(a) The first term of an Arithmetic Progression(AP) is 3 and the common difference is 4. Find the sum of the first 28 terms.
(b) Given that \(x = \frac{2m}{1 - m^{2}}\) and \(y = \frac{2m}{1 + m}\), express 2x - y in terms of m in the simplest form.
(c) The angles of pentagon are x°, 2x°, 3x°, 2x° and (3x - 10)°. Find the value of x.
(b) Given that \(x = \frac{2m}{1 - m^{2}}\) and \(y = \frac{2m}{1 + m}\), express 2x - y in terms of m in the simplest form.
(c) The angles of pentagon are x°, 2x°, 3x°, 2x° and (3x - 10)°. Find the value of x.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
(a) \(a = 3; d = 4 ; n = 28\)
\(S_{n} = \frac{n}{2}(2a + (n - 1)d)\)
\(S_{28} = \frac{28}{2}(2(3) + (28 - 1) 4)\)
= \(14(6 + 108)\)
= \(14(114)\)
= \(1596\)
(b) \(x = \frac{2m}{1 - m^{2}} ; y = \frac{2m}{1 + m}\)
\(2x - y = 2(\frac{2m}{1 - m^{2}}) - (\frac{2m}{1 + m})\)
= \(\frac{4m}{1 - m^{2}} - \frac{2m}{1 + m}\)
= \(\frac{4m - 2m(1 - m)}{1 - m^{2}}\)
= \(\frac{4m - 2m + 2m^{2}}{1 - m^{2}}\)
= \(\frac{2m^{2} + 2m}{1 - m^{2}}\)
= \(\frac{2m(m + 1)}{(1 + m)(1 - m)}\)
= \(\frac{2m}{1 - m}\)
(c) Sum of interior angles = (2n - 4)90°
When n = 5 (pentagon),
= \((2(5) - 4) \times 90°\)
= \(540°\)
\(x + 2x + 3x + 2x + (3x - 10) = 11x - 10\)
= \(11x - 10 = 540°\)
= \(11x = 550°\)
= \(x = 50°\)
(a) \(a = 3; d = 4 ; n = 28\)
\(S_{n} = \frac{n}{2}(2a + (n - 1)d)\)
\(S_{28} = \frac{28}{2}(2(3) + (28 - 1) 4)\)
= \(14(6 + 108)\)
= \(14(114)\)
= \(1596\)
(b) \(x = \frac{2m}{1 - m^{2}} ; y = \frac{2m}{1 + m}\)
\(2x - y = 2(\frac{2m}{1 - m^{2}}) - (\frac{2m}{1 + m})\)
= \(\frac{4m}{1 - m^{2}} - \frac{2m}{1 + m}\)
= \(\frac{4m - 2m(1 - m)}{1 - m^{2}}\)
= \(\frac{4m - 2m + 2m^{2}}{1 - m^{2}}\)
= \(\frac{2m^{2} + 2m}{1 - m^{2}}\)
= \(\frac{2m(m + 1)}{(1 + m)(1 - m)}\)
= \(\frac{2m}{1 - m}\)
(c) Sum of interior angles = (2n - 4)90°
When n = 5 (pentagon),
= \((2(5) - 4) \times 90°\)
= \(540°\)
\(x + 2x + 3x + 2x + (3x - 10) = 11x - 10\)
= \(11x - 10 = 540°\)
= \(11x = 550°\)
= \(x = 50°\)