An arithmetic progression or ap is a sequence where the difference between two successive terms is always a constant.The sum of 3 consecutive terms of an ap is 27 and the product of these 3 terms is 704.The first term of this ap is
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0
Let 3 consecutive terms in an A.P are and, a, a+d
By 1st condin
a - d + a + a + d = 27
3 a = 27
a = 27/3
a = 9
By 2nd condin
(a-d) (a) (a+d) = 504
(g-d) x g x (g+d) = 504
(g-d) (g+d) = 504/g
(g-d) (g+d) = 56
(g)² - (d)² = 56
d² = 56 - 81
d² = 25
d² = 25
Taking square root
d = + - 5
abhi178:
plz correct your answer
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Let (a - d ) , a , (a + d) are three terms in an AP.
A/C to question,
sum of three terms = 27
(a - d) + a + (a + d) = 27
3a = 27
a = 9............(1)
product of these three terms = 704
(a - d)a(a + d) = 704
(a² - d²)a = 704
(a³ - ad²) = 704
from equation (1),
(9³ - 9d²) = 704
729 - 9d² = 704
729 - 704 = 9d²
25/9 = d²
d = ±5/3
so, the first term of this ap = (a - d)
= (9 - 5/3) or, (9 + 5/3)
= 22/3 or, 32/3
A/C to question,
sum of three terms = 27
(a - d) + a + (a + d) = 27
3a = 27
a = 9............(1)
product of these three terms = 704
(a - d)a(a + d) = 704
(a² - d²)a = 704
(a³ - ad²) = 704
from equation (1),
(9³ - 9d²) = 704
729 - 9d² = 704
729 - 704 = 9d²
25/9 = d²
d = ±5/3
so, the first term of this ap = (a - d)
= (9 - 5/3) or, (9 + 5/3)
= 22/3 or, 32/3
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