₂ - a₁ = a₄ - a₃ ≠ a₃ - a₂.

Thus the difference of any term from its preceding term is not a fixed number.

Hence, the given series does not form an A.P.

(iii) Given, 2, 4, 8, 16, ....

Here, a₂ - a₁ = 4 - 2 = 2,     a₃ - a₂ = 8 - 4 = 4,

          a₄ - a₃ = 16 - 8 = 8

⇒ a₂ - a₁ ≠ a₃ - a₂ ≠ a₄ - a₃.

Thus the difference of any term from its preceding term is not a fixed number.

Hence, the given series does not form an A.P.

(iv) Given, $2, \dfrac{5}{2}, 3, \dfrac{7}{2}, ....$

Here, a₂ - a₁ = $\dfrac{5}{2} - 2 = \dfrac{1}{2}$,     a₃ - a₂ = $3 - \dfrac{5}{2} = \dfrac{1}{2}$,

          a₄ - a₃ = $\dfrac{7}{2} - 3 = \dfrac{1}{2}$

i.e. any term - preceding term = $\dfrac{1}{2}$, a fixed number.
Hence, the given list of numbers forms an A.P. with common difference = d = $\dfrac{1}{2}$.

For the next three terms, we have:

    a₅ = a₄ + d = $\dfrac{7}{2} + \dfrac{1}{2} = \dfrac{8}{2} = 4$,

    a₆ = a₅ + d = $4 + \dfrac{1}{2} = \dfrac{9}{2}$,

    a₇ = a₆ + d = $\dfrac{9}{2} + \dfrac{1}{2} = \dfrac{10}{2} = 5$.

Hence, the given series is in A.P. with common difference d = $\dfrac{1}{2}$ and the next three terms : 4, $\dfrac{9}{2}$, 5.

(v) Given, -10, -6, -2, 2, ...

Here, a₂ - a₁ = -6 - (-10) = 4,     a₃ - a₂ = -2 - (-6) = 4,

          a₄ - a₃ = 2 - (-2) = 4

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

For the next three terms, we have:
    a₅ = a₄ + d = 2 + 4 = 6,
    a₆ = a₅ + d = 6 + 4 = 10,
    a₇ = a₆ + d = 10 + 4 = 14.

Hence, the given series is in A.P. with common difference d = 4 and the next three terms : 6, 10, 14.

(vi) Given, 1², 3², 5², 7², ....

or, 1, 9, 25, 49 ....

Here, a₂ - a₁ = 9 - 1 = 8,     a₃ - a₂ = 25 - 9 = 16,

          a₄ - a₃ = 49 - 25 = 24

⇒ a₂ - a₁ ≠ a₃ - a₂ ≠ a₄ - a₃.

Thus the difference of any term from its preceding term is not a fixed number.

Hence, the given series does not form an A.P.

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, ....

Find the sum of the following A.P.s :

(i) 2, 7, 12 ... to 10 terms

(ii) $\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10}, ...$ to 11 terms

Answer

(i) Here, a = 2, d = 7 - 2 = 5 and n = 10.

We know that Sum = $\dfrac{n}{2}[2a + (n - 1)d]$

$\therefore \text{ Sum } = \dfrac{10}{2}[2 \times 2 + (10 - 1)5] \\[1em] = 5[4 + 9 \times 5] \\[1em] = 5 \times 49 \\[1em] = 245.$

Hence, the sum of the A.P. 2, 7, 12, ... upto 10 terms is 245.

(ii) Here, a = $\dfrac{1}{15}, d = \dfrac{1}{12} - \dfrac{1}{15} = \dfrac{5 - 4}{60} = \dfrac{1}{60}$ and n = 11.

We know that Sum = $\dfrac{n}{2}[2a + (n - 1)d]$

$\therefore \text{ Sum } = \dfrac{11}{2}[2 \times \dfrac{1}{15} + (11 - 1)\dfrac{1}{60}] \\[1em] = \dfrac{11}{2}\Big[\dfrac{2}{15} + \dfrac{10}{60}\Big] \\[1em] = \dfrac{11}{2} \times \Big[\dfrac{2 \times 4 + 10}{60}\Big] \\[1em] = \dfrac{11}{2} \times \dfrac{18}{60}$

Dividing numerator and denominator by 6,

$= \dfrac{198}{120} \\[1em] = \dfrac{33}{20}. \\[1em]$

Hence, the sum of the A.P. $\dfrac{1}{15}, \dfrac{1}{12}, \dfrac{1}{10}, ...$ upto 11 terms is $\dfrac{33}{20}$.

Find the sums given below:

(i) 34 + 32 + 30 + .... + 10

(ii) -5 + (-8) + (-11) + .... + (-230)

Answer

(i) The given numbers form an A.P. with a = 34 and d = 32 - 34 = -2.

Let nth term be 10,
as, a_n = a + (n - 1)d

∴ 10 = 34 + (n - 1)(-2)
⇒ 10 = 34 - 2n + 2
⇒ 10 = 36 - 2n
⇒ 2n = 36 - 10
⇒ 2n = 26
⇒ n = 13.

$\text{Sum } = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \therefore \text{ Sum } = \dfrac{13}{2}[2 \times 34 + (13 - 1)(-2)] \\[1em] = \dfrac{13}{2}[68 + 12(-2)] \\[1em] = \dfrac{13}{2}[68 - 24] \\[1em] = \dfrac{13}{2} \times 44 \\[1em] = 13 \times 22 \\[1em] = 286.$

Hence, the sum of the series 34 + 32 + 30 + .... + 10 is 286.

(ii) The given numbers form an A.P. with a = -5 and d = -8 - (-5) = -3.

