Question

in an ap the sum of second and third term is 22 and the product of first and fourth term is 85 ,find the sum of first 10 terms​

Answer 1

Answer:

Let the numbers be

a-3d, a-d, a+d, a+3d

(a-d) + (a+d) = 22

a = 11

(a-3d) (a+3d) = 35

$a^{2} - 9d^{2} = 85$

$9d^{2} = 11^{2} - 85 = 35$

$d = \sqrt{4} = 2$

or -2

So the numbers are

5, 9, 13, 17

or

17, 13, 9, 5

Answer 2

In the Ap, The required terms are 5, 9, 13, 17.

Step-by-step explanation:

When the consecutive terms of series differ by a common number, then the series is said to be Arithmetic Progression.

Let:

• a be the first term of the AP
• d be the common difference of the AP
• nth term of AP ⇒ a + ( n - 1) d

Given:

• S (2nd term + 3rd term) is 22

⇒ (a + d) + (a + 2d) =22

⇒ 2a + 3d = 22

⇒ d = (22- 2a) × 1/3

• The product of the first and fourth term is 85

⇒ a ( a + 3d) = 85

⇒ a² + 3ad = 85

Substituting the value of d gives,

⇒ a ( a + 3 ( 1/3 * (22 - 2a) )) = 85

⇒ a ( a + 22 - 2a) = 85

⇒ a ( - a + 22) = 85

⇒ - a² + 22a = 85

⇒ a² - 22a + 85 = 0

⇒ a² - 17a - 5a + 85 = 0

⇒ a ( a - 17) - 5 ( a - 17)= 0

⇒ (a-5)(a-17)= 0

⇒ a = 5 or a = 17

• If a = 5,

d = 1/3 ( 22 - 10) = 1/3 ( 12) = 4

• If a = 17,

d = 1/3 ( 22 - 34) = 1/3 ( - 12) = - 4

• a = 5, d = 4.

Then Arithmetic Progression is 5, 9, 13, 17

• a = 17, d = - 4

Then Arithmetic Progression is, 17, 13, 9, 5

Therefore, The required terms in the AP are 5, 9, 13, 17.