Question

Find 3 consecutive numbers in AP whose sum is 30 and product is 640
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Answer 1

Answer:

numbers are 4, 10, 16

Step-by-step explanation:

Let (a-d),(a) and (a+d) be three consecutive terms in A.P.

As given,

(a - d )+ a+ (a + d)= 30

(a-d) a (a+d)  = 640

From first equation,

3a = 30

a =10

Substitute value of a in the equation two we get,

(a-d) a (a+d)  = 640

(10-d) 10 (10+d) = 640

(10 - d) (10 + d) = 64

100 - d² = 64

d² = 36

d = 6

numbers are 4, 10, 16

Answer 2

Hello,Buddy!!

Given:-

• 3 Consecutive Numbers in AP.
• Sum of the numbers is 30 & their product is 640.

To Find:-

• Value of the three numbers.

Required Solution:-

Let,

Three Numbers be (a-d),a & (a+d) which are in AP

According To Sum

$(a - d) + a + (a + d) = 30 \\ a \cancel{ - d} + a + a \cancel{ + d} = 30 \\ 3a = 30 \\ a = \cancel{ \frac{30}{3} } \\ a = 10$

• Value of a ➪ 10

$(a - d).a.(a + d) = 640 \\ {a}^{2} - {d}^{2} = \frac{640}{a} \\ {(10)}^{2} - {d}^{2} = \cancel \frac{640}{10} \\ - {d}^{2} = 64 - 100 \\ - {d}^{2} = - 36 \\ d = \sqrt{36} \\ d = 6$

• Value of d ➪ 6.

By Substituting Values of a & d

→ (a-d) = 10-6 ☞ 4

→ a = 10

→ (a+d) = 10+6 ☞ 16

• Required Sequence ➪ 4,10,16.

$\boxed{\tt{@MrSovereign}}$

Hope This Helps!!