BPSC 70th CCE Prelims General Studies · Question 109 of 143

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) < 0 requires n > 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.

• (a)– 34 — 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)– 42 — 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)– 46 — 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.

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.

The examinable content of this question is one habit: never count steps in your head when a formula will do it for you. Write down a, write down d, and substitute. The temptation to extend the sequence term by term is strong when n is small, and it is not wrong — with n = 14 the list is short enough to write out — but it is slower and it is where the miscount happens. A useful discipline is to verify the index rather than the arithmetic. Test the formula on a term you can see: a + (4 − 1)d = 14 + 3 × (−4) = 2, which matches the printed fourth term, so the formula and the count are both right before you use them at n = 14. Notice also what the option set is telling you. When four options are equally spaced by exactly the common difference, the setter has deliberately built the neighbouring terms into the choices, which is a signal that the question is about indexing and not about the subtraction. On BPSC's paper this matters twice over, because a wrong answer costs one-third of a mark, and an AP question is one where a thirty-second verification eliminates the risk entirely.

• For the sequence 14, 10, 6, 2, …: first term a = 14 and common difference d = −4, since each term is 4 less than the one before
• nth term of an arithmetic progression: aₙ = a + (n − 1)d; here a₁₄ = 14 + 13 × (−4) = 14 − 52 = −38
• General term of this sequence: 14 − 4(n − 1) = 18 − 4n, so the sequence first goes negative at n = 5, where the term is −2
• The three wrong options are the 13th (−34), 15th (−42) and 16th (−46) terms of the same progression, all spaced by the same difference of 4
• Sum of the first n terms: Sₙ = n/2 × [2a + (n − 1)d] = n/2 × (first term + last term); here S₁₄ = 7 × (14 + (−38)) = −168
• BPSC's marking is +1 for a correct answer and −1/3 for a wrong one, so a thirty-second check of the index is worth making on every such question

• Using a + nd instead of a + (n − 1)d, which shifts every answer one term along
• Counting from the last printed term without noticing it is already the fourth term, not the first
• Losing a sign when the common difference is negative, and computing 14 + 52 instead of 14 − 52

BPSC's aptitude block sets short, fully determinate questions — an AP term, a missing number in a grid, a coded word — where a correct method always produces exactly one of the printed options, and the distractors are near-misses generated by the standard slips rather than random numbers. UPSC asked the same material before 2011, when Paper I still carried a quantitative section, and it now lives in CSAT Paper II, so the skill has moved rather than disappeared.

• practice — not a real PYQ

  The 20th term of the arithmetic progression 7, 11, 15, 19, … is

  • (a)79
  • (b)83
  • (c)87
  • (d)91

  Answer(b) 83 — here a = 7 and d = 4, so the 20th term is 7 + 19 × 4 = 7 + 76 = 83; 79 is the 19th term and 87 the 21st.

• practice — not a real PYQ

  Which term of the sequence 14, 10, 6, 2, … is the first to be negative ?

  • (a)4th
  • (b)5th
  • (c)6th
  • (d)7th

  Answer(b) 5th — the general term is 18 − 4n, which is first negative at n = 5, giving −2; the 4th term is still 2.