Let nth term be -230, then as, a_n = a + (n - 1)d

∴ -230 = -5 + (n - 1)(-3)
⇒ -230 = -5 -3n + 3
⇒ -230 = -2 - 3n
⇒ -230 + 2 = -3n
⇒ -228 = -3n
⇒ 3n = 228
⇒ n = 76.

$\text{Sum } = \dfrac{n}{2}[2a + (n - 1)d] \\[1 em] \therefore \text{Sum } = \dfrac{76}{2}[2 \times -5 + (76 - 1)(-3)] \\[1em] = 38[-10 + 75(-3)] \\[1em] = 38[-10 - 225] \\[1em] = 38 \times -235 \\[1em] = -8930.$

Hence, the sum of the series -5 + (-8) + (-11) + .... + (-230) is -8930.

In an A.P. (with usual notations):

(i) given a = 5, d = 3, a_n = 50, find n and S_n

(ii) given a = 7, a₁₃ = 35, find d and S₁₃

(iii) given d = 5, S₉ = 75, find a and a₉

(iv) given a = 8, a_n = 62, S_n = 210, find n and d

(v) given a = 3, n = 8, S = 192, find d.

Answer

(i) a = 5, d = 3, a_n = 50.

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

⇒ 50 = 5 + (n - 1)3
⇒ 50 = 5 + 3n - 3
⇒ 50 = 2 + 3n
⇒ 50 - 2 = 3n
⇒ 48 = 3n
⇒ n = 16.

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

$S_n = \dfrac{16}{2}[2 \times 5 + (16 - 1)3] \\[1em] = 8[10 + 15 \times 3] \\[1em] = 8 \times 55 \\[1em] = 440.$

Hence, n = 16 and S_n = S₁₆ = 440.

(ii) a = 7, a₁₃ = 35

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

⇒ 35 = 7 + (13 - 1)d
⇒ 35 = 7 + 12d
⇒ 35 - 7 = 12d
⇒ 28 = 12d
⇒ d = $\dfrac{28}{12}$ (Dividing by 4)

⇒ d = $\dfrac{7}{3}.$

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

$S_{13} = \dfrac{13}{2}[2 \times 7 + (13 - 1)\Big(\dfrac{7}{3}\Big)] \\[1em] = \dfrac{13}{2}[14 + 12 \times \dfrac{7}{3}] \\[1em] = \dfrac{13}{2} \times [14 + 28] \\[1em] = \dfrac{13}{2} \times 42 \\[1em] = 13 \times 21 \\[1em] = 273.$

Hence, d = $\dfrac{7}{3}$ and S_n = S₁₃ = 273.

(iii) d = 5, S₉ = 75

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

$S_9 = \dfrac{9}{2}[2 \times a + (9 - 1)5] \\[1em] \Rightarrow 75 = \dfrac{9}{2}[2a + 8 \times 5] \\[1em] \Rightarrow 75 = \dfrac{9}{2} \times [2a + 40] \\[1em] \Rightarrow 75 = 9(a + 20) \\[1em] \Rightarrow 75 = 9a + 180 \\[1em] \Rightarrow 9a = 75 - 180 \\[1em] \Rightarrow 9a = -105 \\[1em] \Rightarrow a = -\dfrac{105}{9} \\[1em] \Rightarrow a = -\dfrac{35}{3}$

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

$\Rightarrow a_9 = -\dfrac{35}{3} + (9 - 1)5 \\[1em] = -\dfrac{35}{3} + 40 \\[1em] = \dfrac{-35 + 120}{3} \\[1em] = \dfrac{85}{3}.$

Hence, a = $-\dfrac{35}{3} \text{ and } a_9 = \dfrac{85}{3}.$

(iv) a = 8, a_n = 62, S_n = 210.

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

⇒ 62 = 8 + (n - 1)d
⇒ 62 - 8 = (n - 1)d
⇒ (n - 1)d = 54          (Eq 1)

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

$\Rightarrow 210 = \dfrac{n}{2}[2 \times 8 + (n - 1)d] \\[1em]$

Putting value of (n - 1)d from Eq 1 in above equation,

$\Rightarrow \dfrac{n}{2}[16 + 54] = 210 \\[1em] \Rightarrow \dfrac{n}{2} \times 70 = 210 \\[1em] \Rightarrow 35n = 210 \\[1em] \Rightarrow n = \dfrac{210}{35} \\[1em] \Rightarrow n = 6 \\[1em] \text{From Eq 1, } (n - 1)d = 54 \\[1em] \Rightarrow (6 - 1)d = 54 \\[1em] \Rightarrow 5d = 54 \\[1em] \therefore d = \dfrac{54}{5} \text{ and } n = 6.$

Hence, the value of n = 6 and d = $\dfrac{54}{5}.$

(v) a = 3, n = 8, S = 192.

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

⇒ 192 = $\dfrac{8}{2}[2 \times 3 + (8 - 1)d]$
⇒ 192 = 4[6 + 7d]

⇒ $\dfrac{192}{4}$ = 6 + 7d
⇒ 48 = 6 + 7d
⇒ 7d = 42
⇒ d = 6.

Hence, the value of d is 6.

The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Answer

Let nth be the last term so,
a = 5, l = a_n = 45, S_n = 400.

By formula a_n = a + (n - 1)d
⇒ 45 = 5 + (n - 1)d
⇒ 45 - 5 = (n - 1)d
⇒ (n - 1)d = 40      (Eq 1)

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

$\Rightarrow 400 = \dfrac{n}{2}[2 \times 5 + (n - 1)d]$

Putting value of (n - 1)d from Eq 1 in above equation,

$\Rightarrow 400 = \dfrac{n}{2}[10 + 40] \\[1em] \Rightarrow \dfrac{n}{2} \times 50 = 400 \\[1em] \Rightarrow 25n = 400 \\[1em] \Rightarrow n = 16$

Putting value of n in Eq 1 we get,

$\Rightarrow (16 - 1)d = 40 \\[1em] \Rightarrow 15d = 40 \\[1em] \Rightarrow d = \dfrac{40}{15} = \dfrac{8}{3}. \\[1em]$

Hence, number of terms = 16 and common difference = $\dfrac{8}{3}.$

The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.

