Determine whether 97 and 172 are terms of the sequence t_n = 5n − 3 for n ≥ 1.

Step-by-step solution

Idea: Set 5n − 3 equal to the number and solve. If n is a natural number, the number is a term at that position.

1. For 97: 5n − 3 = 97 ⇒ 5n = 100 ⇒ n = 20.1 mark
2. 20 is a natural number, so 97 is a term (the 20th).½ mark
3. For 172: 5n − 3 = 172 ⇒ 5n = 175 ⇒ n = 35.1 mark
4. 35 is a natural number, so 172 is a term (the 35th).½ mark

Check: 5 × 20 − 3 = 97 ✓; 5 × 35 − 3 = 172 ✓. Also every term ends in 2 or 7, and so do 97 and 172 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Solving 5n − 3 = 97 as 5n = 94 (subtracting 3 instead of adding).
• Deciding without solving, e.g. ‘odd numbers cannot be terms’. The terms are 2, 7, 12, 17, … which include odd numbers.
• Giving only ‘yes’ without the position or a reason. Show n and say it is a natural number.

How this can come in the exam

Try one yourself

Determine whether 85 and 120 are terms of t_n = 4n + 1.

More questions like this

• Which term of the sequence t_n = 5n − 3 for n ≥ 1 is 607?
• A sequence is given by the recursive rule t₁ = −5, t_n+1 = t_n + 3 for n ≥ 1. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?
• Let T₁ = 1, T₂ = 2, T₃ = 4, and T_n = T_n−1 + T_n−2 + T_n−3 for n ≥ 4. Find T₄, T₅, T₆, T₇, and T₈.
• 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?

All Sequences and progressions questions · All maths questions