The sequence $a_1,a_2,a_3\cdots\cdots$ is a geometric sequence. The sequence $b_1,b_2,b_3\cdots\cdots$ is a geometric sequence. $$b_1=1$$ $$b_2=7^{\frac{1}{4}}-28^{\frac{1}{4}}+1$$ $$a_1=28^{\frac{1}{4}}$$
$\sum_{n=1}^{\infty}\dfrac{1}{a_n}=\sum_{n=1}^{\infty}{b_n}$
If the area of the triangle with sides lengths $a_1,a_2,a_3$ can be expressed in the form of $\dfrac{p}{q}$ where p and q are relatively prime, find $(p + q)$.
My attempt is as follows:-
$$b_2=7^{\frac{1}{4}}-\sqrt{2}\cdot7^{\frac{1}{4}}+1$$ $$b_2=7^{\frac{1}{4}}(1-\sqrt{2})+1$$
$$\sum_{n=1}^{\infty}\dfrac{1}{a_n}=\dfrac{b_1}{1-\dfrac{b_2}{b_1}}$$ $$\sum_{n=1}^{\infty}\dfrac{1}{a_n}=\dfrac{1}{1-7^{\frac{1}{4}}(1-\sqrt{2})-1}$$ $$\sum_{n=1}^{\infty}\dfrac{1}{a_n}=\dfrac{1}{7^{\frac{1}{4}}(\sqrt{2}-1)}$$
$$\dfrac{\dfrac{1}{a_1}}{1-\dfrac{a_1}{a_2}}=\dfrac{1}{7^{\frac{1}{4}}(\sqrt{2}-1)}$$
$$\dfrac{1}{a_1}\cdot7^\frac{1}{4}(\sqrt{2}-1)=1-\dfrac{a_1}{a_2}$$ $$\dfrac{\sqrt{2}-1}{\sqrt{2}}=1-\dfrac{a_1}{a_2}$$ $$\dfrac{a_1}{a_2}=1-\dfrac{\sqrt{2}-1}{\sqrt{2}}$$ $$\dfrac{a_1}{a_2}=\dfrac{1}{\sqrt{2}}$$
$$\dfrac{a_2}{a_1}=\sqrt{2}$$ $$a_2=7^\frac{1}{4}\cdot2$$ $$a_3=7^\frac{1}{4}\cdot2\sqrt{2}$$
Applying heron's formula to find the area:-
$$s=\dfrac{a_1+a_2+a_3}{2}$$ $$s=\dfrac{7^\frac{1}{4}(3+\sqrt{2})}{\sqrt{2}}\tag{1}$$
$$(s-a)=7^\frac{1}{4}\left(\dfrac{3+\sqrt{2}}{\sqrt{2}}-\sqrt{2}\right)$$ $$(s-a)=7^\frac{1}{4}\left(\dfrac{1+\sqrt{2}}{\sqrt{2}}\right)\tag{2}$$
$$(s-b)=7^\frac{1}{4}\left(\dfrac{3+\sqrt{2}}{\sqrt{2}}-2\right)$$ $$(s-b)=7^\frac{1}{4}\left(\dfrac{3-\sqrt{2}}{\sqrt{2}}\right)\tag{3}$$
$$(s-c)=7^\frac{1}{4}\left(\dfrac{3+\sqrt{2}}{\sqrt{2}}-2\sqrt{2}\right)$$ $$(s-c)=7^\frac{1}{4}\left(\dfrac{\sqrt{2}-1}{\sqrt{2}}\right)\tag{4}$$
Multiplying $1,2,3,4$
$$s(s-a)(s-b)(s-c)=\dfrac{7\cdot(9-2)\cdot(2-1)}{4}$$ $$s(s-a)(s-b)(s-c)=\dfrac{49}{4}$$
$$\triangle=\sqrt{s(s-a)(s-b)(s-c)}$$ $$\triangle=\dfrac{7}{2}$$
So $p$ will be $7$ and $q$ will be $2$, hence $p+q=9$, But actual answer is $21$. I checked multiple times, but didn't get what is wrong with this. Please help me in this.