Advertisements
Advertisements
Question
The sum of (x + 5) observations is x4 – 625. Find the mean of the observations.
Advertisements
Solution
We have, the sum of (x + 5) observations = x4 – 625
We know that, the mean of the n observations x1, x2, ... xn is given by `(x_1 + x_2 + ... x_n)/n`.
∴ The mean of (x + 5) observations
= `("Sum of" (x + 5) "observations")/(x + 5)`
= `(x^4 - 625)/(x + 5)`
= `((x^2)^2 - (25)^2)/(x + 5)`
= `((x^2 + 25)(x^2 - 25))/(x + 5)` ...[∵ a2 – b2 = (a + b)(a – b)]
= `((x^2 + 25)[(x)^2 - (5)^2])/(x + 5)`
= `((x^2 + 25)(x + 5)(x - 5))/((x + 5))`
= (x2 + 25)(x – 5)
APPEARS IN
RELATED QUESTIONS
Using the identity (a + b)(a – b) = a2 – b2, find the following product
(1 + 3b)(3b – 1)
(5 + 20)(–20 – 5) = ?
Factorise the following using the identity a2 – b2 = (a + b)(a – b).
x2 – 9
Factorise the following using the identity a2 – b2 = (a + b)(a – b).
3a2b3 – 27a4b
Factorise the following using the identity a2 – b2 = (a + b)(a – b).
`x^2/8 - y^2/18`
Factorise the following using the identity a2 – b2 = (a + b)(a – b).
`x^2 - y^2/100`
Factorise the expression and divide them as directed:
(x3 + x2 – 132x) ÷ x(x – 11)
The sum of first n natural numbers is given by the expression `n^2/2 + n/2`. Factorise this expression.
Verify the following:
(a2 – b2)(a2 + b2) + (b2 – c2)(b2 + c2) + (c2 – a2) + (c2 + a2) = 0
Find the value of a, if pq2a = (4pq + 3q)2 – (4pq – 3q)2
