Question

answer is option (3)

need explanation​

Answer 1

First let us find standard deviation of $n$ terms in an AP.

$n^{th}$ term of the AP of first term $a$ and common difference $d$ is,

• $a_n=a+(n-1)d$

Mean of the terms is,

$\longrightarrow \bar a=\dfrac{1}{n}\displaystyle\sum_{i=1}^na_i$

$\longrightarrow \bar a=\dfrac{1}{n}\cdot\dfrac{n}{2}[2a+(n-1)d]$

$\longrightarrow \bar a=a+\left(\dfrac{n-1}{2}\right)d$

$\longrightarrow \bar a=a+\left(\dfrac{n+1}{2}-1\right)d$

$\longrightarrow \bar a=a_{\frac{n+1}{2}}$

I.e., $\left(\dfrac{n+1}{2}\right)^{th}$ term is the mean.

Deviation of each term will be,

$\longrightarrow a_i-\bar a=a_i-a_{\frac{n+1}{2}}$

$\longrightarrow a_i-\bar a=[a+(i-1)d]-\left[a+\left(\dfrac{n-1}{2}\right)d\right]$

$\longrightarrow a_i-\bar a=\left(i-\dfrac{n+1}{2}\right)d$

Then, standard deviation,

$\displaystyle\longrightarrow\sigma=\sqrt{\dfrac{1}{n}\sum_{i=1}^n(a_i-\bar a)^2}$

$\displaystyle\longrightarrow\sigma=\sqrt{\dfrac{1}{n}\sum_{i=1}^n\left[\left(i-\dfrac{n+1}{2}\right)^2d^2\right]}$

$\displaystyle\longrightarrow\sigma=\sqrt{\dfrac{d^2}{n}\sum_{i=1}^n\left(i^2-i(n+1)+\dfrac{(n+1)^2}{4}\right)}$

$\displaystyle\longrightarrow\sigma=\sqrt{\dfrac{d^2}{n}\sum_{i=1}^ni^2-\dfrac{d^2(n+1)}{n}\sum_{i=1}^ni+\dfrac{d^2}{n}\sum_{i=1}^n\dfrac{(n+1)^2}{4}}$

$\displaystyle\longrightarrow\sigma=\sqrt{\dfrac{d^2}{n}\cdot\dfrac{n(n+1)(2n+1)}{6}-\dfrac{d^2(n+1)}{n}\cdot\dfrac{n(n+1)}{2}-\dfrac{d^2}{n}\cdot\dfrac{n(n+1)^2}{4}}$

$\displaystyle\longrightarrow\sigma=\sqrt{\dfrac{d^2(n+1)(2n+1)}{6}-\dfrac{d^2(n+1)^2}{2}+\dfrac{d^2(n+1)^2}{4}}$

$\displaystyle\longrightarrow\sigma=\sqrt{\dfrac{d^2(n+1)(2n+1)}{6}-\dfrac{d^2(n+1)^2}{4}}$

$\displaystyle\longrightarrow\sigma=|d|\sqrt{(n+1)\left(\dfrac{2n+1}{6}-\dfrac{n+1}{4}\right)}$

$\displaystyle\longrightarrow\sigma=|d|\sqrt{(n+1)\cdot\dfrac{8n+4-6n-6}{24}}$

$\displaystyle\longrightarrow\underline{\underline{\sigma=|d|\sqrt{\dfrac{n^2-1}{12}}}}$

Consider two APs consisting of $n$ terms each, having standard deviation $\sigma_1$ and $\sigma_2$ and common difference $d_1$ and $d_2$ respectively.

Since both APs contain same no. of terms,

$\longrightarrow\sigma\propto|d|$

Therefore,

$\longrightarrow\underline{\underline{\dfrac{\sigma_1}{\sigma_2}=\left|\dfrac{d_1}{d_2}\right|}}$

According to the question, the standard deviation of the AP 5, 7, 9,..., 151 is $\sigma.$

Here,

• $a=5$

• $d=7-5=2$

• $n=\dfrac{151-5}{2}+1=74$

We've to find standard deviation of the AP 32, 44, 56,..., 908.

Here,

• $a=32$

• $d=56-44=12$

• $n=\dfrac{908-32}{12}+1=74$

Both APs contain same no. of terms. Therefore,

$\longrightarrow\dfrac{\sigma_1}{\sigma_2}=\left|\dfrac{d_1}{d_2}\right|$

$\longrightarrow\dfrac{\sigma}{\sigma'}=\left|\dfrac{2}{12}\right|$

$\longrightarrow\dfrac{\sigma}{\sigma'}=\dfrac{1}{6}$

$\longrightarrow\underline{\underline{\sigma'=6\sigma}}$

Hence (3) is the answer.

Answer 2

Answer:

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