Question

Find the variance of the numbers 8, 21, 34, ..., 320.

• A. 352.08
• B. 358.80
• C. 1244
• D. 1324

Answer 1

The Correct Option is A

The given numbers form an arithmetic progression (AP) with the first term \( a = 8 \) and common difference \( d = 21 - 8 = 13 \). The nth term of an AP is given by: \[ T_n = a + (n-1) \cdot d. \] We are given the last term as 320, so we can find the number of terms \( n \) in the sequence by setting \( T_n = 320 \): \[ 320 = 8 + (n-1) \cdot 13. \] Solving for \( n \): \[ 320 - 8 = (n-1) \cdot 13, \] \[ 312 = (n-1) \cdot 13, \] \[ n-1 = \frac{312}{13} = 24, \] \[ n = 25. \] So, there are 25 terms in the sequence. Next, we calculate the mean \( \mu \) of the numbers: \[ \mu = \frac{\text{Sum of terms}}{n}. \] The sum of the terms of an AP is given by: \[ S_n = \frac{n}{2} \cdot (a + T_n). \] Substitute the known values: \[ S_{25} = \frac{25}{2} \cdot (8 + 320) = \frac{25}{2} \cdot 328 = 4100. \] Thus, the mean is: \[ \mu = \frac{4100}{25} = 164. \] Now, to find the variance, we use the formula for the variance of an AP: \[ \text{Variance} = \frac{n}{12} \cdot d^2. \] Substitute \( n = 25 \) and \( d = 13 \): \[ \text{Variance} = \frac{25}{12} \cdot 13^2 = \frac{25}{12} \cdot 169 = 352.08. \] Thus, the variance of the numbers is approximately \( \boxed{352.08} \).