Question

plzz.give me the answer of this question.right now I will brainist u​

Answer 1

Since  $a_n$  is the n'th term of our AP, we have  $a_n=a_1+(n-1)d.$

Simply let  $a_1=a.$

Now we have two sums  α  and  β.

$\begin{aligned}&\sum_{r=1}^{100}a_{2r}=\alpha\\ \\ \Longrightarrow\ \ &\sum_{r=1}^{100}[a+(2r-1)d]=\alpha\\ \\ \Longrightarrow\ \ &\sum_{r=1}^{100}a+d\sum_{r=1}^{100}(2r-1)=\alpha\\ \\ \Longrightarrow\ \ &\sum_{r=1}^{100}a+2d\sum_{r=1}^{100}r-d\sum_{r=1}^{100}1=\alpha\\ \\ \Longrightarrow\ \ &100a+2d\cdot\dfrac{100\cdot101}{2}-100d=\alpha\quad\quad\left[\because\ \sum_{r=1}^{n}r=\dfrac{n(n+1)}{2}\right]\\ \\ \Longrightarrow\ \ &100(a+100d)=\alpha\end{aligned}$

$\Longrightarrow\ \ a+100d=\dfrac{\alpha}{100}\\ \\ \\ \Longrightarrow\ \ a_{101}=\dfrac{\alpha}{100}\quad\quad\quad\quad\quad\longrightarrow\quad\quad\quad\quad\quad(1)$

And,

$\begin{aligned}&\sum_{r=1}^{100}a_{2r-1}=\beta\\ \\ \Longrightarrow\ \ &\sum_{r=1}^{100}[a+(2r-2)d]=\beta\\ \\ \Longrightarrow\ \ &\sum_{r=1}^{100}a+d\sum_{r=1}^{100}(2r-2)=\beta\\ \\ \Longrightarrow\ \ &\sum_{r=1}^{100}a+2d\sum_{r=1}^{100}(r-1)=\beta\\ \\ \Longrightarrow\ \ &\sum_{r=1}^{100}a+2d\sum_{r=1}^{100}r-2d\sum_{r=1}^{100}1=\beta\\ \\ \Longrightarrow\ \ &100a+2d\cdot\dfrac{100\cdot101}{2}-200d=\beta\\ \\ \Longrightarrow\ \ &100(a+99d)=\beta\end{aligned}$

$\Longrightarrow\ \ a+99d=\dfrac{\beta}{100}\\ \\ \\ \Longrightarrow\ \ &a_{100}=\dfrac{\beta}{100}\quad\quad\quad\quad\quad\longrightarrow\quad\quad\quad\quad\quad(2)$

Now,

$\begin{aligned}&(1)-(2)\\ \\ \Longrightarrow\ \ &a_{101}-a_{100}-\dfrac{\alpha}{100}-\dfrac{\beta}{100}\\ \\ \Longrightarrow\ \ &\boxed{d=\dfrac{\alpha-\beta}{100}}\end{aligned}$

[We know the common difference is the difference between the consecutive terms of an AP.]

Hence the answer is (C).

Answer 2

Answer:

(C)

Step-by-step explanation:

Since    is the n'th term of our AP, we have  

Simply let  

Now we have two sums  α  and  β.

And,

Now,

[We know the common difference is the difference between the consecutive terms of an AP.]

Hence the answer is (C).