Answer

Given, a = 15, S₁₅ = 750

By formula, S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow S_{15} = \dfrac{15}{2}[2 \times 15 + (15 - 1)d] \\[1em] \Rightarrow 750 = \dfrac{15}{2}[30 + 14d] \\[1em] \Rightarrow 750 \times 2 = 450 + 210d \\[1em] \Rightarrow 450 + 210d = 1500 \\[1em] \Rightarrow 210d = 1500 - 450 \\[1em] \Rightarrow 210d = 1050 \\[1em] \Rightarrow d = \dfrac{1050}{210} \\[1em] \Rightarrow d = 5.$

By formula, a_n = a + (n - 1)d
⇒ a₂₀ = 15 + (20 - 1)5
⇒ a₂₀ = 15 + 95
⇒ a₂₀ = 110.

Hence, the 20th term of the A.P. is 110.

The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Answer

Given, a = 17, l = 350 and d = 9.

Let nth be the last term so,
l = a_n = 350

By formula, a_n = a + (n - 1)d
⇒ 350 = 17 + (n - 1)9
⇒ 350 = 17 + 9n - 9
⇒ 350 = 9n + 8
⇒ 9n = 350 - 8
⇒ 9n = 342
⇒ n = 38.

By formula, S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

⇒ S₃₈ = $\dfrac{38}{2}[2 \times 17 + (38 - 1)9]$
⇒ S₃₈ = 19[34 + 37(9)]
⇒ 19[34 + 333]
⇒ 19(367)
⇒ 6973.

Hence, number of terms = 38 and the sum of the terms = 6973.

Solve for x : 1 + 4 + 7 + 10 + ... + x = 287.

Answer

The above series is in A.P. because,
any term - preceding term = 3 = common difference.

Let there be n terms so, x = a_n.

Given, a = 1, d = 3 and S_n = 287.

By formula, S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow S_n = \dfrac{n}{2}[2 \times 1 + (n - 1)3] \\[1em] \Rightarrow 287 = \dfrac{n}{2}[2 + 3n - 3] \\[1em] \Rightarrow 287 \times 2 = n[3n - 1] \\[1em] \Rightarrow 3n^2 - n = 574 \\[1em] \Rightarrow 3n^2 - n - 574 = 0 \\[1em] 3n^2 - 42n + 41n - 574 = 0 \\[1em] 3n(n - 14) + 41(n - 14) = 0 \\[1em] (3n + 41)(n - 14) = 0 \\[1em] 3n + 41 = 0 \text{ or } n - 14 = 0 \\[1em] n = -\dfrac{41}{3} \text{ or } n = 14.$

Since number of terms cannot be negative so n ≠ $-\dfrac{41}{3}$

∴ n = 14.

We know x = a_n so,

⇒ x = a + (n - 1)d
⇒ x = 1 + (14 - 1)3
⇒ x = 1 + 13(3)
⇒ x = 1 + 39
⇒ x = 40.

Hence, the value of x = 40.

How many terms of the A.P. 25, 22, 19, .... are needed to give the sum 116 ? Also find the last term.

Answer

Let the number of terms required be n.

For the given A.P.,
a = 25, d = 22 - 25 = -3, S_n = 116

By formula, S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow 116 = \dfrac{n}{2}[2 \times 25 + (n - 1)(-3)] \\[1em] 116 \times 2 = n[50 - 3n + 3] \\[1em] 232 = n[53 - 3n] \\[1em] 232 = 53n - 3n^2 \\[1em] 3n^2 - 53n + 232 = 0 \\[1em] 3n^2 - 24n - 29n + 232 = 0 \\[1em] 3n(n - 8) - 29(n - 8) = 0 \\[1em] (3n - 29)(n - 8) = 0 \\[1em] 3n - 29 = 0 \text{ or } n - 8 = 0 \\[1em] n = \dfrac{29}{3} \text{ or } n = 8.$

Since, number of terms will be a natural number so n ≠ $\dfrac{29}{3}$

∴ n = 8.

By formula S_n = $\dfrac{n}{2}[a + l]$

⇒ 116 = $\dfrac{8}{2}[25 + l]$
⇒ 116 = 4[25 + l]
⇒ 29 = 25 + l
⇒ 29 - 25 = l
⇒ l = 4.

Hence, number of terms = 8 and the last term = 4.

How many terms of the A.P. 24, 21, 18, ... must be taken so that the sum is 78 ? Explain the double answer.

Answer

Let numbers of terms be n.
Given, a = 24, d = 21 - 24 = -3 and S_n = 78.

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

⇒ 78 = $\dfrac{n}{2}[2 \times 24 + (n - 1)(-3)]$
⇒ 78 × 2 = n[48 - 3n + 3]
⇒ 156 = n[51 - 3n]
⇒ 156 = 51n - 3n²
⇒ 3n² - 51n + 156 = 0
⇒ 3n² - 12n - 39n + 156 = 0
⇒ 3n(n - 4) - 39(n - 4) = 0
⇒ (3n - 39)(n - 4) = 0
⇒ 3n - 39 = 0 or n - 4 = 0
⇒ 3n = 39 or n = 4
⇒ n = 13 or n = 4.

Let's check sum of 4 terms
⇒ a₄ = a₃ + d
⇒ a₄ = 18 + (-3) = 15

∴ S₄ = 24 + 21 + 18 + 15 = 78.

