Question

find four consecutive term in A P such that their sum is 66 and product of middle term and end term is the ratio 15:14(assume that four consecutive term in AP are a-3d,a-d,a+d,a+3d )​

Answer 1

The four consecutive terms in A.P. are  12, 15, 18 & 21

Step-by-step explanation:

We are given to assume the 4 consecutive terms in the A.P. as (a - 3d), (a - d), (a + d) & (a + 3d).

The sum of the 4 consecutive terms are given as 66

∴ (a-3d) + (a-d) + (a+d) + (a+3d) = 66

⇒ (a + a + a + a) + (-3d - d + d +3d) = 66

⇒ 4a = 66

⇒ a = 16.5

Also given that product of middle term and end term is the ratio 15:14, therefore we can write as,

$\frac{(a-d) * (a+d)}{(a-3d) * (a+3d)} = \frac{15}{14}$

⇒  $\frac{[a^2 - d^2]}{[a^2 - 9d^2]} = \frac{15}{14}$

⇒ 14a² - 14d² = 15a² - 135d²

⇒ a² = 121d²

taking square roots on both sides

⇒ a = 11d

substituting a = 16.5

⇒ d = 16.5/11

⇒ d = 1.5

Now, by substituting the value of a and d, we get

a - 3d = 16.5 - (3×1.5) = 12

a - d = 16.5 - 1.5 = 15

a + d = 16.5 + 1.5 = 18

a + 3d = 16.5 + (3×1.5) = 21

Thus, 12, 15, 18 & 21 are 4 consecutive terms in A.P.

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Find the common difference of an A.P. whose first term is 5 and the sum of first four terms is half the sum of next four terms.

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If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 terms is

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Answer 2

Step-by-step explanation:

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