Question

If $$a_1,a_2,a_3,...a_n$$ are in the arithmetic progression and $$a_1=0$$. Then what is the value of $$\left({\displaystyle \frac{a_3}{a_2}}+{\displaystyle \frac{a_4}{a_3}}+...+{\displaystyle \frac{a_n}{a_{n-1}}}\right)-a_2\left({\displaystyle \frac{1}{a_2}}+{\displaystyle \frac{1}{a_3}}+...+{\displaystyle \frac{1}{a_{n-2}}}\right)$$ ?
A. $$\left(n-2\right)+{\displaystyle \frac{1}{\left(n-2\right)}}$$
B. $${\displaystyle \frac{1}{\left(n-2\right)}}$$
C. $$\left(n-2\right)$$
D. $$\left(n+2\right)$$

Answer 1

Hint: First, use the given first term and find the common difference in the given arithmetic progression. Then use the common difference and calculate the values of all terms present in the arithmetic progression. In the end, substitute the values in the given expression and solve the expression to get the required answer.

Formula used:
The formula of the $$n^{th}$$ term in an arithmetic progression: $$a_n=a+\left(n-1\right)d$$, where $$a$$ is the first term and $$d$$ is the common difference.
Formula of the common difference: $$d=a_n-a_{n-1}$$

Complete step by step solution:
Given:
$$a_1,a_2,a_3,...a_n$$ are in the arithmetic progression and $$a_1=0$$.
Let’s calculate the common difference of the above arithmetic progression.
$$d=a_2-a_1$$
Since $$a_1=0$$.
So, $$a_2=d$$

Now apply the formula of the $$n^{th}$$ term in an arithmetic progression.
We get,
$$a_3=a_1+2d=2d$$
$$a_4=a_1+3d=3d$$
And so on.
$$a_{n-1}=a_1+\left(n-2\right)d=\left(n-2\right)d$$
$$a_n=a_1+\left(n-1\right)d=\left(n-1\right)d$$

Let $$V$$ be the value of $$\left({\displaystyle \frac{a_3}{a_2}}+{\displaystyle \frac{a_4}{a_3}}+...+{\displaystyle \frac{a_n}{a_{n-1}}}\right)-a_2\left({\displaystyle \frac{1}{a_2}}+{\displaystyle \frac{1}{a_3}}+...+{\displaystyle \frac{1}{a_{n-2}}}\right)$$.
Then,
$$V=\left({\displaystyle \frac{a_3}{a_2}}+{\displaystyle \frac{a_4}{a_3}}+...+{\displaystyle \frac{a_n}{a_{n-1}}}\right)-a_2\left({\displaystyle \frac{1}{a_2}}+{\displaystyle \frac{1}{a_3}}+...+{\displaystyle \frac{1}{a_{n-2}}}\right)$$
Substitute the values in the above equation.
$$V=\left({\displaystyle \frac{2d}{d}}+{\displaystyle \frac{3d}{2d}}+...+{\displaystyle \frac{\left(n-1\right)d}{\left(n-2\right)d}}\right)-d\left({\displaystyle \frac{1}{d}}+{\displaystyle \frac{1}{2d}}+...+{\displaystyle \frac{1}{\left(n-3\right)d}}\right)$$
Simplify the above equation.
$$V=\left(2+{\displaystyle \frac{3}{2}}+...+{\displaystyle \frac{\left(n-1\right)}{\left(n-2\right)}}\right)-\left(1+{\displaystyle \frac{1}{2}}+...+{\displaystyle \frac{1}{\left(n-3\right)}}\right)$$
$$\Rightarrow V=\left(\left(1+1\right)+\left(1+{\displaystyle \frac{1}{2}}\right)+...+\left(1+{\displaystyle \frac{1}{\left(n-2\right)}}\right)\right)-\left(1+{\displaystyle \frac{1}{2}}+...+{\displaystyle \frac{1}{\left(n-3\right)}}\right)$$
$$\Rightarrow V=\left(1+1+1+....\left(n-2\right)times\right)+\left(1+{\displaystyle \frac{1}{2}}+...+{\displaystyle \frac{1}{\left(n-3\right)}}+{\displaystyle \frac{1}{\left(n-2\right)}}\right)-\left(1+{\displaystyle \frac{1}{2}}+...+{\displaystyle \frac{1}{\left(n-3\right)}}\right)$$
$$\Rightarrow V=\left(n-2\right)+{\displaystyle \frac{1}{\left(n-2\right)}}$$
Thus, the value of $$\left({\displaystyle \frac{a_3}{a_2}}+{\displaystyle \frac{a_4}{a_3}}+...+{\displaystyle \frac{a_n}{a_{n-1}}}\right)-a_2\left({\displaystyle \frac{1}{a_2}}+{\displaystyle \frac{1}{a_3}}+...+{\displaystyle \frac{1}{a_{n-2}}}\right)$$ is $$\left(n-2\right)+{\displaystyle \frac{1}{\left(n-2\right)}}$$.
Hence the correct option is A.

Note: Students often do a common mistake that is they are using the formula for $$n^{th}$$ term is $$a_n=a+nd$$ which is a wrong formula. The correct formula is $$a_n=a+\left(n-1\right)d$$.