By formula, a_n = a + (n - 1)d
⇒ a₅ = 24 + (5 - 1)(-3)
⇒ a₅ = 24 + 4(-3)
⇒ a₅ = 24 - 12 = 12.

a₆ = a₅ + d = 12 + (-3) = 9
a₇ = a₆ + d = 9 + (-3) = 6
a₈ = a₇ + d = 6 + (-3) = 3
a₉ = a₈ + d = 3 + (-3) = 0
a₁₀ = a₉ + d = 0 + (-3) = -3
a₁₁ = a₁₀ + d = -3 + (-3) = -6
a₁₂ = a₁₁ + d = -6 + (-3) = -9
a₁₃ = a₁₂ + d = -9 + (-3) = -12.

Taking sum of from 5th term to 13th,

12 + 9 + 6 + 3 + 0 - 3 - 6 - 9 - 12 = 0

Since, the sum from 5tn term to 13th is zero hence, it will not add a value.

∴ n = 13 and 4.

Hence, the number of terms that can be taken for sum to be 78 can be 4 or 13.

Find the sum of first 22 terms of an A.P. in which d = 7 and a₂₂ is 149.

Answer

Given, d = 7 and a₂₂ = 149.

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

⇒ 149 = a + (22 - 1)7
⇒ 149 = a + 21(7)
⇒ 149 = a + 147
⇒ a = 149 - 147
⇒ a = 2.

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

⇒ S_n = $\dfrac{22}{2}[2 \times 2 + (22 - 1)7]$
⇒ S_n = 11[4 + 147]
⇒ S_n = 11 × 151
⇒ S_n = 1661.

Hence, the sum of first 22 terms of the A.P. is 1661.

In an A.P., the 5^th and 9^th term are 4 and -12 respectively. Find :

(a) the first term

(b) common difference

(c) sum of the first 20 terms.

Answer

If first term is a and common difference is d of the A.P.

By formula,

n^th term = a_n = a + (n - 1)d

Given,

⇒ 5^th term = 4

⇒ a₅ = 4

⇒ a + (5 - 1)d = 4

⇒ a + 4d = 4 .........(1)

⇒ 9^th term = -12

⇒ a₉ = -12

⇒ a + (9 - 1)d = -12

⇒ a + 8d = -12 .........(2)

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

⇒ (a + 8d) - (a + 4d) = -12 - 4

⇒ a - a + 8d - 4d = -16

⇒ 4d = -16

⇒ d = $-\dfrac{16}{4}$

⇒ d = -4.

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

⇒ a + 4d = 4

⇒ a + 4(-4) = 4

⇒ a - 16 = 4

⇒ a = 4 + 16 = 20.

(a) Hence, first term of A.P. = 20.

(b) Common difference of A.P. = -4.

(c) By formula,

Sum of n terms of A.P. = $\dfrac{n}{2}(a + l)$

Sum of first 20 terms of A.P. = $\dfrac{n}{2}(a + a_{20})$

$= \dfrac{20}{2}[a + a + (20 - 1)d] \\[1em] = 10(2a + 19d) \\[1em] = 10 \times (2 \times 20 + 19 \times -4) \\[1em] = 10 \times (40 - 76) \\[1em] = 10 \times -36 \\[1em] = -360.$

Hence, sum of first 20 terms of A.P. = -360.

Find the sum of first 51 terms of the A.P. whose second and third terms are 14 and 18 respectively.

Answer

Given, a₂ = 14 and a₃ = 18.

common difference = d = any term - preceding term = a₃ - a₂ = 18 - 14 = 4.

By formula, a_n = a + (n - 1)d
⇒ a₂ = a + 4(2 - 1)
⇒ 14 = a + 4
⇒ a = 10.

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

$\Rightarrow S_{51} = \dfrac{51}{2}[2 \times 10 + (51 - 1)4] \\[1em] = \dfrac{51}{2}[20 + 50 \times 4] \\[1em] = \dfrac{51}{2}[20 + 200] \\[1em] = \dfrac{51}{2} \times 220 \\[1em] = 51 \times 110 \\[1em] = 5610.$

Hence, the sum of first 51 terms of the A.P. is 5610.

The 4th term of an A.P. is 22 and 15th term is 66. Find the first term and the common difference. Hence, find the sum of first 8 terms of the A.P.

Answer

Given, a₄ = 22 and a₁₅ = 66.

By formula, a_n = a + (n - 1)d
⇒ a₄ = a + (4 - 1)d
⇒ 22 = a + 3d
⇒ a = 22 - 3d         (Eq 1)

⇒ a₁₅ = a + (15 - 1)d
⇒ 66 = a + 14d
Putting value of a from Eq 1 in above equation
⇒ 66 = 22 - 3d + 14d
⇒ 66 = 22 + 11d
⇒ 11d = 66 - 22
⇒ 11d = 44
⇒ d = 4.

Putting value of d in Eq 1,
⇒ a = 22 - 3(4)
⇒ a = 22 - 12
⇒ a = 10.

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

⇒ S₈ = $\dfrac{8}{2}[2 \times 10 + (8 - 1)4]$
⇒ S₈ = 4[20 + 28]
⇒ S₈ = 4 × 48
⇒ S₈ = 192.

Hence, first term = a = 10, common difference = d = 4 and sum of first 8 terms = S₈ = 192.

If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

Answer

Given, S₆ = 36 and S₁₆ = 256.

