Long Questions

The product of the first five terms of a geometric series is 243. If the third term of the geometric series is equal to the tenth term of an arithmetic series, find the sum of the first 19 terms of the arithmetic series.

Let, a be the first term, r be the common ratio, and d be the common difference.

As the product of the first five terms of a geometric series is 243, t₁ × t₂ × t₃ × t₄ × t₅ = 243 or, a × ar × ar² × ar³ × ar⁴ = 243
or, (ar²)⁵ = 3⁵
so, ar² = 3 --- (i)

Third term of the geometric series is equal to the tenth term of an arithmetic series. so, ar² = t₁₀ or, 3 = t₁₀ (from 'i')
so, 3 = a + 9d --- (ii)

For the sum of the first 19 terms of the arithmetic series,
S₁₉ = $\frac{19}{2}$ [2a + (19- 1)d] = $\frac{19}{2}$ (2a + 18d)
= $\frac{19}{2}$ × 2(a + 9d)
= 19(a + 9d)
= 19 × 3
= 57

Hence, the sum of the first 19 terms of the arithmetic series is 57.

There are five geometric means between a and b. If second and last mean are 2 and 16 respectively, find the value of a and b.

Solution,
Let, p, q, r, s, t are five geometric means between a and b.
Given, last mean = 16 = t
And, second mean = 2 = q

So, the geometric sequence is: a, p, 2, r, s, 16, b

From above,
Common ratio (r) = b/16 ---- (A)

Again, t₃ = 2 or, ar^{3 - 1} = 2
or, a $\left(\frac{b}{16}\right)^2$ = 2 (From 'A')
or, a = $\frac{512}{b^2}$ ---- (B)

Again, we have, r = $\left(\frac{b}{a}\right)^{\frac{1}{n \text{ + }1}}$ or, $\frac{b}{16}$ = $\left(b \text{ × } \frac{b^2}{512}\right)^{\frac{1}{6}}$ [From (B)]
or, $\frac{b}{16}$ = $\left(\frac{b^3}{512}\right)^{\frac{1}{6}}$
or, $\frac{b}{16}$ = $\left(\frac{b}{8}\right)^{\frac{1}{2}}$
Squaring on both side, we get,
or, $\frac{b^2}{16^2}$ = $\frac{b}{8}$
∴ b = 32

From (B),
∴ a = $\frac{512}{32^2}$ = $\frac{1}{2}$
Hence, a = 1/2 and b = 32

If the sum of the first 9 terms of an arithmetic series is 72 and the sum of the first 17 term is 289 and find the sum of the first 25 terms.

Here, In arithmetic series,
Sum of the first 9 terms (S₉) = 72
Sum of the first 17 terms (S₁₇) = 289

Now, S₉ = 72 or, $\frac{n}{2}$ {2a + (n - 1)d} = 72
or, $\frac{9}{2}$ {2a + (n - 1)d} = 72
or, a + 4d = 8
or, a = 8 - 4d ---- (A)

Again, S₁₇ = 289 or, $\frac{n}{2}$ {2a + (n - 1)d} = 289
or, $\frac{17}{2}$ {2a + 16d} = 289
or, a + 8d = 17
or, 8 - 4d + 8d = 17 (From 'A')
or, 4d = 9
So, d = 9/4

Finally, sum of the first 25 terms,
S₂₅ = $\frac{n}{2}$ {2a + (n - 1)d} = $\frac{25}{2}$ {-2 + 54}
= 650

If the second term and fifth term of the geometric series are 15 and 405 respectively, find the sum of first 6 terms of the series.

Solution,
Given, In geometric series,
Second term (t₂) = 15
Fifth term (t₅) = 405

Now, t₂ = 15 or, ar^{n - 1} = 15
or, ar^{2 - 1} = 15
or, ar = 15 ---- (A)

Again,
t₅ = 405 or, ar^{5 - 1} = 405
or, ar⁴ = 405
or, ar × r³ = 405
or, 15 × r³ = 405 (From 'A')
So, r = 3
And, a = 5 (From 'A')

Finally, sum of first six terms of the series,
S₆ = $\frac{a(r^n \text{ - }1)}{\text{r - 1}}$ = $\frac{5(r^6 \text{ - }1)}{\text{r - 1}}$
= $\frac{5(3^6 \text{ - }1)}{\text{3 - 1}}$
= $\frac{5 \times 728}{2}$
= 1820

Find the sum of the first 9 terms of a geometric series whose third term and seventh term are 20 and 320 respectively.

