Search any question & find its solution
Question:
Answered & Verified by Expert
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:
Solution:
1183 Upvotes
Verified Answer
The correct answer is:
-4,2
Let the last three numbers in A.P. be a, a +6 , a $+12,$ then the first term is alsoa +12 . But $a+12, a, a+6$ are in G.P.
$\therefore \mathrm{a}^{2}=(\mathrm{a}+12)(\mathrm{a}+6) \Rightarrow \mathrm{a}^{2}=\mathrm{a}^{2}+18 \mathrm{a}+72$
$\therefore \mathrm{a}=-4$
$\therefore$ The numbers are 8,-4,2,8.
$\therefore \mathrm{a}^{2}=(\mathrm{a}+12)(\mathrm{a}+6) \Rightarrow \mathrm{a}^{2}=\mathrm{a}^{2}+18 \mathrm{a}+72$
$\therefore \mathrm{a}=-4$
$\therefore$ The numbers are 8,-4,2,8.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.