Three numbers in increasing|Sequence and Series|JEEMAIN 2021

JEE MAIN 2021

31-08-2021 S1

Question

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:

$7-7 \sqrt{3}$

$7+\sqrt{3}$

$7-\sqrt{3}$

$7+3 \sqrt{3}$

✓ 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

Question Tags

JEE Main

Mathematics

Medium

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