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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2020 · Mathematics · Sequences and Series

There are four numbers of which the first three are in G.P. and the...

There are four numbers of which the first three are in G.P. and the last three are in A.P., whose common difference is 6. If the first and the last numbers are equal then two other numbers are

Options

  1. A-2,4
  2. B-4,2
  3. C2,6
  4. Dnone

Correct answer

B. -4,2

Step-by-step solution

Let the last three numbers in A.P. be a, a +6 , a +12,+12, then the first term is alsoa +12. But a+12,a,a+6a+12, a, a+6 are in G.P. ∴\mathrma2=(a+12)(a+6)⇒\mathrma2=\mathrma2+18a+72∴a=−4∴\therefore \mathrma^{2}=(\mathrm{a}+12)(\mathrm{a}+6) \Rightarrow \mathrma^{2}=\mathrma^{2}+18 \mathrm{a}+72\therefore \mathrm{a}=-4\therefore The numbers are 8,-4,2,8.
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