If b is the first term of an infinite G.P whose sum is five, then b lies in the interval:

Let the common ratio of the infinite geometric progression be denoted by $$r.$$ For an infinite G.P. to have a finite sum, it is necessary that the absolute value of the common ratio satisfy $$|r| < 1.$$

The standard formula for the sum of an infinite G.P. whose first term is $$b$$ and whose common ratio is $$r$$ (with $$|r| < 1$$) is

$$S \;=\; \frac{b}{1-r}.$$

In the present problem we are told that the sum is $$5,$$ so we substitute $$S = 5$$ into the formula and write

$$5 \;=\; \frac{b}{1-r}.$$

Now we isolate $$b$$ by multiplying both sides by $$(1-r).$$ This gives

$$5(1-r) \;=\; b.$$

Next we expand the left‐hand side:

$$5 \times 1 \;-\; 5 \times r \;=\; b,$$

so that

$$b \;=\; 5 \;-\; 5r.$$

To find the possible numerical values of $$b,$$ we must remember the condition on $$r.$$ Because $$|r| < 1,$$ we have

$$-1 \;<\; r \;<\; 1.$$

We now study the expression $$b = 5 - 5r$$ as $$r$$ varies in the open interval $$(-1,\,1).$$ Notice that the coefficient of $$r$$ is negative, so as $$r$$ increases, $$b$$ decreases.

• When $$r$$ approaches $$-1$$ from the right (that is, $$r \to -1^{+}$$), we have

$$b \;=\; 5 - 5(-1) \;=\; 5 + 5 \;=\; 10.$$

Because $$r$$ can never actually equal $$-1,$$ the value $$b = 10$$ is never attained; $$b$$ merely gets arbitrarily close to $$10$$ from below. Therefore we have $$b < 10.$$

• When $$r$$ approaches $$1$$ from the left (that is, $$r \to 1^{-}$$), we obtain

$$b \;=\; 5 - 5(1) \;=\; 5 - 5 \;=\; 0.$$

Again, since $$r$$ cannot actually reach $$1,$$ the value $$b = 0$$ is never reached; $$b$$ only approaches $$0$$ from above. Hence $$b > 0.$$

Combining the two strict inequalities, we finally have

$$0 \;<\; b \;<\; 10.$$

This interval is exactly $$(0,\,10),$$ which corresponds to Option C.

Hence, the correct answer is Option C.