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

⇒ S₆ = $\dfrac{6}{2}[2a + (6 - 1)d]$
⇒ 36 = 3[2a + 5d]
⇒ 2a + 5d = 12
⇒ 2a = 12 - 5d     (Eq 1)

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

⇒ S₁₆ = $\dfrac{16}{2}[2a + (16 - 1)d]$
⇒ 256 = 8[2a + 15d]
⇒ 2a + 15d = 32
⇒ 2a = 32 - 15d

Putting value of 2a from Eq 1 in above equation,

⇒ 12 - 5d = 32 - 15d
⇒ -5d + 15d = 32 - 12
⇒ 10d = 20
⇒ d = 2.

∴ From Eq 1,
⇒ 2a = 12 - 5d
⇒ 2a = 12 - 5(2)
⇒ 2a = 2
⇒ a = 1.

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

⇒ S₁₀ = $\dfrac{10}{2}[2 \times 1 + (10 - 1)2]$
⇒ S₁₀ = 5[2 + 18]
⇒ S₁₀ = 5 × 20
⇒ S₁₀ = 100.

Hence, the sum of first 10 terms of the A.P. is 100.

Show that a₁, a₂, a₃, ..... form an A.P. where a_n is defined as a_n = 3 + 4n. Also find the sum of first 15 terms.

Answer

a_n = 3 + 4n
a₁ = 3 + 4 × 1 = 3 + 4 = 7
a₂ = 3 + 4 × 2 = 3 + 8 = 11
a₃ = 3 + 4 × 3 = 3 + 12 = 15
a₄ = 3 + 4 × 4 = 3 + 16 = 19.

Since, a₄ - a₃ = a₃ - a₂ = a₂ - a₁ = 4, i.e any term - preceding term = fixed number = 4.

∴ a₁, a₂, a₃ ..... form an A.P.

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

$\Rightarrow S_{15} = \dfrac{15}{2}[2 \times 7 + (15 - 1)4] \\[1em] = \dfrac{15}{2}[14 + 14 \times 4] \\[1em] = \dfrac{15}{2}[14 + 56] \\[1em] = \dfrac{15}{2} \times 70 \\[1em] = 15 \times 35 \\[1em] = 525.$

Hence, the sum of first 15 terms of the A.P. is 525.

The sum of first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is 1 : 3. Calculate the first and the thirteenth term.

Answer

Given, S₆ = 42 and a₁₀ : a₃₀ = 1 : 3

$\Rightarrow a_{10} : a_{30} = 1 : 3 \\[1em] \Rightarrow \dfrac{a_{10}}{a_{30}} = \dfrac{1}{3} \\[1em] \Rightarrow 3a_{10} = a_{30}$

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

$\Rightarrow 3[a + (10 - 1)d] = a + (30 - 1)d \\[1em] \Rightarrow 3(a + 9d) = a + 29d \\[1em] \Rightarrow 3a + 27d = a + 29d \\[1em] \Rightarrow 3a - a = 29d - 27d \\[1em] \Rightarrow 2a = 2d \\[1em] \Rightarrow a = d.$

S₆ = 42 or,

⇒ $\dfrac{6}{2}[2a + (6 - 1)d]$ = 42

Since, a = d
⇒ 3[2d + 5d] = 42
⇒ 3 × 7d = 42
⇒ 21d = 42
⇒ d = 2.

Since, a = d hence a = 2.

By formula, a_n = a + (n - 1)d
⇒ a₁₃ = 2 + (13 - 1)(2)
⇒ a₁₃ = 2 + 24
⇒ a₁₃ = 26.

Hence, the first term of the A.P. is 2 and the thirteenth term is 26.

In an A.P., the sum of its first n terms is 6n - n². Find its 25th term.

Answer

Given, S_n = 6n - n².

So, sum of (n -1) terms will be

⇒ S_{n - 1} = 6(n - 1) - (n - 1)²
⇒ S_{n - 1} = 6n - 6 -(n² + 1 - 2n)
⇒ S_{n - 1} = 6n - 6 - n² - 1 + 2n
⇒ S_{n - 1} = 8n - n² - 7.

By formula, a_n = S_n - S_{n - 1}
⇒ a_n = 6n - n² - (8n - n² - 7)
⇒ a_n = 6n - 8n - n² + n² + 7
⇒ a_n = -2n + 7.

∴ a₂₅ = -2(25) + 7 = -50 + 7 = -43.

Hence, the 25th term of the A.P. is -43.

If S_n denotes the sum of first n terms of an A.P., prove that S₃₀ = 3(S₂₀ - S₁₀).

Answer

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

We need to prove S₃₀ = 3(S₂₀ - S₁₀).

$\Rightarrow \text{L.H.S.} = S_{30} = \dfrac{30}{2}[2a + (30 - 1)d] \\[1em] = 15[2a + 29d] \\[1em] = 30a + 335d \\[1em] \Rightarrow \text{R.H.S.} = 3(S_{20} - S_{10}) \\[1em] = 3\Big(\dfrac{20}{2}[2a + (20 - 1)d] - \dfrac{10}{2}[2a + (10 - 1)d]\Big) \\[1em] = 3\Big(10[2a + 19d] - 5[2a + 9d]\Big) \\[1em] = 3(20a + 190d - 10a - 45d) \\[1em] = 3(10a + 145d) \\[1em] = 30a + 335d.$

∴ L.H.S. = R.H.S. = 30a + 335d.

Hence, proved that S₃₀ = 3(S₂₀ - S₁₀).

Find the sum of first positive 1000 integers.

Answer

We need to find 1 + 2 + 3 + 4 + ...... + 1000

The above series is an A.P. with first term a = 1, d = 1 and n = 1000.

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

⇒ S₁₀₀₀ = $\dfrac{1000}{2}[2 \times 1 + (1000 - 1)1]$
⇒ S₁₀₀₀ = 500[2 + 999]
⇒ S₁₀₀₀ = 500 × 1001
⇒ S₁₀₀₀ = 500500.

Hence, the sum of first 1000 positive integers is 500500.

