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Mathematics Working Solutions

Arithmetic Progressions (A.P.) & Series — Clean Typed Notes

Example 2.24

Question: Write an A.P. whose first term is 20 and common difference is 8.

Solution:

Given first term a = 20, common difference d = 8.

The general form of an Arithmetic Progression is: a, a + d, a + 2d, a + 3d, ...

1st term (t1):	a = 20
2nd term (t2):	a + d = 20 + 8 = 28
3rd term (t3):	a + 2d = 20 + 2(8) = 20 + 16 = 36

Required A.P. series: 20, 28, 36, ...

Example 2.26

Question: Find the number of terms in the A.P. 3, 6, 9, 12, ..., 111.

Solution:

Given sequence: 3, 6, 9, 12, ..., 111

First term (a):	3
Common diff (d):	6 - 3 = 3
Last term (l):	111

Formula for number of terms (n):

n = ^{(l - a)}/_d + 1

n = ^{(111 - 3)}/₃ + 1 = ¹⁰⁸/₃ + 1 = 36 + 1 = 37

Number of terms (n) = 37

Problem 4

Question: Find the 19^th term of an A.P. -11, -15, -19, ...

Solution:

Formula for n^th term: t_n = a + (n - 1)d

First term (a):	-11
Common diff (d):	-15 - (-11) = -15 + 11 = -4
Term number (n):	19

t₁₉ = -11 + (19 - 1)(-4)
t₁₉ = -11 + 18(-4)
t₁₉ = -11 + (-72) = -83

19^th term (t₁₉) = -83

Problem 5

Question: Which term of an A.P. 16, 11, 6, 1, ... is -54?

Solution:

First term (a):	16
Common diff (d):	t2 - t1 = 11 - 16 = -5
Last term (l):	-54

Formula for number of terms (n):

n = ^{(l - a)}/_d + 1
n = ^{(-54 - 16)}/_-5 + 1
n = ^-70/_-5 + 1 = 14 + 1 = 15

-54 is the 15^th term of the A.P.

Problem 11

Question: The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.

Solution:

Let the three consecutive terms be (a - d), a, (a + d).

1. From the sum condition:

(a - d) + a + (a + d) = 27
3a = 27 ⇒ a = 9

2. From the product condition:

(a - d) × a × (a + d) = 288
a(a² - d²) = 288

Substitute a = 9:

9(9² - d²) = 288
81 - d² = ²⁸⁸/₉ = 32
d² = 81 - 32 = 49 ⇒ d = ±7

Taking a = 9, d = 7:

Terms: (9 - 7), 9, (9 + 7) ⇒ 2, 9, 16

(Taking d = -7 gives terms: 16, 9, 2)

The three terms are: 2, 9, 16 (or 16, 9, 2)

Problem 12

Question: The ratio of 6^th and 8^th term of an A.P. is 7 : 9. Find the ratio of 9^th term to 13^th term.

Solution:

Given: t₆ : t₈ = 7 : 9

^{(a + 5d)}/_{(a + 7d)} = ⁷/₉
9(a + 5d) = 7(a + 7d)
9a + 45d = 7a + 49d
9a - 7a = 49d - 45d
2a = 4d ⇒ a = 2d

Now find ratio of 9^th term to 13^th term (t₉ : t₁₃):

^t₉/_t13 = ^{(a + 8d)}/_{(a + 12d)}

Substitute a = 2d:

^t₉/_t13 = ^{(2d + 8d)}/_{(2d + 12d)} = ^10d/_14d = ⁵/₇

Ratio of 9^th term to 13^th term = 5 : 7

Example 2.36

Question: Find the sum of all natural numbers between 300 and 600 which are divisible by 7.

Solution:

Natural numbers between 300 and 600 divisible by 7 form an A.P.:

301, 308, 315, ..., 595

First term (a):	301
Common diff (d):	7
Last term (l):	595

1. Find the number of terms (n):

n = ^{(l - a)}/_d + 1 = ^{(595 - 301)}/₇ + 1 = ²⁹⁴/₇ + 1 = 42 + 1 = 43

2. Find the sum (S_n):

S_n = ⁿ/₂ [a + l]
S₄₃ = ⁴³/₂ [301 + 595] = ⁴³/₂ × 896 = 43 × 448 = 19,264

Sum of natural numbers = 19,264

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