There are 5 numbers in geometric progression, sum of first three terms is 14 and last two terms is 48 find the numbers


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sorry for delayed answer
next time we would make sure ur questions are answered fast
madan can you explain me how you got r=2?
i just sub. 2 in the found equation
hit and trial

Answers

  • Brainly User
2014-03-25T15:23:16+05:30
Let the terms be a,ar,ar^2,ar^3,ar^4
a+ar+ar^2=a( \frac{r^3-1}{r-1} )=14
ar^3+ar^4=48
a+ar+ar^2+ar^3+ar^4= a(\frac{r^5-1}{r-1})=14+48= 62
 \frac{r^5-1}{r^3-1} = \frac{62}{14}
14r^5-62r^3+48=0
u get real r=2  (just sub. 2 and saw)
which satisfies and further a=2
therefore the terms are 2,4,8,16,32


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