Question

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Answer 1

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$\huge\underline\mathfrak{Question ; }$ The sum of the four consecutive numbers in an ap is 32 and the ratio of the product of the first and the last term of the product of two middle term is 7 : 15 . find the numbers .

$\huge\underline\mathfrak\blue{Answer ; }$
$<i><b>$

Let the four consecutive numbers of AP be :- ( a - 3d ) , ( a- d ) , ( a +d) and (a+3d)

according to question .

a - 3d + a - d + a +d + a +3d = 32

4a =32

$a \: \: \: = \: \frac{32}{4} = 8$

Now :-

$\frac{(a \: - 3d)(a + 3d)}{(a - d)(a + d) } = \frac{7}{15}$

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 = 4

d = 2

So the four consecutive numbers are :-

8-(3×2)

8-6=2

8-2=6

8+2=6

8+(3×2)

8+6 = 14

show the four consecutive number are 2 6 , 10 , and 14 .

$\huge\mathfrak\orange{thanks }$

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Answer 2

Answer:

Step-by-step explanation: