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Mathematics

The mean of numbers in A.P. 2, 4, 6, 8, ……, 40 is :

  1. 40+22\dfrac{40 + 2}{2}

  2. 202(2+40)\dfrac{20}{2}(2 + 40)

  3. 102\dfrac{10}{2} (2 + 40)

  4. 840

Statistics

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Answer

Given,

A.P. = 2, 4, 6, 8, ……, 40

First term (a) = 2

Common difference (d) = 4 - 2 = 2

Last term (l) = 40

Let no. of terms in A.P. be n.

⇒ 40 = a + (n - 1)d

⇒ 40 = 2 + 2(n - 1)

⇒ 40 = 2 + 2n - 2

⇒ 2n = 40

⇒ n = 402\dfrac{40}{2} = 20.

By formula,

Sum of first n terms of an A.P. = n2(a+l)=202(2+40)=420\dfrac{n}{2}(a + l) = \dfrac{20}{2}(2 + 40) = 420.

Mean = Sum of observationsNo. of observations=42020\dfrac{\text{Sum of observations}}{\text{No. of observations}} = \dfrac{420}{20} = 21.

Solving,

⇒40+22=422\Rightarrow \dfrac{40 + 2}{2} = \dfrac{42}{2} = 21.

Hence, option 1 is the correct option.

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