U .Finding the nth term(with no GP or AP)/General term
Find the general term Tn for each of the following sequences: 1,4,9,16,25
1,4,9,16,25 1,2², 3², 4², 5² This means that: Tn = n²
Find the general term: 9, 6, 3
ANS: an = Tn = 12 -3n 9, 6, 3 a = 9 d = t2 - t1 d = 6 - 9 d = -3 an = a + (n - 1) d an = 9 + (n - 1) -3 an = 9 + -3n + 3 an = 9 + 3 -3n an = 12 -3n an = Tn = 12 -3n
Find the nth term of the sequence 1, 3, 5, 7, 9
To find formula: 1, 3, 5, 7, 9 d = 3 - 1 d = 2 a = 1 1, 3, 5, 7, 9 an = a + (n - 1) * d an = 1 + (n - 1) * 2 an = 1 + 2n - 2 an = 2n - 1 tn = 2n - 1 n = 1 tn = 2n - 1 t2 = 2(2) - 1 t2 = 3 n = 1 tn = 2n - 1 t1 = 2(1) - 1 = n = 2 tn = 2n - 1 t2 = 2(2) - 1 t2 = 4 - 1 = 3 n = 3 tn = 2n - 1 t3 = 2(3) - 1 t3 = 6 - 1 = 5 n = 4 tn = 2n - 1 t4 = 2(4) - 1 t4 = 8 - 1 = 7 n = 5 tn = 2n - 1 t5 = 2(5) - 1 t5 = 10 - 1 = 9
Find the general term: 3, 5, 7, 9.....
an = Tn = 1 + 2n 3, 5, 7, 9..... a = 3 d = t2 - t1 d = 5 - 3 d = 2 an = a + (n - 1) d an = Tn = 3 + (n - 1) 2 an = Tn = 3 + 2n - 2 an = Tn = 3 - 2 + 2n an = Tn = 1 + 2n
What is the general term of a sequence?
aₙ = 4ₙ + 2 This formula was given in the notes, however it's not the general term of a sequence formula It would either be an = a + (n - 1) d or ar^n-1 aₙ = 4ₙ + 2 would have to be given, so (perhaps it was given apart of a question that I missed) because otherwise, there is no occassion where you would go "I must use aₙ = 4ₙ + 2 to derive this answer" unless that specific formula is actually given NB: The aₙ = 4ₙ + 2 type of formula would most likely be: Directly provided in the problem statement Part of a specific problem where the formula is given
Find the nth term of the sequence 1/1 x 2, 1/2 x 3, 1/3 x 4, 1/4 x 5........
nth term of 1/1 x 2, 1/2 x 3, 1/3 x 4, 1/4 x 5........ This means the nth term is ANS = tn = 1/n(n + 1) We know this since the numerator is always 1, and the denominator is always multiplied and goes up by one eg: 1 x 2, 2 x 3, 3 x 4 the term right after remains the same, while the far right term is added by 1 so it would be tn = n(n + 1) eg n = 1 tn = n(n + 1) 1(1 + 2) 1 x 2 n = 2 tn = n(n + 1) 2(2 + 1) 2 x 3 PROOF 1, 2, 3, 4.... n 2, 3, 4, 5..... (n + 1) n = 1 tn = 1/n(n + 1) t1 = 1/1(1 + 1) t1 = 1/1 x 2 n = 2 t2 = 1/2(2 + 1) t2 = 1(3)
Find the nth term of the sequence 2, 4, 6, 8, 10
tn = 2n 2, 4, 6, 8, 10 d = 4 - 2 d = 2 we can use this since the difference is constant an = a + (n - 1) * d an = 2 + (n - 1) * 2 an = 2 + 2n - 2 an = 2n 2, 4, 6, 8, 10 t1 = 2 x 1 = 2 t2 = 2 x 2 = 4 t3 = 2 x 3 = 6
Find the general term: 1/2, 2/3, 3/4
1/2, 2/3, 3/4 numerator = 1, 2, 3 denominator = 2, 3, 4 an = numerator/numerator + 1 an = n/n + 1
Find the nth term of the sequence 1/1, 1/3, 1/9, 1/27, 1/81
1/3ⁿ NB: left is trial and error? right is correct answer (in picture) 1/1 1/3, 1/9, 1/27, 1/81 Simplify as much as possible, try to find common bases (here we can recognize that all the numbers are multiples of 3... so we know that we can simplify all to three with different exponents) 1/3⁰, 1/3¹, 1/3², 1/3³, 1/3⁴ identify exponents t0, t1, t2, t3, t4 Find nth term nth term= tn = 1/3ⁿ We found this from the pattern of: 1/3⁰, 1/3¹, 1/3², 1/3³, 1/3⁴ tn = 1/3ⁿ Proof: n = 0 tn = 1/3ⁿ t1 = 1/3⁰ t1 = 1/1 t1 = 1 n = 1 tn = 1/3ⁿ t1 = 1/3¹ t1 = 1/3 n = 2 tn = 1/3ⁿ t1 = 1/3² t1 = 1/9
Find the general term of: 3, 6, 12, 24....
3, 6, 12, 24.... a = 3 r = 6/3 r = 2 aₙ = arⁿ ⁻ ¹ aₙ = Tₙ = 3 x 2ⁿ ⁻ ¹
