http://prntscr.com/amimft @imqwerty
here the x coordinate of the points represent "n" and the y coordinates represent "\(a_n\)" we will solve this question by trying out the options by "trying out the options" i meant that we will select any of the 3 given points and substitute the value of n ( the x coordinate) into the options the option which gives the correct value of corresponding \(a_n\) to that "n" is correct it may happen that 2 options give correct value of "\(a_n\)" for a particular "n" in that case try another "n" ( i mean another point :) )
Uhh ok tell me how to start it lol im slow on this one
okay well the point (1,3) looks cute so we will start off with it so the correct option will give "\(a_n=3\)" when we put "\(n=1\)" in it \(\color{blue}{option~1}\) \(a_n=3 \large \left(\frac{1}{5}\right)^{n-1}\) putting n=1 \(a_n=3 \large \left(\frac{1}{5}\right)^{1-1}\) \(a_n=3 \large \left(1\right)\) \(a_n=3\) \(\color{blue}{option~2}\) \(a_n=3(-5)^{n-1}\) \(a_n=3(-5)^{1-1}\) \(a_n=3(1)\) \(a_n=3\) \(\color{blue}{option~3}\) \(a_n=0.3(5)^{n-1}\) \(a_n=0.3(5)^{1-1}\) \(a_n=0.3(1)\) \(a_n=0.3\) similarly you can try out the 4th option now here we see that options 1 and 2 are giving correct value for \(a_n\) so they might be correct but option 3 is giving wrong value of \(a_n\) so it is wrong now we need to find which one of options1 or 2 is correct so wee will take another point (for example (2,0.6)) and then check
How did you get the 3(1) each time for the first 2? Like where did the -5 go?
anything raised to the power "0" is "1" \(\color{red}{except~this~case- 0^0}\) so like in 1st equation we had this- \(a_n=3 \large \left(\frac{1}{5}\right)^{1-1}\) \(a_n=3 \large \left(\frac{1}{5}\right)^{0}\) \(a_n=3 \large \left(1\right)\)
Yeaa uh I can't process this one so ima guess the second option is right xD
noo thats wrong cx
1st option
e.e thats correct
xD
:)
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