Find four consecutive terms in an A.P. such that the sum of the middle two terms is 18 and product of the two end terms is 45 ?
Answers
Four consecutive terms in an A.P. such that the sum of the middle two terms is 18 and product of the two end terms is 45 are: 3, 7, 11, 15
Given:
i) There are 4 consecutive terms in an A.P
ii) Sum of the middle two terms is 18
ii) Product of the two end terms is 45
To find:
Values of those 4 consecutive terms
Solution:
Let the 4 consecutive terms in AP in increasing order be
a - 3d, a - d, a + d, a + 3d
where, (2*d) is the common difference of this AP series and 'a' denote any integer.
It is given that
Sum of the middle two terms is 18
Here, the middle two terms are a - d and a + d
=> (a - d) + (a + d) = 18
=> 2 a = 18
=> a = 9 ---(1)
Also,
Product of the two end terms = 45
=> (a - 3d) * (a + 3d) = 45
=> a^2 - (3d)^2 = 45
Substituting the value of a from equation (1) in this expression, we get
9^2 - (3d)^2 = 45
=> 81 - 9 (d^2) = 45
=> 9 (d^2) = 81 - 45
=> 9 (d^2) = 36
=> d^2 = 4
=> d = √4
=> d = +2, -2
Since we have considered the 4 consecutive terms in increasing order, the value of d should be positive and therefore the value d = -2 is rejected.
Hence, d = +2
Therefore, the 4 terms of AP are:
9 - 3*2, 9 - 2, 9 + 2, 9 + 3*2
=> 9 - 6, 9 - 2, 9 + 2, 9 + 6
=> 3, 7, 11, 15
Hence,
Four consecutive terms in an A.P. such that the sum of the middle two terms is 18 and product of the two end terms is 45 are: 3, 7, 11, 15
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