Question

The sum of the three terms which are in an A.P is 33. If the product of the first and third term exceeds the second term by 29. Find the A.P

Answer 1

Answer:

$\boxed{\red{\bf AP : 2, 11, 20.. \: or \: 20, 11, 2.. }}$

Step-by-step explanation:

Given that -

• The sum of the three terms which are in an AP is 33.
• The product of the first and third term exceeds the second term by 29.

To find :

• The required AP.

Solution :

Let the first term, second term and third term od the AP be (a - d), a and (a + d) respectively.

⇒ Sum of the terms = 33

⇒ a - d + a + a + d = 33

⇒ 3a = 33

⇒ a = $\dfrac{33}{3}$

⇒ a = 11

Also, it is given that :

First term * Third term = Second term + 29

⇒ (a - d)(a + d) = a + 29

⇒ a² - d² = a + 29

⇒ (11)² - d² = a + 29

⇒ 121 - d² = 11 + 29

⇒ - d² = 40 - 121

⇒ - d² = - 81

⇒ d² = 81

⇒ d = ± √81

⇒ d = ± 9

Consider the value of d as positive :

When d = 9.

AP : (a - d), a, (a + d)

⇒ AP : (11 - 9), 11, (11 + 9)

⇒ AP : 2, 11, 20...

Consider the value of d as negative :

When d = - 9.

AP : (a - d), a, (a + d)

⇒ AP : (11 + 9), 11, (11 - 9)

⇒ AP : 20, 11, 2..

Hence, the required AP is 2, 11, 20... or 20, 11, 2....

Answer 2

Given :

• The sum of the three terms which are in an A.P is 33.
• The product of the first and third term exceeds the second term by 29.

To Find :

• The AP.

Solution :

Let the first term be (a-d)

Let the second term be a.

Let the third term be (a + d)

Case 1 :

The sum of three terms is 33.

Equation :

$\sf{(\:a\:-\:d)\:+\:(a)\: +\: (a\:+\:d) \:=\:33}$

→ a - d + a + a + d = 33

→ 3a = 33

→a = 33/3

→ a = 11.... (1)

Case 2 :

The product of 1st term and 3rd term exceeds 2nd term by 29.

Equation

→ (a-d) (a+d) = a + 29

→(11 - d) (11 + d) = 11 + 29

→ 11(11+d) -d (11+d) = 40

→ 121 + 11d - 11d - d² = 40

→121 - d² = 40

→ -d² = 40 - 121

→-d² = - 81

→ d² = 81

→ d = √81

→ d = ±9

Now, we have the value of first term as well as the value of common difference, we can therefore form our required AP.

When, d = 9,

First term,

→ a - d

→ 11 - 9

→ 2

Second tem,

→ a

→ 11

Third term,

→ a + d

→ 11 + 9

→ 20

$\large{\boxed{\red{\sf{AP : 2,11,20...t_n}}}}$

Now, when d = - 9,

First term,

→ a - d

→11 - (-9)

→ 11 + 9

→ 20

Second term,

→ a

→ 11

Third term,

→ a + d

→ 11 + (-9)

→ 11 - 9

→ 2

$\large{\boxed{\purple{\sf{AP : 20,11,2...t_n}}}}$