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

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


Let the terms be a,ar,ar^2,ar^3,ar^4
a+ar+ar^2=a( \frac{r^3-1}{r-1} )=14
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}
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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