Question

Sum of the two means of an A.P with four terms is 44. If the product of the extremes is 340, what is the third term of the A.P?

Answer 1

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Let 4 terms of an AP be a-3d, a-d, a+d, a +3d.

a-d+a+d = 44

2a = 44

$\fbox{a = 22}$

(a-3d)(a+3d) = 340

(22 - 3d)(22+3d) = 340

(22 × 22) - (3d × 3d)

$484 - 9 {d}^{2} = \: 340$

$9 {d}^{2} = 144$

${d}^{2} = 16$

$\fbox{d = +-4}$

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Answer 2

Answer:

The third term of the A.P is 26.

Step-by-step explanation:

Arithmetic Progression (A.P) is a series of numbers in which the consecutive numbers differ by a constant number called common difference.

Consider the 4 terms of A.P as a - 3d, a - d, a + d, a + 3d.

1. Given sum of two means of an A.P = 44

Let mean 1 is the mean of first two terms a - 3d, a - d,

• Mean 1 = (a - 3d + a - d)/2 = (2a - 4d)/2

• Mean 1 = a - 2d

Let mean 2 is the mean of last two terms a + d, a + 3d,

• Mean 2 = (a + d + a + 3d)/2 = (2a + 4d)/2

• Mean 2 = a + 2d

Mean 1 + Mean 2 = 44

a - 2d + a + 2d = 44

2a = 44

a = 22

      2. Given the product of extremes is 340.

(a - 3d)*(a + 3d) = 340

a² - (3d)² = 340               [(a+b)(a-b) = a² - b²]

22² - 9d² = 340

(484 - 340)/9 = d²

d² = 16

d = 4

The third term of the A.P = a + d = 22 + 4 = 26

Therefore, the third term of the A.P is 26.

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