Question-The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and last term to the product of two middle term is 7 : 15.find numbers.
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Question-The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and last term to the product of two middle term is 7 : 15.find numbers.
SOLUTION-
Let the four consecutive terms of an AP are a – 3d, a – d, a + d, a + 3d
According to question that the sum of these term is 32
Therefore a – 3d + a – d + a + d + a + 3d = 32
4a = 32d
a = 8
15a² – 135d² = 7a² – 7d²
15a² – 7a² = 135d² – 7d²
8a² = 128d²
a² = 16d²
Putting the value of a = 8
16d² = 8² = 64
If d = 2 then the terms will be
a – 3d = 8 – 3×2 = 2, a – d = 8 – 2 =6 ,a + d = 8 + 2= 10 ,a + 3d = 8 + 3×2 =14
If d = –2 then a – 3d = 8 – 3 × –2 = 14, a – d = 8 – (–2) = 10, a + d = 8 + (–2) = 6, a + 3d = 8 + 3× –2= 2
Therefore the terms of AP are 2,6,10 and 14 or 14, 10, 6 and 2
Alternative Method-
Let the four terms of AP are a, a + d, a + 2d, a + 3d
As it is given that sum of these terms is 32
Therefore a + a + d + a + 2d + a + 3d = 32
4a + 6d = 32
2a + 3d = 16………….(i)
We are given that the ratio between the product of first and last term is 7 : 15
15a² + 45ad = 7(a² +2ad +ad + 2d²)
15a² + 45ad = 7a² + 14ad + 7ad + 14d²
8a² + 24ad – 14d² = 0
4a² + 12ad – 7d² = 0
Substituting 2a + 3d = 16⇒ a = (16 – 3d)/2 in equation (ii)
256 + 9d² – 96d + 96d – 18d² – 7d² = 0
–16d² + 256 = 0
16d² = 256
d² = 16
d = ±4
Substituting value of d = 4 in equation number (i) 2a + 3d = 16 value of a = 2 and if d =–4 then a = 14
When d = 4
First term a = 2, second term a + d = 2 + 4 = 6, third term = a + 2d = 2 + 2 ×4 = 10, fourth term = a + 3d = 2 + 3 ×4 = 14
When d = –4
First term = 14, second term a +d= 14 –4 = 10, third term a +2d =14 –2×4 =6, fourth term a + 3d = 14 –3×4 = 2
Therefore the reqired term of AP 2,6,10,14 or 14,10,6,2
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