Show that the sum of the first six terms is equal to four times the fifth term

Answer 1

6/2 (2a+5d) = 6a+15d
4 times the 5th term is 4(a+4d) = 4a+16d
so, unless d = 2a, the statement is false.

Answer 2

What kind of sequence?

arithmetic:
3(2a + 5d) = 4(a+4d)
6a + 15d = 4a + 16d
d = 2a
so let a be any number, e.g. a = 5, then d = 10
sequence is: 5,15,25,35,45, .....
4 times 45 = 180
sum of first 6 terms = 3(10 + 5(10)) = 3(60) = 180

pick any other a value and it will work

geometric:
a(r^6 - 1)/(r-1) = 4(a r^4)
divide by a
r^6 - 1 = 4r^4(r-1)
r^6 - 1 = 4r^5 - 4r^4
r^6 - 4r^5 + 4r^4 - 1 = 0
sent this to Wolfram and
Wolfram says r = 1, 1.61803, -.61803, 2.2056
there are also 2 complex solutions, but we should reject these

https://www.wolframalpha.com/input/?i=r%5E6+-+4r%5E5+%2B+4r%5E4+-+1+%3D+0

so a can be anything and r any of the above 4 values to create your sequence.
I tested the r = 2.2056 and it worked