Find the sum of first 15 multiples of 8.

Answer

We need to find 8 + 16 + 24 + ..... + (15th term)

The above series forms an A.P. with a = 8 and d = 8.

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

⇒ S₁₅ = $\dfrac{15}{2}[2 \times 8 + (15 - 1)8]$
⇒ 2S₁₅ = 15[16 + 14(8)]
⇒ 2S₁₅ = 15[128]
⇒ 2S₁₅ = 1920
⇒ S₁₅ = 960

Hence, the sum of first 15 terms is 960.

Find the sum of all two digit natural numbers which are divisible by 4.

Answer

The sum of series of the two digit natural numbers that are divisible by 4 is $12 + 16 + 20 + .... + 96.$

a = 12, d = 16 - 12 = 4 and l = 96.

Let 96 be nth term then,

⇒ 96 = 12 + 4(n - 1)
⇒ 96 - 12 = 4n - 4
⇒ 84 = 4n - 4
⇒ 4n = 84 + 4
⇒ 4n = 88
⇒ n = 22.

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

⇒ S₂₂ = $\dfrac{22}{2}[2 \times 12 + 4(22 - 1)]$
⇒ S₂₂ = 11[24 + 84]
⇒ S₂₂ = 11 × 108
⇒ S₂₂ = 1188.

Hence, the sum of all two digits natural numbers which are divisible by 4 is 1188.

Find the sum of all natural numbers between 100 and 200 which are divisible by 4.

Answer

The sum of natural numbers between 100 and 200 that are divisible by 4 is 104 + 108 + 112 + ..... + 196.

The above series is an A.P. with a = 104, d = 108 - 104 = 4 and l = 196.

Let 196 be nth term then,

⇒ 196 = 104 + 4(n - 1)
⇒ 196 - 104 = 4n - 4
⇒ 92 = 4n - 4
⇒ 4n = 92 + 4
⇒ 4n = 96
⇒ n = 24.

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

⇒ S₂₄ = $\dfrac{24}{2}[2 \times 104 + 4(24 - 1)]$
⇒ S₂₄ = 12[208 + 92]
⇒ S₂₄ = 12 × 300
⇒ S₂₄ = 3600.

Hence, the sum of all two digits natural numbers between 100 and 200 which are divisible by 4 is 3600.

Find the sum of all multiples of 9 lying between 300 and 700.

Answer

The sum of all multiples of 9 lying between 300 and 700 is 306 + 315 + ..... + 693.

The above series is an A.P. with a = 306, d = 9 and l = 693.

Let 693 be nth term of the series then,

⇒ 693 = 306 + 9(n - 1)
⇒ 693 - 306 = 9n - 9
⇒ 387 = 9n - 9
⇒ 9n = 387 + 9
⇒ 9n = 396
⇒ n = 44.

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

⇒ S₄₄ = $\dfrac{44}{2}[2 \times 306 + 9(44 - 1)]$
⇒ S₄₄ = 22[612 + 387]
⇒ S₄₄ = 22 × 999
⇒ S₄₄ = 21978.

Hence, the sum of all multiples of 9 lying between 300 and 700 is 21978.

Find the sum of all natural numbers less than 100 which are divisible by 6.

Answer

The sum of all natural numbers less than 100 which are divisible by 6 is given as 6 + 12 + 18 + .... + 96.

The above series is an A.P. with a = 6, d = 6 and l = 96.

Let 96 be nth term of the series then,

⇒ 96 = 6 + 6(n - 1)
⇒ 96 - 6 = 6n - 6
⇒ 90 = 6n - 6
⇒ 6n = 90 + 6
⇒ 6n = 96
⇒ n = 16.

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

⇒ S₁₆ = $\dfrac{16}{2}[2 \times 6 + 6(16 - 1)]$
⇒ S₁₆ = 8[12 + 90]
⇒ S₁₆ = 8 × 102
⇒ S₁₆ = 816.

Hence, the sum of all natural numbers less than 100 which are divisible by 6 is 816.

An arithmetic progression (A.P.) has 3 as its first term. The sum of the first 8 terms is twice the sum of the first 5 terms. Find the common difference of the A.P.

Answer

Let common difference be d.

a = 3

Sum of first n terms of an A.P. = $\dfrac{\text{n}}{2}(\text{a } + \text{ l})$

Given,

The sum of the first 8 terms is twice the sum of the first 5 terms.

$\therefore \dfrac{8}{2}(a + a_8) = 2 \times \dfrac{5}{2}(a + a_5) \\[1em] \Rightarrow 4[a + a + (8 - 1)d] = 5[a + a + (5 - 1)d] \\[1em] \Rightarrow 4[2a + 7d] = 5[2a + 4d] \\[1em] \Rightarrow 4[2 \times 3 + 7d] = 5[2 \times 3 + 4d] \\[1em] \Rightarrow 4[6 + 7d] = 5[6 + 4d] \\[1em] \Rightarrow 24 + 28d = 30 + 20d \\[1em] \Rightarrow 28d - 20d = 30 - 24 \\[1em] \Rightarrow 8d = 6 \\[1em] \Rightarrow d = \dfrac{6}{8} = \dfrac{3}{4}.$

Hence, common difference = $\dfrac{3}{4}$.

Find the next term of the list of numbers $\dfrac{1}{6}, \dfrac{1}{3}, \dfrac{2}{3}, ...$

Answer

The given list of numbers $\dfrac{1}{6}, \dfrac{1}{3}, \dfrac{2}{3}, ...$ is a G.P. with first term a = $\dfrac{1}{6}$ and common ratio = r = 2.

By formula, a_n = ar^{n - 1}

∴ a₄ = $\dfrac{1}{6}(2)^3 = \dfrac{8}{6} = \dfrac{4}{3}.$

Hence, the next term of the G.P. is $\dfrac{4}{3}$.

