Question

Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their ${{10}^{\text{th}}}$ terms is the same as the difference between their ${{21}^{\text{st}}}$ terms, which is the same as the difference between any two corresponding terms, why? $$$$

Answer 1

We denote the common difference of first AP as ${{d}_{1}}$ and second AP is ${{d}_{2}}$. Similarly we denote the first terms of first and second AP as ${{a}_{1}}$ and ${{a}_{2}}$. We find the difference of ${{n}^{\text{th}}}$ term both the AP and see that the difference is constant and is equal to the difference between their first terms. $$$$

Complete step by step answer:
Arithmetic sequence otherwise known as arithmetic progression, abbreviated as AP is a type sequence where the difference between any two consecutive numbers is constant. If $\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...$ is an AP, then ${{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$ . The difference between two terms is called common difference and conventionally denoted as $d$ where$d={{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$. . The first term ${{x}_{1}}$ is conventionally denoted as $a$.
We know that the ${{n}^{\text{th}}}$ term of an AP is given by the formula
${{x}_{n}}=n+a\left( n-1 \right)d$
We are given in the question that two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. Let us denote the common difference of first AP as ${{d}_{1}}$ and second AP is ${{d}_{2}}$. Similarly we denote the first terms of first and second AP as ${{a}_{1}}$ and ${{a}_{2}}$. So in accordance with the question we have;

& {{d}_{1}}={{d}_{2}} \\\ & {{a}_{1}}=2,{{a}_{2}}=7 \\\ \end{aligned}$$ We find the ${{n}^{\text{th}}}$ term of first AP as ${{x}_{n}}$using formula for ${{n}^{\text{th}}}$ with $d={{d}_{1}}$ and ${{a}_{1}}=2$as; $$\begin{aligned} & \Rightarrow {{x}_{n}}={{a}_{1}}+\left( n-1 \right){{d}_{1}} \\\ & \Rightarrow {{x}_{ n}}=2+\left( n-1 \right){{d}_{1}} \\\ \end{aligned}$$ We find the ${{n}^{\text{th}}}$ term of second AP as $x_{n}^{'}$using formula for ${{n}^{\text{th}}}$ with $d={{d}_{2}}$ and ${{a}_{1}}=7$as; $$\begin{aligned} & \Rightarrow x_{n}^{'}={{a}_{2}}+\left( n-1 \right){{d}_{2}} \\\ & \Rightarrow x_{n}^{'}=7+\left( n-1 \right){{d}_{2}} \\\ \end{aligned}$$ So let us find the difference between ${{n}^{\text{th}}}$ first and second AP. We have; $$\begin{aligned} & x_{n}^{'}-{{x}_{n}}=7+\left( n-1 \right)d-\left\\{ 2+\left( n-1 \right)d \right\\} \\\ & \Rightarrow x_{n}^{'}-{{x}_{n}}=7+\left( n-1 \right)d-2-\left( n-1 \right)d \\\ & \Rightarrow x_{n}^{'}-{{x}_{n}}=5 \\\ \end{aligned}$$ We see that difference between ${{n}^{\text{th}}}$ terms of both APs is constant and is equal to the difference between the first two terms ${{a}_{2}}-{{a}_{1}}=5$. It means whatever the value of $n$ the difference between ${{n}^{\text{th}}}$ terms will be constant. So if we take $n=10$ we get the difference between ${{10}^{\text{th}}}$ terms of the APs and if we take $n=21$ we get the difference between ${{21}^{\text{st}}}$ terms of the APs which will be equal to 5. $$$$ **Note:** We can also find the value of ${{10}^{\text{th}}}\text{ and }{{21}^{\text{st}}}$ terms of the APs in terms of $d$ and then calculate difference to verify our result. We note that the terms of the second AP will be always greater than corresponding terms of first AP in order. We should also note that the sum up to ${{n}^{\text{th}}}$ terms $\left( \dfrac{n}{2}\left\\{ a+\left( n-1 \right)d \right\\} \right)$ the two APs are not same because of the unequal firs terms.