Animated Solution for Mathematics - Sequence and Series: If $a_1,a_2,a_3,\dots$ and $b_1,b_2,b_3,\dots$ are A.P. and $a_1=2,a_{10}=3,a_1{b}_1=1=a_{10}{b}_{10}$ then $a_4{b}_4$ is equal to

Imagine you are standing at the start of two parallel paths. Each path represents an Arithmetic Progression (A.P.), a sequence where every step you take is of a constant, predictable size.

In our problem, we have two such sequences, $a_n$ and $b_n$. Our mission is to find the product of their fourth terms, $a_4{b}_4$.

To do this, we must first decode the "DNA" of these sequences. The DNA of any A.P. is defined by two parameters: its starting point (the first term) and its stride (the common difference).

Let us look at the first sequence, $a_n$. We are given the starting point $a_1=2$ and a milestone at the tenth step $a_{10}=3$.

The general formula for any term in an A.P. is $a_n=a_1+(n-1)d_1$. By plugging in our known values for the tenth term, we get:

This simplifies to $3=2+9d_1$. Subtracting $2$ from both sides gives $1=9d_1$, which reveals the stride of our first path:

With this, we can find any term in the sequence. We need the fourth term, $a_4$:

Now, we turn our attention to the second path, $b_n$. The problem provides a bridge between the two sequences: the product of their first terms is $1$, and the product of their tenth terms is also $1$.

This implies $a_1{b}_1=1$ and $a_{10}{b}_{10}=1$. Since we know $a_1=2$, it follows that $2b_1=1$, so $b_1=\frac{1}{2}$.

Similarly, since $a_{10}=3$, we have $3b_{10}=1$, so $b_{10}=\frac{1}{3}$. We have successfully identified the start and the tenth milestone of our second sequence.

Just as we did for the first sequence, we need the common difference $d_2$ for $b_n$. Using the general term formula $b_{10}=b_1+9d_2$, we substitute our values:

Rearranging gives $9d_2=\frac{1}{3}-\frac{1}{2}$. Finding a common denominator of $6$, we get:

Dividing by $9$, we find $d_2=-\frac{1}{54}$. Now, we calculate the fourth term $b_4$:

Converting to a common denominator of $18$, we get:

We have arrived at the final stage. We have $a_4=\frac{7}{3}$ and $b_4=\frac{4}{9}$. The problem asks for their product, $a_4{b}_4$.

Multiplying these fractions, we get:

The final answer is $\frac{28}{27}$.