the sum of four consecutive term in an ap is 32 and the ratio of the product of first and the last term to the product of two middle term is 7:15.find the number
Answers
Answer:
→ 2, 6, 10, 14 .
Step-by-step explanation:
Note :- This question is come in CBSE class 10th board 2018 .
Solution:-
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
⇒ 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.
Four consecutive numbers are 2, 6, 10 and 14
Hence, it is solved .
THANKS .
Answer:
Step-by-step explanation:
Solution :-
Let the four consecutive terms of A.P. be (a - 3d), (a - d), (a + d) and (a + 3d).
According to the given conditions,
⇒ a - 3d + a - d + a + d + a + 3d = 32
⇒ 4a + 32
⇒ a = 32/4
⇒ a = 8 ..... (i)
Ratio of the product of first and the last term to the product of two middle term is 7 : 15.
And,
⇒ (a - 3d) (a + 3d)/(a - d) (a + d) = 7/15
⇒ a² - 9d²/a² - d² = 7/15
⇒ 15(a² - 9d²) = 7(a² - d²)
⇒ 15a² - 135d² = 7a² - 7d²
⇒ 15a² - 7a² = 135d² - 7d²
⇒ 8a² = 128d²
⇒ 8(8)² = 128d² (From (i))
⇒ 128d² = 512
⇒ d² = 512/128
⇒ d² = 4
⇒ d = ± 2
Hence, the numbers are 2, 6, 10 and 14 or 14, 10, 6 and 2.
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