Solution,
Given, In geometric series,
Third term (t₃) = 20
Seventh term (t₇) = 320

Now, t₃ = 20 or, ar^{n - 1} = 20
or, ar^{3 - 1} = 20
or, ar² = 20 ---- (A)

Again,
t₇ = 320 or, ar^{7 - 1} = 320
or, ar⁶ = 320
or, ar² × r⁴ = 320
or, 20 × r⁴ = 320 (From 'A')
or, r⁴ = 16
So, r = 2
And, a = 5 (From 'A')

Finally, Sum fo first nine terms of the series,
S₉ = $\frac{a(r^n \text{ - }1)}{\text{r - 1}}$ = $\frac{5(2^9 \text{ - }1)}{\text{2 - 1}}$
= 2555

Insert four arithmetic means between 5 and 25.

Here,
Let, p, q, r, s be four arithmetic means between 5 and 25.
Where, first term = a = 5
And, last mean = b = 25

So, the arithmetic series are: 5, p, q, r, s, 25

From above,
Common difference (d) = $\frac{\text{b - a}}{\text{n + 1}}$ = $\frac{\text{25 - 5}}{\text{4 + 1}}$ = 4

Finally,
p = t₂ = a + (n - 1)d = 5 + (2 - 1)4
= 5 + 4
= 9

q = t₃ = a + (n - 1)d = a + 2d
= 5 + 8
= 13

r = t₄ = a + 3d = 5 + 12
= 17

s = t₅ = a + 4d = 5 + 16
= 21

Hence, required arithmetic means are 9, 13, 17, 21.

Find the sum of the first 8 terms of a geometric series whose fourth term is 54 and common ratio is 3.

Here,
Give, fourth term (t₄) = 54
And, common ratio (r) = 3

Now, t₄ = 54 or, ar^{4 - 1} = 54
or, ar³ = 54
∴ a = 2

Finally,
S₈ = $\frac{a(r^8 \text{ - }1 )}{\text{r - 1}}$ = $\frac{2(3^8 \text{ - }1 )}{\text{3 - 1}}$
= 6560

There are n arithmetic means between 4 and 24. If the ratio of third means to the last mean is 4:5 then find the number of terms in the series.

Here,
First term (a) = 4
Last term (b) = 24
Ratio of third mean (m₃) to last mean (m_n) = 4:5

Now, for common difference,
d = $\frac{\text{b - a}}{\text{n + 1}}$
= $\frac{\text{24 - 4}}{\text{n + 1}}$
= $\frac{20}{\text{n + 1}}$ ---- (A)

Again,
m₃ = t₄ = a + 3d = 4 + $\frac{60}{\text{n + 1}}$ (From 'A')
= $\frac{\text{4n + 64}}{\text{n + 1}}$

Again,
m_n = b - d

= 24 - d
=

\frac{\text{24n + 24 - 20}}{\text{n + 1}}

\frac{\text{24n + 4}}{\text{n + 1}}

Finally, according to question, m₃ : m_n = 4 : 5 or, $\frac{\text{4(n + 16)}}{\text{n + 1}}$ × $\frac{\text{n + 1}}{\text{4(6n+1)}}$ = $\frac{4}{5}$
or, $\frac{\text{n + 16}}{\text{6n + 1}}$ = $\frac{4}{5}$
or, 5n + 80 = 24n + 4
or, 76 = 19n
∴ n = 4

Here, total number of mean = 4
Hence, total number of terms in series = 4 + 2 = 6

The third and seventh terms of an arithmetic series are 18 and 30 respectively, find the sum of the first 20 terms.

Here,
Given, third term (t₃) = 18
And, seventh terms (t₇) = 30

Now, t₃ = 18 or, a + 2d = 18
or, a = 18 - 2d ---- (A)

Again, t₇ = 30 or, a + 6d = 30
or, 18 - 2d + 6d = 30 (From 'A')
or, 4d = 12
So, d = 3
And, a = 12 (From 'A')

Finally,
S₂₀ = $\frac{n}{2}$ {2a + (n - 1)d} = 10 {24 + 19(3)}
= 10(24 + 57)
= 810

The sum of three numbers in an arithmetic progression is 51 and the sum of their square is 939. Find the numbers.

Here,
Let three term in arithmetic progression is a - d, a, a + d

According to question,
(a - d) + a + (a + d) = 51
So, a = 17

Again, (a - d)² + a² + (a + d)² = 939 or, (17 - d)² + 17² + (17 + d)² = 939
or, 17² - 34d + d² + 17² + 17² + 34d + d² = 939
or, 867 + 2d² = 939
or, 2d² = 72
So, d = ±6

If d = 6,
a - d = 11
a = 17
a + d = 23 If d = -6
a - d = 23
a = 17
a + d = 11

Hence, required numbers are 11, 17, 23 or 23, 17, 11

Three terms in an arithmetic progression have sum 21 and product 315. Find the terms.