Find the next term of the list of numbers $\dfrac{3}{16}, -\dfrac{3}{8}, \dfrac{3}{4}, -\dfrac{3}{2}, ...$

Answer

The given list of numbers,

$\dfrac{3}{16}, -\dfrac{3}{8}, \dfrac{3}{4}, -\dfrac{3}{2}, ...$ is a G.P with first term a = $\dfrac{3}{16}$ and common ratio = r = -2.

By formula, a_n = ar^{n - 1}

∴ a₅ = $\dfrac{3}{16}(-2)^4 = \dfrac{3}{16} \times 16 = 3.$

Hence, the next term of the G.P. is 3.

Find the 15th term of the series $\sqrt{3} + \dfrac{1}{\sqrt{3}} + \dfrac{1}{3\sqrt{3}} + ....$

Answer

The given list of numbers,

$\sqrt{3} + \dfrac{1}{\sqrt{3}} + \dfrac{1}{3\sqrt{3}} + ....$ is a G.P. with first term a = $\sqrt{3}$ and common ratio = r = $\dfrac{1}{3}.$

By formula, a_n = ar^{n - 1}

∴ a₁₅ = $\sqrt{3}\Big(\dfrac{1}{3}\Big)^{14} = 3^{1/2}\Big(\dfrac{1}{3^{14}}\Big) = 3^{1/2}\times 3^{-14} = 3^{1/2 - 14} = 3^{-27/2} .$

Hence, the 15th term of the G.P. is $3^{-27/2}$.

Find the 10th and nth terms of the list of numbers 5, 25, 125, ...

Answer

The given list of numbers 5, 25, 125 .... is a G.P. with first term a = 5 and r = 5.

By formula, a_n = ar^{n - 1}

$a_{10} = 5(5)^{10 - 1} = 5(5)^9 = 5^{10}. \\[1em] a_n = 5(5)^{n - 1} = 5 \times 5^n \times 5^{-1} = 5^n.$

Hence, the 10th term is 5¹⁰ and nth term is 5ⁿ.

Find the 6th and the nth terms of the list of numbers $\dfrac{3}{2}, \dfrac{3}{4}, \dfrac{3}{8}, ...$

Answer

The given list of numbers

$\dfrac{3}{2}, \dfrac{3}{4}, \dfrac{3}{8}, ...$ is a G.P. with first term a = $\dfrac{3}{2}$ and common ratio = r = $\dfrac{1}{2}.$

By formula, a_n = ar^{n - 1}

$a_6 = \dfrac{3}{2}\Big(\dfrac{1}{2}\Big)^{6 - 1} = \dfrac{3}{2}\Big(\dfrac{1}{2}\Big)^5 = \dfrac{3}{2} \times \dfrac{1}{32} = \dfrac{3}{64}. \\[1em] a_n = \dfrac{3}{2}\Big(\dfrac{1}{2}\Big)^{n - 1} = \dfrac{3}{2 \times 2^n \times 2^{-1}} = \dfrac{3}{2^n}.$

Hence, the 6th term is $\dfrac{3}{64} \text{ and nth term is } \dfrac{3}{2^n}.$

Find the 6th term from the end of the list of numbers 3, -6, 12, -24, .... , 12288.

Answer

The given list of numbers 3, -6, 12, -24, .... , 12288 is a G.P. with last term = l = 12288 and the common ratio = r = -2.

By formula, nth term from end = $l\Big(\dfrac{1}{r}\Big)^{n - 1}$

$\text{7th term from end } = 12288\Big(-\dfrac{1}{2}\Big)^{6 - 1} \\[1em] = 12288\Big(-\dfrac{1}{2}\Big)^5 \\[1em] = -\dfrac{12288}{32} \\[1em] = -384.$

Hence, the 6th term from end of the G.P. is -384.

Which term of the G.P.

(i) 2, $2\sqrt{2}$, 4, .... is 128?

(ii) $1, \dfrac{1}{3}, \dfrac{1}{9}, .... \text{ is } \dfrac{1}{243}$ ?

Answer

(i) Given a = 2, r = $\sqrt{2}$.

Let nth term be 128.

By formula, a_n = ar^{n - 1}.

$\Rightarrow 128 = 2(\sqrt{2})^{n - 1} \\[1em] \Rightarrow \dfrac{128}{2} = (2)^{(n - 1)/2} \\[1em] \Rightarrow 64 = (2)^{(n - 1)/2} \\[1em] \Rightarrow 2^6 = (2)^{(n - 1)/2} \\[1em] \Rightarrow \dfrac{n - 1}{2} = 6 \\[1em] \Rightarrow n - 1 = 12 \\[1em] \Rightarrow n = 13.$

Hence, 128 is 13th term of the G.P.

(ii) Given a = 1, r = $\dfrac{1}{3}$.

Let nth term be $\dfrac{1}{243}$.

By formula, a_n = ar^{n - 1}.

$\Rightarrow \dfrac{1}{243} = 1\Big(\dfrac{1}{3}\Big)^{n - 1} \\[1em] \Rightarrow \dfrac{1}{3^5} = \dfrac{1}{3^{(n - 1)}} \\[1em] \Rightarrow n - 1 = 5 \\[1em] \Rightarrow n = 6.$

Hence, $\dfrac{1}{243}$ is 6th term of the G.P.

Determine the 12th term of a G.P. whose 8th term is 192 and common ratio is 2.

Answer

Given, a₈ = 192 and r = 2.

By formula, a_n = ar^{n - 1}.

