Three numbers are in an increasing geometric progression with common ratio $r$. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference $d$. If the fourth term of GP is $3 r_{m, 2}^2$ then $r^2-d$ is equal
Select the correct option:
A
$7-7 \sqrt{3}$
B
$7+\sqrt{3}$
C
$7-\sqrt{3}$
D
$7+3 \sqrt{3}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Let the numbers be: a/r, a, ar (in G.P.)
Given:
a/r + 2a + ar = 4
⇒ a(1/r + 2 + r) = 4
⇒ (r + 1/r + 2) = 4/a
For G.P. condition:
r + 1/r = 2 ± √3
⇒ r = 2 ± √3
4th term of G.P. = ar² = 3r²
⇒ a = 3
Take r = 2 + √3
Then,
2a − a/r = 6 − 3/(2 + √3) = 3√3
Now,
r² − d = (2 + √3)² − 3√3
= (4 + 4√3 + 3) − 3√3
= 7 + 4√3 − 3√3
= 7 + √3
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