Here,
Let three term in arithmetic progression is a - d, a, a + d

According to question, (a - d) + a + (a + d) = 21 So, a = 7

Again, (a - d) × a × (a + d) = 315 or, (a² - d²)a = 315
or, (49 - d²)7 = 315
or, 4 = d²
So, d = ±2

If d = 2,
a - d = 5
a = 7
a + d = 9 If d = -2
a - d = 9
a = 7
a + d = 5

Hence, required numbers are 5, 7, 9 or 9, 7, 5

The ninth and nineteenth terms of an arithmetic series are 40 and 60 respectively. Find the sum of the first 25 terms of the series.

Here,
Given, ninth term (t₉) = 40
And, nineteenth term (t₁₉) = 60

Now, t₉ = 40 or, a + (9 - 1)d = 40
or, a + 8d = 40
or, a = 40 - 8d ---- (A)

Again, t₁₉ = 60 or, a + (19 - 1)d = 60
or, 40 - 8d + 18d = 60 (From 'A')
or, 40 + 10d = 60
or, 10d = 20
So, d = 2
And, a = 24 (From 'A')

Finally, Sum of first 25 terms,
S₂₅ = $\frac{n}{2}$ {2a + (n - 1)d} = $\frac{25}{2}$ {48 + 24(2)}
= 1200

In an arithmetic sequence, the sixth term is equal to three times the fourth term and the sum of the first three terms is -12. Find the sum of the first ten terms.

Given,
In arithmetic series, Sixth term (t₆) = 3 × fourth term (t₄) or, a + 5d = 3(a + 3d)
or, a + 5d = 3a + 9d
or, -2a = 4d
∴ a = -2d ---- (A)

Again, Sum of first three term = -12
or, (a - d) + a + (a + d) = -12
or, 3a = -12
So, a = -4
And, d = 2 (From 'A')

Finally, sum of first ten terms,
S₁₀ = $\frac{n}{2}$ {2a + (n - 1)d} = $\frac{10}{2}$ {-8 + (10 - 1)2}
= 5 ( -8 + 18)
= 50

The sum of first eight terms of a arithmetic series is 180 and its fifth term is five times of the first term then find the first term and common difference.

Given,
In arithmetic series, fifth term (t₅) = 5 × first term (a) or, a + 4d = 5a
So, a = d ---- (A)

Also, Sum of first eight terms (S₈) = 180 or, $\frac{n}{2}$ {2a + (n - 1)d} = 180
or, 4(2a + 7d) = 180
or, 9a = 45
So, d = 5
∴ a = 5 (From 'A')

Hence, a = 5 and b = 5

Find the number of geometric means inserted between 1 and 64 in which the ratio of first mean to the last mean is 1:16.

Here,
First term (a) = 1
Last term (b) = 64
Total geometric mean = n
first mean (m₁) : last mean (m_n) = 1 : 16

So, the geometric series is a, m₁, m₂, ... ..., m_n, b

From above,
Common ratio (r) = m₁/a = m₁

Similarly,
m_n = 64/r (b/m_n = r) = 64/m₁

According to question, $\frac{m_1}{m_n}$ = $\frac{1}{16}$ or, m₁ × $\frac{m_1}{64}$ = $\frac{1}{16}$
or, (m₁)² = 4
So, m₁ = 2
And, r = 2 (From 'A')

Finally, we have, r = $\left(\frac{b}{a}\right)^{\frac{1}{n \text{ + }1}}$ or, 2 = $\text{(64)}^{\frac{1}{n \text{ + }1}}$
or, (2)¹ = $(2)^{\frac{6}{ \text{n + 1}}}$
or, 1 = $\frac{6}{\text{n + 1}}$
∴ n = 5

In a geometric, the sum of the first three terms is 1 and the sum of its first 6 term is 28. Find the common ratio and first term.

Here,
In a geometric series,
Sum of first three terms (S₃) = 1
Sum of first six terms (S₆) = 28

Now, S₃ = 1 or, $\frac{a(r^n \text{ - }1)}{\text{r - 1}}$ = 1
or, $\frac{a(r^3 \text{ - }1)}{\text{r - 1}}$ = 1 ---- (A)

Again, S₆ = 28 or, $\frac{a(r^n \text{ - }1)}{\text{r - 1}}$ = 28
or, $\frac{a(r^6 \text{ - }1)}{\text{r - 1}}$ = 28
or, $\frac{a(r^3 \text{ +}1)(r^3 { - 1})}{\text{r - 1}}$ = 28 ---- (B)

Dividing (A) by (B), we get, $\frac{a\cancel{(r^3 - 1)}}{\cancel{(r - 1)}}$ × $\frac{\cancel{(r - 1)}}{a(r^3 +1)\cancel{(r^3 - 1)}}$ = $\frac{1}{28}$
or, r³ + 1 = 28
or, r³ = 27
So, r = 3
And, a = 1/13 (From 'A')

Hence, common ratio (r) is 3 and first term (a) is 1/13