Question

Let a1, a2, .....a11: are in increasing AP. and if variance of these number is 90 then value of common differenuation of A.P. is.

Answer 1

Let the n'th term of the AP be,

$\longrightarrow a_n=a+(n-1)d$

Mean of these 11 terms is,

$\longrightarrow \bar a=\dfrac{a_1+a_{11}}{2}$

$\longrightarrow \bar a=\dfrac{2a+10d}{2}$

$\longrightarrow \bar a=a+5d$

$\longrightarrow \bar a=a_6$

Square deviation of each term is,

$\longrightarrow(a_1-\bar a)^2=(a_{11}-\bar a)^2=25d^2$

$\longrightarrow(a_2-\bar a)^2=(a_{10}-\bar a)^2=16d^2$

$\longrightarrow(a_3-\bar a)^2=(a_9-\bar a)^2=9d^2$

$\longrightarrow(a_4-\bar a)^2=(a_8-\bar a)^2=4d^2$

$\longrightarrow(a_5-\bar a)^2=(a_7-\bar a)^2=d^2$

$\longrightarrow(a_6-\bar a)^2=0$

Mean square deviation,

$\displaystyle\longrightarrow\dfrac{1}{11}\sum_{i=1}^{11}(a_i-\bar a)^2=\dfrac{2(d^2+4d^2+9d^2+16d^2+25d^2)}{11}$

$\displaystyle\longrightarrow\dfrac{1}{11}\sum_{i=1}^{11}(a_i-\bar a)^2=10d^2$

This is the variance which is equal to 90 as per the question.

$\displaystyle\longrightarrow90=10d^2$

$\displaystyle\longrightarrow\underline{\underline{d=\pm 3}}$

Answer 2

Answer:

REFER TO THE ATTACHMENT

HOPE IT HELPS :)