Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P. with common difference six. If first and last terms of this sequence are equal, then the last term is :

Let the four numbers in the sequence be $$a, b, c, d$$. The first three are in G.P., so $$b^2 = ac$$. The last three are in A.P. with common difference 6, so $$c = b + 6$$ and $$d = b + 12$$. We are also told that the first and last terms are equal, i.e., $$a = d = b + 12$$.

Substituting into the G.P. condition: $$b^2 = a \cdot c = (b + 12)(b + 6) = b^2 + 18b + 72$$. Simplifying gives $$0 = 18b + 72$$, so $$b = -4$$.

Therefore $$c = -4 + 6 = 2$$, $$d = -4 + 12 = 8$$, and $$a = 8$$. The sequence is $$8, -4, 2, 8$$. We can verify: the first three terms $$8, -4, 2$$ form a G.P. with common ratio $$-\frac{1}{2}$$, the last three terms $$-4, 2, 8$$ form an A.P. with common difference $$6$$, and the first and last terms are both $$8$$.

The last term is $$8$$.