Question

The sum of an AP is 21 and product of th first and third term exceed the second term by 6.Find 3terms

Answer 1

Solution :

let the three term of an AP be (a - d), a and a + d

Then

(a - d) + a + (a + d) = 21

3a = 21

a = 21/3

a = 7

(a - d) (a + d) = a + 6

a² - d² = a + 6

7² - d² = 7 + 6

49 - d² = 13

d² = 49 - 13

d² = 36

d = ±√36

d = ± 6

therefore 3 terms are (7 - 6), 7 and (7 + 6)

I. e. 1 , 7 , 13 or 13 , 7 , 1

Answer 2

Step-by-step explanation:

Given : -

• The sum of an AP is 21 and product of th first .

• third term exceed the second term by 6.

To Find : -

• Find the 3 terms

Solution : -

Let a = 1st term,

d = common difference 

Then the 1st 3 terms are a, a + d, a + 2d 

Their sum is 3a + 3d = 21, so a + d = 7, and 

thus d = 7 - a. 

The problem says a(a + 2d) = a + d + 6 

Substitute all values :

a² + 2ad = a + d + 6

a²+ 2a(7 - a) = a + (7 - a) + 6 

a² + 14a - 2a² = 13

a² - 14a + 13 = 0 

(a - 1)(a - 13) = 0

a = 1 or a = 13 

If a = 1, d = 6; if a = 13, d = - 6.

So there  are 2 possible sequences: 1, 7, 13..or 13, 7, 1