If a₁, a₂, a₃, ... are the terms of an increasing geometric progression such that

a_1+a_3+a_5=21

then a₁ + a₂ + a₃ is

The correct option is 7.

Let the terms of the increasing geometric progression (GP) be $a_1,a_2,a_3,...$ with first term $a_1=a$ and common ratio $r>1$ (since it is an increasing progression).

The terms can be represented as:
$a_1=a$
$a_3=a{r}^2$
$a_5=a{r}^4$

We are given two equations:
1) $a_1+a_3+a_5=21$ which translates to:
$a+a{r}^2+a{r}^4=21$
$a(1+r^2+r^4)=21$ (Equation 1)

2) $a_1{a}_3{a}_5=64$ which translates to:
$a\cdot a{r}^2\cdot a{r}^4=64$
$a^3{r}^6=64$
Taking the cube root on both sides:
$a{r}^2=4$ (Equation 2)

From Equation 2, we get $a=\frac{4}{r^2}$.
Substitute this value of $a$ into Equation 1:
$\frac{4}{r^2}$ $(1+r^2+r^4)=21$
$4(1+r^2+r^4)=21r^2$
$4+4r^2+4r^4=21r^2$
$4r^4-17r^2+4=0$

Let $x=r^2$. The equation becomes:
$4x^2-17x+4=0$
Factoring the quadratic equation:
$4x^2-16x-x+4=0$
$4x(x-4)-1(x-4)=0$
$(4x-1)(x-4)=0$

This gives:
$x=4$ or $x=\frac{1}{4}$

Since the geometric progression is increasing, the common ratio must satisfy $r>1$, which implies $r^2>1$. Therefore, we choose:
$r^2=4$
Since $r>1$, we have $r=2$.

Substitute $r^2=4$ back into Equation 2:
$a\cdot 4=4$
$a=1$

Now, we can find the first three terms of the GP:
$a_1=a=1$
$a_2=ar=1\cdot 2=2$
$a_3=a{r}^2=1\cdot 4=4$

We are asked to calculate $a_1+a_2+a_3$:
$a_1+a_2+a_3=1+2+4=7$