Arithmetic and geometric sequence integrated problem?

Question

Three numbers form an arithmetic sequence having a common difference of 4. If the first number is increased by 2, the second by 3, and the third by 5, they form a geometric sequence. Find the original numbers.

Answer 1

The 3 numbers are x- 4, x and x + 4

Then x - 2, x + 3 and x + 9 are a geometric sequence

then (x + 3) / (x - 2) = (x + 9) / (x + 3)

(x + 3)² = (x + 9)(x - 2)

x² + 6x + 9 = x² + 7x - 18

x = 27

the 3 numbers are 23, 27 and 31

Answer 2

The arithmetic sequence is 4, 4+a, 4+2a. with the intention to variety a geometic sequence, 18/(4+2a) ought to = (4+2a)/(4+a) subsequently 36 + 9a = 8 + 8a + 2a², or 2a² - a - 28 = 0. using quadratic, a = 4 [or -3.5] The sequence is for this reason: 4, 8, 12, 18. in accordance to Zeno, the bob on no account does come to kick back. even however, the cut back of its commute is: fifty six + (.ninety 3)×fifty six + (.ninety 3)²×fifty six + (.ninety 3)³×fifty six + ... = fifty six × Sigma[(i=0, N), .ninety 3^i] = fifty six × [(a million - .ninety 3^(N+a million))/(a million - .ninety 3)] as N procedures infinity. .ninety 3^{infinity} is 0, so the sum is fifty six × (a million/.07) = 800 cm.