Question

The sum of first three numbers in an arithmetic progression is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers.
using the term a-d,a,a+d

Want to check my answer in todays board exam ​

Answer 1

Answer:

(a-d) + a + (a+d) = 18

So, 3a =18

a=6

Thus,

(a-d) (a+d) =5d

(6-d)(6+d) = 5d

d^2 + 5d - 36 = 0

d^2 +9d - 4d - 36 = 0

d(d+9) - 4(d+9) = 0

(d+9)(d-4) = 0

d = - 9 or 4

So, the three nos are 2,6,10 or 15,6,-3

Answer 2

Answer:

When a = 6 and d = 4 , A.P. : 2 , 6 , 10

When a = 6 and d = -9 , A.P. : 15 , 6 , -3

Step-by-step explanation:

Let the three numbers in an A.P. are a - d, a , a + d

First term = a

Common difference = d

Now, sum of three numbers is 18

⇒ a - d + a + a + d = 18

⇒ 3a = 18

⇒ a = 6

Therefore, The firs term is 6

Now, product of the first and the third term is 5 times the common difference

⇒ (a - d)·(a + d) = 5d

⇒ a² - d² = 5d

⇒ d² + 5d - 36 = 0

⇒ d² + 9d - 4d - 36 = 0

⇒ (d - 4)·(d + 9) = 0

⇒ d = 4 or d= -9

Hence, the possible cases of three numbers are :

When a = 6 and d = 4 , A.P. : 2 , 6 , 10

When a = 6 and d = -9 , A.P. : 15 , 6 , -3