Write down the choice of three consecutive numbers in A.P.
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Answers
Answer:
Given:
The consecutive terms in A.P. whose sum is 9 and the product of their cubes is 3375.
To find:
The three consecutive terms in A.P.
Explanation:
Let the three consecutive terms in an A.P. be a-d, a, a+d.
Given,
The sum of these three consecutive numbers is 9.
→ a - d + a + a + d = 9
→ 3a = 9
→ a = \cancel{\frac{9}{3} }
3
9
→ a = 3
The product of their cube is 3375.
⇒ (a-d)³ × a³ × (a+d)³ = 3375
Putting the value of a in this product,we get;
⇒ (3-d)³ × (3)³ × (3+d)³ = 3375
⇒ (3-d)³ × 27 × (3+d)³ = 3375
⇒ (3-d)³ × (3+d)³ = \cancel{\frac{3375}{27} }
27
3375
⇒ (9+3d -3d -d²)³ = 125
⇒ (9 -d²)³ = (5)³
⇒ (9-d^{2} )^\cancel{{3} }\:=(5)^\cancel{{3} }
⇒ 9 - d² = 5
⇒ -d² = 5 - 9
⇒ -d² = -4
⇒ d² = 4
⇒ d= √4
⇒ d = 2
∴ Three consecutive terms in Arithmetic Progression;
1st term, a-d = 3 -2= 1
2nd term, a= 3
3rd term, a+d= 3+2= 5
Step-by-step explanation: