The sum of 4 consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of the middle terms is 7:15. find the numbers.
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Answered by
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Let three consecutive no. be a-3d,a-d,a+d,a+3d
Sum of these no. is 32
Therefore,a-3d+a-d+a+d+a+3d=32
4a=32
a=8
Product if first and last: product of middle terms=7:15
Therefore, (a-3d)(a+3d):(a-d)(a+d)=7:15
Using the identity, (p+q)(p-q)=p²-q²
We get, a²-3d²:a²-d²=7:15
And hence d=±2
Sum of these no. is 32
Therefore,a-3d+a-d+a+d+a+3d=32
4a=32
a=8
Product if first and last: product of middle terms=7:15
Therefore, (a-3d)(a+3d):(a-d)(a+d)=7:15
Using the identity, (p+q)(p-q)=p²-q²
We get, a²-3d²:a²-d²=7:15
And hence d=±2
toxicantthakur:
thnqq
I know this question can come in today's board exam I m too class 10 student
aap cnfrm kaise ho
itne
Question 4 consecutive no aur tume 3 liye q
4 hai bro
And I said this question can come I m not confirm lol
This question confirmly comes in exam
Answered by
0
Answer:
Let the four consecutive term of the AP be ,
a - 3 d , a - d , a + d , a + 3 d
A.T.Q.
a - 3 d + a - d + a + d + a + 3 d = 32
4 a = 32
a = 8
and
( a - 3 d ) ( a + 3 d ) / ( a - d ) ( a + d ) = 7 / 15
15 ( a² - 9 d² ) = 7 ( a² - d² )
8 a² = 128 d² [ a = 8 ]
d = ± 2
Therefore , numbers are 2 , 6 , 10 , 14 or 14 , 10 , 6 , 2 .
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