Question

Consider the sequence: 3,6,9,12,......,99 (a) how many terms are there in the given sequence? (B) find the mean of the sequence. (C)find the sum of squares of each term of the given sequence.(d) find the variance of the sequence

Answer 1

Answer:

The given sequence is Arithmetic Progression.

(a) For finding the number of terms, we use formula

aₙ = a + (n - 1) d

where, aₙ = value of nth term

a = First term

n = number of term

d = difference

Now, In given sequence: 3,6,9,12,......,99

a = 3, d = 3, aₙ = 99

∴ 99 = 3 + (n - 1) × (3)

⇒ 96 = (n - 1) × (3)  

⇒ n = 33

(b) For finding the mean,

Mean = $\frac{3 + 6 +9+...+99}{33}$

⇒  $Mean = \frac{3(1 + 2 +3+...+33}{33}$

⇒  $Mean = \frac{(1 + 2 +3+...+33}{11}$

and sum of first n-terms can be determined by $\frac{n(n+1)}{2}$

⇒  $\frac{n(n+1)}{2} = \frac{33(33+1)}{2} = 561$

Thus, ⇒  $Mean = \frac{561}{11}=51$

(c) Formula for sum of square is: $\frac{n(n+1)(2n+1)}{6}$

Thus,

Sum of square =  $3^\times\frac{33(33+1)(66+1)}{6}=112761$

(d) Variance = 9, since, the difference between all numbers is 3.

⇒ Standard deviation = 3

⇒ Variance = (Standard deviation)² = 9