⇒ a₈ = a(2)^{(8 - 1)}
⇒ 192 = a(2)⁷
⇒ a = $\dfrac{192}{2^7} = \dfrac{192}{128} = \dfrac{3}{2}.$

12th term of the G.P. is a₁₂,

$\Rightarrow a_{12} = \dfrac{3}{2}(2)^{12 - 1} \\[1em] \Rightarrow a_{12} = \dfrac{3}{2} \times 2^{11} \\[1em] \Rightarrow a_{12} = 3 \times 2^{10} \\[1em] \Rightarrow a_{12} = 3 \times 1024 \\[1em] \Rightarrow a_{12} = 3072.$

Hence, the 12th term of the G.P. is 3072.

In a G.P., the third term is 24 and 6th term is 192. Find the 10th term.

Answer

Given, a₃ = 24, a₆ = 192.

By formula, a_n = ar^{n - 1}.

⇒ a₃ = ar^{(3 - 1)}
⇒ ar² = 24.             (Eq 1)

⇒ a₆ = ar^{(6 - 1)}
⇒ ar⁵ = 192.            (Eq 2)

Dividing Eq 2 by Eq 1

$\Rightarrow \dfrac{ar^5}{ar^2} = \dfrac{192}{24} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r^3 = (2)^3 \\[1em] \Rightarrow r = 2$

Putting value of r in Eq 1,

$\Rightarrow a(2)^2 = 24 \\[1em] \Rightarrow 4a = 24 \\[1em] \Rightarrow a = 6 \\[1em] \text{10th term of G.P.} = a_{10} \\[1em] \Rightarrow a_{10} = 6(2)^{10 - 1} \\[1em] = 6(2)^9 = 6 \times 512 = 3072.$

Hence, the 10th term of the G.P. is 3072.

Find the number of terms of a G.P. whose first term is $\dfrac{3}{4},$ common ratio is 2 and the last term is 384.

Answer

Let the number of terms be n.

Given, a = $\dfrac{3}{4}$, r = 2 and a_n = 384.

By formula, a_n = ar^{n - 1}.

$\Rightarrow 384 = \dfrac{3}{4}(2)^{n - 1} \\[1em] \Rightarrow \dfrac{384 \times 4}{3} = (2)^{n - 1} \\[1em] \Rightarrow (2)^{n - 1} = 512 \\[1em] \Rightarrow (2)^{n - 1} = (2)^9 \\[1em] \Rightarrow n - 1 = 9 \\[1em] \Rightarrow n = 10.$

Hence, the number of terms in the G.P. are 10.

Find the value of x such that

(i) $-\dfrac{2}{7}, x, -\dfrac{7}{2}$ are three consecutive terms of a G.P.

(ii) x + 9, x - 6 and 4 are three consecutive terms of a G.P.

(iii) x, x + 3, x + 9 are first three terms of a G.P.

Answer

(i) Since, $-\dfrac{2}{7}, x, -\dfrac{7}{2}$ are three consecutive terms of a G.P.

So,

$\Rightarrow \dfrac{x}{-\dfrac{2}{7}} = r = \dfrac{-\dfrac{7}{2}}{x} \\[1em] \Rightarrow \dfrac{x}{-\dfrac{2}{7}} = \dfrac{-\dfrac{7}{2}}{x} \\[1em] \Rightarrow x^2 = -\dfrac{7}{2} \times -\dfrac{2}{7} \\[1em] \Rightarrow x^2 = 1 \\[1em] \Rightarrow x^2 - 1 = 0 \\[1em] \Rightarrow (x - 1)(x + 1) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x + 1 = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = -1.$

Hence, the value of x = 1 or -1.

(ii) Since, x + 9, x - 6 and 4 are three consecutive terms of a G.P.

So,

$\Rightarrow \dfrac{x - 6}{x + 9} = r = \dfrac{4}{x - 6} \\[1em] \Rightarrow \dfrac{x - 6}{x + 9} = \dfrac{4}{x - 6} \\[1em] \Rightarrow (x - 6)^2 = 4(x + 9) \\[1em] \Rightarrow x^2 + 36 - 12x = 4x + 36 \\[1em] \Rightarrow x^2 - 12x - 4x + 36 - 36 = 0 \\[1em] \Rightarrow x^2 - 16x = 0 \\[1em] \Rightarrow x(x - 16) = 0 \\[1em] \Rightarrow x = 0 \text{ or } x - 16 = 0 \\[1em] \Rightarrow x = 0 \text{ or } x = 16.$

Hence, the value of x = 0 or 16.

(iii) Since, x, x + 3 and x + 9 are first three terms of a G.P.

So,

$\Rightarrow \dfrac{x + 3}{x} = r = \dfrac{x + 9}{x + 3} \\[1em] \Rightarrow \dfrac{x + 3}{x} = \dfrac{x + 9}{x + 3} \\[1em] \Rightarrow (x + 3)^2 = x(x + 9) \\[1em] \Rightarrow x^2 + 9 + 6x = x^2 + 9x \\[1em] \Rightarrow x^2 - x^2 + 9 + 6x - 9x = 0 \\[1em] \Rightarrow 9 - 3x = 0 \\[1em] \Rightarrow 3x = 9 \\[1em] \Rightarrow x = 3. \\[1em]$

Hence, the value of x = 3.

If the fourth, seventh and tenth terms of a G.P. are x, y, z respectively, prove that x, y, z are in G.P.

Answer

Given, a₄ = x, a₇ = y and a₁₀ = z.

By formula, a_n = ar^{n - 1}.

⇒ x = a₄ = a(r)³

⇒ y = a₇ = a(r)⁶

⇒ z = a₁₀ = a(r)⁹

From above equations,

$\Rightarrow \dfrac{y}{x} = \dfrac{ar^6}{ar^3} = r^3. \\[1em] \Rightarrow \dfrac{z}{y} = \dfrac{ar^9}{ar^6} = r^3.$