Find the terms. 6. In an A.P.. the sum of 3 terms is 15 and the sum of squares of first and last term is 58. Find the terms. 7. The sum of 3 consecutive terms of an A.P. is 9 and their product is -48. Find the terms
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Answer:
Let the three terms of the AP be a-d, a, and a+d (in increasing order). Then, we have:
a-d + a + a+d = 15 (sum of 3 terms is 15)
3a = 15
a = 5
(a-d)^2 + (a+d)^2 = 58 (sum of squares of first and last term is 58)
2a^2 + 2d^2 = 58
d^2 = 8
Substituting a=5 and d^2=8 into the equation for the three terms, we get:
5-d + 5 + 5+d = 15
d = 1
Therefore, the three terms are 4, 5, and 6.
Let the three consecutive terms of the AP be a-d, a, and a+d. Then, we have:
(a-d) + a + (a+d) = 3a = 9 (sum of 3 terms is 9)
a = 3
(a-d)(a)(a+d) = -48 (product of 3 terms is -48)
(a^2 - d^2) a = -48
(a^2 - d^2) = -16
Substituting a=3 and solving for d, we get:
d = ±2
Therefore, the three consecutive terms of the AP are either 1, 3, 5 or 5, 3, 1.
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Answer:
Step-by-step explanation:
Let the three terms of the AP be a-d, a, and a+d (in increasing order). Then, we have:
a-d + a + a+d = 15 (sum of 3 terms is 15)
3a = 15
a = 5
(a-d)^2 + (a+d)^2 = 58 (sum of squares of first and last term is 58)
2a^2 + 2d^2 = 58
d^2 = 8
Substituting a=5 and d^2=8 into the equation for the three terms, we get:
5-d + 5 + 5+d = 15
d = 1
Therefore, the three terms are 4, 5, and 6.
Let the three consecutive terms of the AP be a-d, a, and a+d. Then, we have:
(a-d) + a + (a+d) = 3a = 9 (sum of 3 terms is 9)
a = 3
(a-d)(a)(a+d) = -48 (product of 3 terms is -48)
(a^2 - d^2) a = -48
(a^2 - d^2) = -16
Substituting a=3 and solving for d, we get:
d = ±2
Therefore, the three consecutive terms of the AP are either 1, 3, 5 or 5, 3, 1.
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