SAT (Stage-2) Mock Test - 30

109, 119,129,– – – – – – –  999

999 = 109 + (n+1) ×10

n = 90

Let p(x) = 9x² + 6xy + y²

= (3x)² + 2(3x)(y) + y²

= (3x + y)²

The identity used above is (a + b)² = a² + b² + 2ab

g(x) = x + 3
Now, put x + 3 = 0, x = - 3.
The remainder can be found by using p(- 3).
Now, p(-3) = (- 3)³ + 2 (- 3)² - 3(- 3) - 4
= - 27 + 18 + 9 - 4
= - 31 + 27
= - 4
So, remainder is - 4.

Let the zeroes of the polynomial be α, β and γ

Then according to the condition of cubic polynomial formula:

 α.β +  α.γ + β.γ = c/a .......(1)

Since one of the zeroes is zero

Let γ = 0

Then from equation (1)

Product of the other two zeroes = α.β = c/a

The pair of equations x = a and y = b graphically represents the lines which are intersecting at (a, b).

The point at which a line meets y-axis, the value of x is 0.
Substituting 0 for x in the equation 2x + 3y = 6, we get y = 2. So, the coordinates of that point is (0, 2).

Given, 1/2 is a solution of the equation
So, (1/2)² + k/2 - 5/4 = 0
or, 1/4 - 5/4 = -k/2
or, -k/2 = -1
or k = 2

a = 21
d = 21
Applying the formula for the n^th term of an AP, we get
210 = 21 + (n - 1) x 21
On solving for n, we get n = 10.
Hence, 210 is the 10^th term of the given AP.