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Question

If the variance of the terms in an increasing A.P., b1,b2,b3,...,b11 is 90, then the common difference of this A.P. is


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Solution

Step 1. Define the terms of the A.P :

Let the first term and the common difference of the A.P. be a and d. And as it is an increasing A.P. so d>0. The terms of the A.P. are defined as:

b1=a,b2=a+d,b3=a+2d,...b11=a+10d

Step 2. Find the mean of the terms

The mean of the 11 terms is given as:

X¯=b1+b2+b3+...+b1111=a+(a+d)+(a+2d)+...+(a+10d)11=11a+(1+2+...+10)d11=11a11+55d11=a+5d

Step 3. Find the common difference of the A.P.

Now the variance of the terms is 90, so

Variance=i=111X-bi21190=X-b12+X-b22+...X-b1121190×11=a+5d-a2+a+5d-(a+d)2+....+a+5d-a+10d2990=5d2+4d2+3d2+2d2+d2+-d2+-2d2+-3d2+-4d2+-5d2990=25+16+9+4+1+1+4+9+16+25d2990=110d2990110=d29=d2±9=d±3=d

But the value of d must be greater than 0, so d=-3 is rejected and the solution d=3 is accepted.

Hence, the common difference of the A.P. is d=3.


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