The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the
last term to the product of two middle terms is 7:15. Find the numbers.
(March, 2018)
Answers
Let the four consecutive numbers in AP be (a−3d),(a−d),(a+d) and (a+3d)
So, according to the question.
a−3d+a−d+a+d+a+3d=32
4a=32
a=32/4
a=8......(1)
Now, (a−3d)(a+3d)/(a−d)(a+d)=7/15
15(a²−9d²)=7(a²−d²)
15a²−135d²=7a²−7d²
15a²−7a²=135d²−7d²
8a²=128d²
Putting the value of a=8 in above we get.
8(8)²=128d²
128d²=512
d²=512/128
d²=4
d=2
So, the four consecutive numbers are
8−(3×2)
8−6=2
8−2=6
8+2=10
8+(3×2)
8+6=14
Four consecutive numbers are 2,6,10and14
hope this helped you
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Answer :
The required numbers are 2 , 6 , 10 and 14.
Step-by-step explanation :
Given :
- The sum of four consecutive numbers in an AP is 32
- the ratio of the product of the first and the last term to the product of two middle terms is 7:15.
To find :
the numbers
Solution :
Let the four consecutive numbers in AP be
a - 3d , a - d , a + d , a + 3d
Their sum = 32
➙ a - 3d + a - d + a + d + a + 3d = 32
➙ 4a = 32
➙ a = 32/4
➙ a = 8
⇾ The product of the first and last term
= (a - 3d) (a + 3d)
= a(a + 3d) - 3d(a + 3d)
= a² + 3ad - 3ad - 9d²
= a² - 9d²
= 8² - 9d²
= 64 - 9d²
⇾ The product of two middle terms
= (a - d) (a + d)
= a(a + d) - d(a + d)
= a² + ad - ad - d²
= a² - d²
= 8² - d²
= 64 - d²
the ratio of the product of the first and the last term to the product of two middle terms is 7 : 15
15(64 - 9d²) = 7(64 - d²)
960 - 135d² = 448 - 7d²
960 - 448 = 135d² - 7d²
128d² = 512
d² = 512/128
d² = 4
d = √4
d = +2 , -2
If d = 2,
a - 3d = 8 - 3(2) = 8 - 6 = 2
a - d = 8 - 2 = 6
a + d = 8 + 2 = 10
a + 3d = 8 + 3(2) = 8 + 6 = 14
The four consecutive numbers are 2, 6, 10 and 14
If d = -2,
a - 3d = 8 - 3(-2) = 8 + 6 = 14
a - d = 8 - (-2) = 8 + 2 = 10
a + d = 8 + (-2) = 8 - 2 = 6
a + 3d = 8 + 3(-2) = 8 - 6 = 2
The four consecutive numbers are 14, 10 , 6 and 2.