Let ${a_1},{\text{ }}{a_2},{\text{ }}{a_3}...................$ be in A.P. And ${q_1},{\text{ }}{q_2},{\text{ }}{q_3}................$ be in G.P, such that ${a_1} = {q_1} = 2$ and ${a_{10}} = {q_{10}} = 3$ then:
${\text{a}}{\text{. }}{{\text{a}}_1}{q_{19}}$ Is not an integer
${\text{b}}{\text{. }}{{\text{a}}_{19}}{q_7}$ Is an integer
${\text{c}}{\text{. }}{{\text{a}}_7}{a_{18}} = {{\text{a}}_{19}}{q_{10}}$
${\text{d}}{\text{. }}$ None

Hint – In this question first calculate the common difference and common ratio of A.P and G.P respectively using the formula of ${n^{th}}$ term of an A.P and G.P later on using these values calculate all the given terms so, use these concepts to reach the solution of the question.

Complete step-by-step solution -
As you know the general formula of an A.P is ${a_n} = {a_1} + \left( {n - 1} \right)d$ and the general formula of G.P is ${q_n} = {q_1}{r^{n - 1}}$ where ${a_1}$ and ${q_1}$be the first term of A.P and G.P respectively, while, $d$ and $r$ are the common difference and common ratio respectively.
It is given that ${a_{10}} = {q_{10}} = 3$
\[
  {a_n} = {a_1} + \left( {n - 1} \right)d \\
   \Rightarrow {a_{10}} = {a_1} + \left( {10 - 1} \right)d = {a_1} + 9d = 3 \\
  {q_n} = {q_1}{r^{n - 1}} \\
   \Rightarrow {q_{10}} = {q_1}{r^{10 - 1}} = {q_1}{r^9} = 3 \\
\]
Now it is also give that ${a_1} = {q_1} = 2$
$
   \Rightarrow {a_1} + 9d = 3 \\
   \Rightarrow 2 + 9d = 3 \Rightarrow d = \dfrac{1}{9} \\
   \Rightarrow {q_1}{r^9} = 3 \\
   \Rightarrow 2{r^9} = 3 \Rightarrow {r^9} = \dfrac{3}{2} \\
$
Now from A.P general equation ${a_n} = {a_1} + \left( {n - 1} \right)d$, the value of
\[
  {a_{19}} = {a_1} + \left( {19 - 1} \right)d \\
   \Rightarrow {a_{19}} = {a_1} + 18d = 2 + 18 \times \dfrac{1}{9} = 4 \\
  {a_7} = {a_1} + \left( {7 - 1} \right)d \\
   \Rightarrow {a_7} = {a_1} + 6d = 2 + 6 \times \dfrac{1}{9} = \dfrac{8}{3} \\
  {a_{18}} = {a_1} + \left( {18 - 1} \right)d \\
   \Rightarrow {a_{18}} = {a_1} + 17d = 2 + 17 \times \dfrac{1}{9} = \dfrac{{35}}{9} \\
\]
Now from G.P general equation ${q_n} = {q_1}{r^{n - 1}}$, the value of
\[
  {q_{19}} = {q_1}{r^{19 - 1}} \\
   \Rightarrow {q_{19}} = {q_1}{r^{18}} = 2{\left( {{r^9}} \right)^2} = 2{\left( {\dfrac{3}{2}} \right)^2} = \dfrac{9}{2} \\
  {q_7} = {q_1}{r^{7 - 1}} \\
   \Rightarrow {q_7} = {q_1}{r^6} \\
  {r^9} = \dfrac{3}{2} \Rightarrow r = {\left( {\dfrac{3}{2}} \right)^{\dfrac{1}{9}}} \Rightarrow {r^6} = {\left( {\dfrac{3}{2}} \right)^{\dfrac{6}{9}}} \\
   \Rightarrow {q_7} = 2{\left( {\dfrac{3}{2}} \right)^{\dfrac{6}{9}}} = 2{\left( {\dfrac{3}{2}} \right)^{\dfrac{2}{3}}} \\
\]
Now check out option (a) which is ${{\text{a}}_1}{q_{19}}$
$ \Rightarrow {{\text{a}}_1}{q_{19}} = 2 \times \dfrac{9}{2} = 9$
Which is an integer so option (a) is ruled out.
Now check out option (b) which is ${{\text{a}}_{19}}{q_7}$
$ \Rightarrow {{\text{a}}_{19}}{q_7} = 4 \times 2{\left( {\dfrac{3}{2}} \right)^{\dfrac{2}{3}}} = 8{\left( {\dfrac{3}{2}} \right)^{\dfrac{2}{3}}}$
Which is not an integer so option (b) is also ruled out.
Now check out option (c) which is ${{\text{a}}_7}{a_{18}} = {{\text{a}}_{19}}{q_{10}}$
$
  {{\text{a}}_7}{a_{18}} = {{\text{a}}_{19}}{q_{10}} \\
   \Rightarrow \dfrac{8}{3}\left( {\dfrac{{35}}{9}} \right) = 4\left( 3 \right) \\
$
Which is also not true
Hence, option (d) is correct none of these.

Note: - In such types of question the key concept we have to remember is that always remember the general formulas of A.P and G.P, then according to given conditions calculate the values of common difference and common ratio, then calculate the terms which is given in options, then check out every option we will get the required answer.