1. The sum of five numbers in AP is 40. The product of the first and the last is 28. Find then
find the numbers
Answers
Let those five terms be a - 2d, a - d, a, a + d and a + 2d.
Given: Sum of first five terms in A.P. is 40.
→ a - 2d + a - d + a + a + d + a + 2d = 40
→ 5a = 40
→ a = 8
Given: Product of last and first term is 28.
→ (a + 2d)(a - 2d) = 28
→ a² - 4d² = 28
→ 8² - 4d² = 28
→ - 4d² = 28 - 64
→ d² = - 36/(- 4)
→ d = √9
→ d = ±3
Formation of A.P
→ a - 2d = 8 - 2(±3) = 2 or 14
→ a - d = 8 - (±3) = 5 or 11
→ a = 8
→ a + d = 8 + (±3) = 11 or 5
→ a + 2d = 8 + 2(±3) = 14 or 2
Hence A.P is either 2, 5, 8, 11, 14 or 14, 11, 8, 5, 2.
Let five terms be a - 2d, a - d, a, a + d and a + 2d
» Sum of first five terms of an A.P. is 40.
→ a - 2d + a - d + a + a + d + a + 2d = 40
→ 5a = 40
→ a = 8 _______ (eq 1)
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» Product of last and first term is 28.
→ (a + 2d)(a - 2d) = 28
Now
(a + b)(a - b) = a² - b²
→ a² - 4d² = 28
→ 8² - 4d² = 28
→ - 4d² = 28 - 64
→ d² = - 36/(- 4)
→ d = √9
→ d = ±3 _______ (eq 2)
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A.P. :
› Take a = 8 and d = + 3
• a - 2d = 8 - 2(3)
=> 8 - 6 = 2
• a - d = 8 - 3
=> 5
• a = 8
• a + d = 8 + 3
=> 11
• a + 2d = 8 + 2(3)
=> 8 + 6 = 14
Similarly
› Take a = 8 and d = - 3
• a - 2d = 8 - 2(-3)
=> 8 + 6 = 14
• a - d = 8 + 3
=> 11
• a = 8
• a + d = 8 - 3
=> 5
• a + 2d = 8 + 2(-3)
=> 8 - 6 = 2
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2, 5, 8, 11, 14 and 14, 11, 8, 5, 2 are the numbers..
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