Let T1 = 1, T2 = 2, T3 = 4, and Tn = Tn−1 + Tn−2 + Tn−3 for n ≥ 4. Find T4, T5, T6, T7, and T8.
Answer: T4 = 7, T5 = 13, T6 = 24, T7 = 44, T8 = 81.
Step-by-step solution
Idea: Each new term is the sum of the three terms just before it.
- T4 = T3 + T2 + T1 = 4 + 2 + 1 = 7.½ mark
- T5 = T4 + T3 + T2 = 7 + 4 + 2 = 13.½ mark
- T6 = 13 + 7 + 4 = 24.½ mark
- T7 = 24 + 13 + 7 = 44.½ mark
- T8 = 44 + 24 + 13 = 81.1 mark
T₄ = 7, T₅ = 13, T₆ = 24, T₇ = 44, T₈ = 81.
Check: Sequence 1, 2, 4, 7, 13, 24, 44, 81: each term from the 4th on equals the sum of the three before it ✓.
Answer to write in the exam
T4 = 4 + 2 + 1 = 7
T5 = 7 + 4 + 2 = 13
T6 = 13 + 7 + 4 = 24
T7 = 24 + 13 + 7 = 44
T8 = 44 + 24 + 13 = 81
∴ T4 = 7, T5 = 13, T6 = 24, T7 = 44, T8 = 81
Common mistakes that cost marks
- Adding only the previous two terms (as in the Virahānka sequence), e.g. T4 = 4 + 2 = 6.
- Adding T1 + T2 + T3 every time. The three terms move along: for T6 use T5, T4, T3.
- Doubling the previous term (1, 2, 4, 8, …) because the first three terms look like powers of 2.
How this can come in the exam
MCQ (1 mark)
A sequence starts 0, 1, 1 and each later term is the sum of the previous three. Its 6th term is
- 4
- 7
- 6
- 5
Show answer
(B) 7
0, 1, 1, 2, 4, 7.
Try one yourself
A sequence starts 1, 1, 1 and each later term is the sum of the previous three. Find the 4th to 8th terms.
Show answer
3, 5, 9, 17, 31.
More questions like this
- Let us look at the growing pattern of squares given in the figure. The first four stages of the pattern are shown. If we count the number of tiny squares at each stage, we get sequence 1, 5, 9, 13. Can you predict the number of squares in Stages 5 and 6 of the sequence? In Stages 10, 11 and 12? In Stage 20? At any stage?
- Consider all the sequences we have discussed so far. Which ones are arithmetic progressions and which ones are not? Can you justify your claim?
- Verify that the following sequences are arithmetic progressions and write their nth terms. What do you observe when you plot the ordered pairs emerging from them?
- Using the formula tn = a + (n − 1) × d, find the nth term of the following arithmetic progressions.
- Consider once again the AP: 1, 5, 9, 13, 17, … Since every term is 4 more than the previous term, another way of writing this sequence is t1 = 1, tn = tn−1 + 4 for n ≥ 2. Verify for yourself that this recursive rule leads to the terms of the sequence.
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