JAMB Mathematics Past Questions & Answers - Page 315

T₂ = 4, T₄ = 16

T_x = ar^n-1

T₂ = ar^2-1 = 4 i.e. ar³ = 16, i.e. ar = 4

T₄ = ar^4-1

therefore, \(\frac{T_4}{T_r}\) = \(\frac{ar^3}{ar}\) = \(\frac{16}{4}\)

r² = 4 and r = 2

but ar = 4

a = \(\frac{4}{r}\) = \(\frac{4}{2}\)

a = 2

Sⁿ = \(\frac{a(r^n - 1)}{r - 1}\)

S⁵ = \(\frac{2(2^5 - 1)}{2 - 1}\)

= \(\frac{2(32 - 1)}{2 - 1}\)

= 2(31)

= 62

3, 6, 9,..., 36.

a = 3, d = 3, i = 36, n = 18

S_n = \(\frac{n}{2}\) [2a + (n - 1)d

S₁₈ = \(\frac{18}{2}\) [2 x 3 + (18 - 1)3]

= 9[6 + (17 x 3)]

= 9 [6 + 51] = 9(57)

= 513

x2 + 2x > 15

x2 + 2x - 15 > 0

(x2 + 5x) - (3x - 15) > 0

x(x + 5) - 3(x + 5) >0

(x - 3)(x + 5) > 0

therefore, x = 3 or -5

i.e. x< 3 or x > -5

-6(x + 3) \(\leq\) 4(x - 2)

-6(x +3) \(\leq\) 4(x - 2)

-6x -18 \(\leq\) 4x - 8

-18 + 8 \(\leq\) 4x +6x

-10 \(\leq\) 10x

10x \(\geq\) -10

x \(\geq\) -1

T \(\alpha \frac{1}{R^3}\)

T = \(\frac{k}{R^3}\)

k = TR³

= \(\frac{2}{81}\) x 3³

= \(\frac{2}{81}\) x 27

dividing 81 by 27

k = \(\frac{2}{2}\)

therefore, T = \(\frac{2}{3}\) x \(\frac{1}{R^3}\)

When R = 2

T = \(\frac{2}{3}\) x \(\frac{1}{2^3}\) = \(\frac{2}{3}\) x \(\frac{1}{8}\)

= \(\frac{1}{12}\)