What is the 14th term of the sequence 14, 10, 6, 2,.... ?

Question

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Correct — C, – 38. The four printed terms fall by a constant amount — 14 to 10 is −4, 10 to 6 is −4, 6 to 2 is −4 — so this is an arithmetic progression with first term a = 14 and common difference d = −4. The nth term of an AP is a + (n − 1)d, and the reason for (n − 1) rather than n is worth holding on to: the first term has had no step applied to it, the second has had one, so the fourteenth has had thirteen. Substituting, the 14th term = 14 + 13 × (−4) = 14 − 52 = −38. If you distrust the formula, count instead: the sequence runs 14, 10, 6, 2, −2, −6, −10, −14, −18, −22, −26, −30, −34, −38, and the fourteenth entry is −38. The option set is unusually well built, and seeing how tells you exactly where the marks are lost. Every one of the three wrong answers is a real term of this same sequence, taken at the wrong index: −34 is the 13th term, −42 is the 15th and −46 is the 16th. The four options are themselves in arithmetic progression with the same difference of −4, so no amount of estimating or eyeballing the magnitude will separate them; only the index count decides. That makes this a pure test of the off-by-one, which is the single most common error in AP questions. A quick sanity check before committing: the sequence turns negative at the 5th term, since 14 − 4(n − 1)  4.5, and the 5th term is indeed −2. Nine further steps of −4 from −2 gives −38 at the 14th, which confirms the answer by a second route.

Why the other options are wrong — (a) The 13th term, and the answer you get by applying the difference twelve times instead of thirteen — that is, by computing 14 + 12 × (−4). It is the error of a candidate who counts the four printed terms and then adds ten more steps to reach 'the fourteenth', forgetting that the fourth term is already three steps in.  (b) The 15th term, obtained by using n rather than n − 1 in the formula: 14 + 14 × (−4) = −42. This is the mirror image of the previous error and the reason the formula carries the (n − 1) at all; a candidate who writes a + nd will land here on every AP question of this type.  (d) The 16th term, two steps beyond the answer. It is what a candidate produces after making both slips at once — using n instead of n − 1 and then adding one more step while counting — and it is placed last so that a hurried elimination that stops at the third option never reaches it.

Concept: An arithmetic progression is a sequence in which each term differs from the one before it by a fixed amount, the common difference d. Everything about it follows from two numbers, the first term a and that difference. The nth term is a + (n − 1)d, which is simply a plus n − 1 steps. The sum of the first n terms has two equivalent forms — n/2 × [2a + (n − 1)d], or n/2 × (first term + last term), the second being the more useful when the last term is already known. Because the terms change by a constant amount, an AP plotted against its index gives a straight line, with d as the slope: a positive d rises without limit, a negative d falls without limit, and the sequence crosses zero exactly once. Here, with a = 14 and d = −4, the general term is 14 − 4(n − 1), which simplifies to 18 − 4n, so every term of this sequence is 2 less than a multiple of 4 and the whole sequence can be regenerated from that one expression.

State PCS 2024 Q109 — Science & Technology PYQ with Answer

State PCS 2024 Q109 — Science & Technology PYQ with Answer