Question

which term of the AP 2,8,32....... is 256

Answer 1

hey mate here is your answer..
here a=first term= 2
An = 256
r = common ration 4
An = a ( R^n-1)
256 = 2 (4^n-1)
128 = 4^n-1
2^7 = 2^2n-1
on comparing the powers
we get 2n-1 = 7
n = 4...
hope it will help you my dear friend..
it's gp

Answer 2

Answer:

the answer is 4.5 so this is not exist okay but the procedure to do is this

Step-by-step explanation:

in the given GP Is A=2 c

common ration r=a$a^{2} /a^{1}$=8/2=4 (cancelation )

nth term of it = 256

now substituting a=2,r=4 we get 2.4n-1 =256=4n-1=256/2 (consolation ) =128=$2 ^{7}$

=4n-1=$2^{2}$ n-1=$2 ^{7}$

so we are taking powers over hear

= $2 ^{2}$n-2=$2 ^{7}$

..2n-2=7 and 2n=7+2=9

.. n=9/2=4.5 but (4.5)  the number will be not exist . why because term number will be only a whole number  256 will not exist  in the given GP

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