Sum of three consecutive terms of an Arithmetic Progression is 42 and their product is 2520.
Find the terms of the Arithmetic Progression.
Answers
GIVEN :
Sum of three consecutive terms in an Arithmetic Progression = 42
Their product = 2520
Let the terms be (a - d), a and (a + d)
(a - d) + a + a + d = 42
a - d + a + a + d = 42
=> 3a = 42
=> a = 42/3
=> a = 14
(a - d) × a × (a + d)
(a - d) × 14 × (a + d) = 2520
(a - d) (a + d) = 2520/14
a² - d² = 180 [ (a - b) (a + b) = a² - b² ]
(14)² - d² = 180
196 - 180 = d²
=> d² = 16
=> d = √16
=> d = ±4
TERMS :
If d = -4,
First term = a - d = 14 - (-4) = 18
Second term = a = 14
Third term = a + d = 14 + (-4) = 10
If d = 4,
First term = a - d = 14 - 4 = 10
Second term = a = 14
Third term = a + d = 14 + 4 = 18
Therefore, the terms are 18, 14 and 10 or 10, 14 and 18.
Answer:
Terms of the Arithmetic progression
= 10 ,14 ,18
Step-by-step explanation:
Given
the Sum of three consecutive terms of an Arithmetic Progression = 42
Let the 1st term be a - d
So,
AP = a - d, a, a + d
ATQ,
a - d + a + a + d = 42
3a = 42
a = 42/3
a = 14
also,
(a - d ) (a)(a + d) = 2520
(a² - d²)14 = 2520
14² - d² = 2520/14
-d² = 180 - 196
-d² = -16
d = 4
1st term = a - d
= 14 - 4
= 10
2nd term = a = 14
3rd term = a + d
= 14 + 4
= 18
So,
Terms of the Arithmetic progression