Consider the set P = {-5, -3, -1, 1, 3, 5 ...} consisting of 1998 numbers. If a be the average of the elements in P and b be twice the average of the first 1998 natural numbers, then which of the...

Consider the set P = {-5, -3, -1, 1, 3, 5 ...} consisting of 1998 numbers. If a be the average of the elements in P and b be twice the average of

the first 1998 natural numbers, then which of the following is equal to (a - b)?

Expert Answers
hala718 eNotes educator| Certified Educator

P = { -5, -3, -1, 1, 3, 5, ....}

we have 1998 elements:

We know that:

an = a1 + (n-1)*r

      = -5 + (n-1)*2

      = -5 + 2n - 2

      = -7 + 2n

==> a1998 = -7 + 2*1998= 3989

==> P = {-5, -3, -1, 1, 3, 5, ..., 3989}

We know that the sum of the all elements is:

S = -5 -3 -1 + S(1,3,...,3989) = -9 + [n(an+1)/2

  = -[9 + (1995*(3990)/2)

   = 3980016

Then the average:

a = S/n = 3980016/1998 = 1992

b = 2*S(1,2,3,,,1998)/1998 = 2*[n(n+1)/2]/1998

                                          = 2*1998*1999/2*1998)

                                          = 1999

==> a-b = 1992-1999 = -7

william1941 | Student

The given set P consists of 1998 natural numbers starting from -5. Each consequent term is the previous plus 2. So the common difference is 2. This is therefore an A. P. with the first term as -5 and a common difference of 2.

The sum of n terms of an AP is (1/2)*n*(2 a + (n − 1)*d). Taking n =1998.

We get (1/2)*n*(2 a + (n − 1)*d) = (1/2)*1998*(-10+1997*2)

= 3980016.

The average of the first 1998 terms is 1992.

Therefore a= 1992

Now the average of first 1998 natural numbers is (1+2+3...1998) / 1998

= [(1/2)*1998*(2+1997)] / 1998

= 999.5

Therefore b=999.5

Hence a-b= 992.5

neela | Student

P = {-5,-3,-1, 1, 3,....} is the given set.

The number of elements in P is 1998.

The starting element  a1= -5, the common iceasing =d = 2 between the consecutive terms.So this is an AP with a = -5 , d =2 and the nth term an = a1+(n-1)d .

Therefore the 1998th term = a1+(n-1)d =  -5+(1998-1)2 = 3989

Therefore  the average of the elements of P = Sum/numberof terms={[a1+a1+(n-1)]n/2 }/n= {[-5+3989]*1998)/2}/1998 =  1992.

So a = 1992

b is twice the average of 1st 1998 natural numbers = 2*(1+2+3...1998)/1998 = 2{(1998)(1998+1)/2}/1998  =  1999.

So a-b = 1992-1999 = -7.