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

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.

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

To show that the sum of the first six terms is equal to four times the fifth term, we need to find an expression for the sum of the first six terms and compare it to four times the fifth term.

Let's assume that the sequence is given by an arithmetic progression, where the first term is 'a' and the common difference is 'd'.

The sum of the first six terms of an arithmetic progression can be calculated using the formula:

Sum(n) = (n/2) * (2a + (n-1)d)

In this case, we want to find the sum of the first six terms, so n = 6.

Sum(6) = (6/2) * (2a + (6-1)d)
= 3 * (2a + 5d) ............. Equation (1)

The fifth term of an arithmetic progression can be found using the formula:

nth term = a + (n-1)d

We want to find the fifth term, so n = 5.

5th term = a + (5-1)d
= a + 4d .................... Equation (2)

To show that the sum of the first six terms is equal to four times the fifth term, we need to compare Equation (1) and Equation (2):

Sum(6) = 3 * (2a + 5d)
4th term = 4 * (a + 4d)

So, if the expression on the right-hand side of Equation (1) is equal to the expression on the right-hand side of Equation (2), then it will be proven that the sum of the first six terms is equal to four times the fifth term.