Question
Sum of three consecutive terms in a GP is 42 and their product is 512. Find the largest of these numbers.
Answer: Option A
:
A
Let the three consecutive terms bear,a,ar
Product =(ar)(a)(ar) = a3= 512, ⇒a= 8
Sum of these numbers=a(1+r+1r)=42⇒8(1+r+1r)=42
4r2−17r+4=0
(4r – 1)(r – 4) = 0
r = 14or r = 4.
Hence the largest number is 32. The series is 32, 8, 2 or 2, 8, 32.
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:
A
Let the three consecutive terms bear,a,ar
Product =(ar)(a)(ar) = a3= 512, ⇒a= 8
Sum of these numbers=a(1+r+1r)=42⇒8(1+r+1r)=42
4r2−17r+4=0
(4r – 1)(r – 4) = 0
r = 14or r = 4.
Hence the largest number is 32. The series is 32, 8, 2 or 2, 8, 32.
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