JANJANG (SOALAN SPM / KLON)

Tajuk janjang (progression) ini sengaja saya post sekali lagi supaya anda tahu bagaimana untuk menjawabnya mengikut fomat SPM. 

Ramai pelajar kata tajuk ini payah. Saya no komen. Hmmm.

Bagi saya, hendak 1000 daya. Tak hendak, 1000 dalih. Kalau dah malas, macam-macam alasan diberi. Pada saya, pelajar macam ini usah dilayan. Biarkan si luncai terjun dengan labu-labunya


Berikut mari kita bicangkan beberapa soalan SPM / klon SPM dan cara menjawabnya.


Q1. Given that  k + 1, k + 5, 2k + 6  is an arithmetic progression, find
(a) the value of k
(b) the sum of the first eight terms

Answer

PERHATIAN : Kalau jujukan arimetik, d =  T2 – T1 atau T3 – T2

(a) k + 1, k + 5, 2k + 6 , … arithmetic progression
d = T2 − T1 = k + 5 − (k + 1) = 4
d = T3 − T2 = 2k + 6 − (k + 5) = k + 1


Since common difference is always the same,
k +1 = 4, so  k = 3
 
(b) When k = 3, we have 4, 8, 12, …
S8 = 8/2 [2(4) + 7(4)] = 144



Q2. Given that the fourth term in the GP is 24 where a = 81, find
a) the common ratio, r
b) the sum of infinity

Answer
PERHATIAN : Tolong guna Tn = ar^n-1 dan S infinity = a/(1 - r)

(a) T4 = 24, so, ar³ = 24.
so we get,       81r³ = 24
r³ = 24 / 81
= 8 / 27
r = 2 / 3 = 0.6667


(b) S∞ = a / (1− r)
 = 81 / (1 – 0.6667)
 = 243



Q3. Given that k, 3, 9/k and m are four consecutive terms of a geometric progression, find the value of  m in term of k

Answer

Since k, 3, 9/k, m are four consecutive terms of a geometric progression,


so 3/k = m/ (9/k)
3/k = mk/9
mk2 = 27
m = 27 / k2



Q4. Given that the arithmetic progression is 5, 9, 13, …Find the three consecutive terms which sum up to 57.

Answer

common difference is 9 – 5 = 4.
Given that S3 = 57
so, 3/2 [2a + (3 − 1)(4)] = 57


3/2 (2a + 8) = 57
3a +12 = 57
a = 15


Hence, the three consecutive terms which sum up to 57 are 15, 19 and 23.



Q5 Given that the volume of water  in the tank increased progressively such as below :
410, 425, 440, …, find the volume of the water at the end of the 8th day


Answer
Volume of water at the end of the 8th day
= T8 = a + 7d
= 410 + 7(15)
= 515 litres



Q6  Write the following 0.848484K in form of fraction

Answer
= 0.84 + 0.0084 + 0.000084 +K
= 0.84 / (1− 0.01), you can use S infinity = a / (1-r) 
= 0.84 /0.99
= 84 /99
= 28 /33


Perhatian : Untuk tujuan memudahkan semakan anda, gunakan sifir di bawah untuk mencari jawapan secara mudah.

(a)    0.84 = 84/99

(b)   0.023 = 23/999

(c)    0.123 = 123/999

Walau bagaimanapun, jawapan anda mesti ditulis seperti di atas. Cara yang ditunjukkan hanya sebagai semakan sahaja.


Q7  Given the squence as follow 3, k, 48….Find the value of k if this sequence is

a)      an arimethic progression

b)      a geometry proggresion

Answer
(a) If 3, k, 48 are in an arithmetic progression,
then
k − 3 = 48 − k
2k = 51
k = 25.5


(b) If 3, k, 48 are in a geometric progression,
then
k/3 = 48/k
k² = 144
k = 12



Q8. Given an arimethic progression 2, 5, 8, …, find .
(a) The common difference of the arithmetic progression.
(b) The sum of all the terms from the 4th term
to the 23rd term.

Answer
PERHATIAN : Soalan ini relative payah. Untuk bahagian (b), guna rumus berikut
     S _{(4 > 23)} = S₂₃ – S₃

(a) The common difference of the arithmetic
progression 2, 5, 8, … is 5 − 2 = 3.


(b) The sum of all the terms from the 4th term
to the 23rd term


= S23 − S3
= 23/2 [2(2) + (23− 1)(3)] − 3/2 [2(2) + (3 − 1)(3)]
= 805 −15
= 790