Question

Two A.P’s have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between and two corresponding terms.
If true then enter 1 and if false then enter 0.

Answer 1

Hint: First, write the terms of first AP. Using a (first term) equal to 2 and common difference ‘d’. then write the terms of second AP, using ${{a}_{2}}$ (first term) equal to 2 and common difference ‘d’, same as that of first AP. After that find the difference between their 10th terms and then between their 21st terms by using the formula for nth term of AP, ${{t}_{n}}=a+\left( n-1 \right)d$ where a = first term of AP and d= common difference of AP.
After that, whether these differences are equal or not, then check whether this difference is equal to the difference between any two corresponding terms.

Complete step-by-step answer:
For first AP:
a, (first term) =2.
Let, common difference =d.
Therefore, terms of this first AP can be written as,
${{a}_{1}},{{a}_{1}}+d,{{a}_{1}}+2d,{{a}_{1}}+3d,{{a}_{1}}+4d..............$
Here, a=2
$\Rightarrow 2,2+d,2+2d,2+3d,2+4d..........$
Now, we will write terms of second AP, for which ${{a}_{2}}$ (first term)=7.
Common differences will be the same as that of first AP i.e. ‘d’.
Therefore, terms of second AP can be written as,
${{a}_{2}},{{a}_{2}}+d,{{a}_{2}}+2d,{{a}_{2}}+3d,{{a}_{2}}+4d............$
Here ${{a}_{2}}$=7
$\Rightarrow 7,7+d,7+2d,7+3d,7+4d..................$
Now, we will find the 10th term of both the AP’s.
Since, nth term of AP, ${{t}_{n}}=a+\left( n-1 \right)d$ .
Where a= first term of the AP,
And d= common difference of the AP.
10th term of the first AP
$\begin{aligned} & \Rightarrow {{t}_{10}}={{a}_{1}}+\left( 10-1 \right)d \\\ & {{t}_{10}}={{a}_{1}}+9d \\\ \end{aligned}$
${{t}_{10}}=2+9d$ ………………(1)
10th term of second AP
$\begin{aligned} & \Rightarrow t{{'}_{10}}={{a}_{2}}+\left( 10-1 \right)d \\\ & t{{'}_{10}}={{a}_{2}}+9d \\\ \end{aligned}$
$t{{'}_{10}}=7+9d$ ………………(2)
Now, we can find the difference between the 10th terms of both AP’s by subtracting equation (1) by (2).
$\begin{aligned} & t{{'}_{10}}-{{t}_{10}}=7+9d-\left( 12+9d \right) \\\ & t{{'}_{10}}-{{t}_{10}}=7+9d-12-9d \\\ \end{aligned}$
$t{{'}_{10}}-{{t}_{10}}=5$ ……..(3)
Now, we will find 21st terms of both the AP’s
$\begin{aligned} & \Rightarrow {{t}_{21}}={{a}_{1}}+\left( 21-1 \right)d \\\ & {{t}_{21}}={{a}_{1}}+20d \\\ \end{aligned}$
${{t}_{21}}=2+20d$ ………………(4)
21st term of second AP
$\begin{aligned} & \Rightarrow t{{'}_{21}}={{a}_{2}}+\left( 21-1 \right)d \\\ & t{{'}_{21}}={{a}_{2}}+21d \\\ \end{aligned}$
${{t}_{21}}=7+20d$ ………………(5)
Now, we will find the difference between the 21st terms of the both AP’s by subtracting equation (4) by equation (5).
$\begin{aligned} & t{{'}_{21}}-{{t}_{21}}=7+20d-\left( 2+20d \right) \\\ & t{{'}_{21}}-{{t}_{21}}=7+20d-2-20d \\\ \end{aligned}$
$t{{'}_{21}}-{{t}_{21}}=5$ ……..(6)
From equation (3) and (6), we can see that
$t{{'}_{10}}-{{t}_{10}}=t{{'}_{21}}-{{t}_{21}}$
Now, we will find the difference between the general terms of the two AP’s.
General term of the first term ${{t}_{n}}={{a}_{1}}+\left( n-1 \right)d$
Since, ${{a}_{1}}=2$ $\therefore {{t}_{n}}=2+\left( n-1 \right)d$ ……….(7)
General term of second term $t{{'}_{n}}={{a}_{2}}+\left( n-1 \right)d$
Since, ${{a}_{2}}=7$ , $\therefore t{{'}_{n}}=7+\left( n-1 \right)d$ ………… (8)
Subtracting equation (7) from equation (8). We get,
t{{'}_{n}}-{{t}_{n}}\Rightarrow \left\\{ 7+\left( n-1 \right)d \right\\}-\left\\{ 2+\left( n-1 \right)d \right\\}
$\begin{aligned} & =7+\left( n-1 \right)d-2-\left( n-1 \right)d \\\ & =5 \\\ \end{aligned}$
We can clearly see that $t{{'}_{n}}-{{t}_{n}}={{t}_{10}}-t{{'}_{10}}$
$={{t}_{21}}-t{{'}_{21}}$
=5
Therefore, the difference between the 10th terms of the two AP’s is equal to the difference between the 21st terms of the two AP’s and is the same as the difference between any two corresponding terms of these two AP’s.
Hence, enter 1 as the given statement is true.

Note: Students can easily check whether the given statement is the true or not by just finding the difference between the general terms of the two given AP’s
General term of first AP $\Rightarrow {{a}_{1}}+\left( n-1 \right)d={{t}_{n}}$
General term of the second AP $\Rightarrow t{{'}_{n}}={{a}_{2}}\left( n-1 \right)d$
$\begin{aligned} & t{{'}_{n}}-{{t}_{n}}={{a}_{2}}+\left( n-1 \right)d-\left[ {{a}_{1}}\left( n-1 \right)d \right] \\\ & ={{a}_{2}}+\left( n-1 \right)d-{{a}_{1}}-\left( n-1 \right)d \\\ & ={{a}_{2}}-{{a}_{1}} \\\ \end{aligned}$
=constant
Because here, given ${{a}_{2}}$ and ${{a}_{1}}$ are constant equal to 7 and 2 respectively.
Therefore, the difference between any two corresponding terms, whether it is 10th term, 21st term or any other terms is equal to difference between their first terms i.e. ${{a}_{2}}$ and ${{a}_{1}}$, which is equal to $\left( 7-2 \right)=5$ here.