Let P, q and 3 be respectively the first, third and fifth terms of an AP. Let d be the common difference. If the product (pa) is minimum, then what is the value of d?
Answers
Answer:
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Step-by-step explanation:
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Given : Let p, q and r be respectively the first, third and fifth terms of an AP. Let d be the common difference. the product |pr| is minimum
( correction in Question)
To Find : the value of d
Solution:
first Term = p
common difference = d
q = p + 2d => p = q - 2d
r = p + 4d => r = q - 2d + 4d => r = q + 2d
pr
= (q - 2d)(q + 2d)
= q² - 4d²
| pr |
= | q² - 4d²|
is minimum when
q² - 4d² = 0
=> 4d² = q²
=> d = ± q/2
Hence Value of d is ± q/2
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