Sequences & Series MCQs for UPSC Prelims

Correct answer: B - 84

Group the terms into blocks: (1), (1, 2), (1, 3, 2), (1, 4, 3, 2), (1, 5, 4, 3, 2), ...; the k-th block has k terms, beginning with 1 and then running k, k-1, ..., 2. Since 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28, the first 28 terms are exactly blocks 1 through 7. The k-th block sums to 1 + (2 + 3 + ... + k) = k(k+1)/2. Total = 1 + 3 + 6 + 10 + 15 + 21 + 28 = 84.

Correct answer: C - Value-I = Value-II

Value-I: the 11 numbers are consecutive integers, each at least -5. The average of 11 consecutive integers is the middle (6th) term, so to minimize it we start as low as allowed: -5, -4, ..., 5, whose average is 0. Any higher starting point raises the average, so the minimum average is 0. Value-II: the 11 numbers are consecutive non-negative integers. Starting at 0 gives 0, 1, ..., 10, whose product is 0 because of the factor 0. Starting at 1 or more gives a large positive product. So the minimum product is 0. Both values equal 0, so Value-I = Value-II. The official UPSC key confirms option (c).

Correct answer: B - Rs. 40

The seven prizes form an arithmetic progression with common difference d = -20 (each prize is Rs. 20 less than the one before it). Let the first (largest) prize be a. The sum of 7 terms is (7/2)(2a + 6d) = 700, so 7(a + 3d) = 700, giving a + 3d = 100. With d = -20, a - 60 = 100, so a = 160. The least (seventh) prize is a + 6d = 160 + 6(-20) = 160 - 120 = 40. The official UPSC 2013 key marks (b) Rs. 40, matching the solution.

Correct answer: A - 0

Every power 9^k with k at least 1 leaves remainder 3 when divided by 6: 9^k is odd and divisible by 3, and any odd multiple of 3 is of the form 6m + 3. The sum runs over exponents 3 through 100, which is 98 terms, each contributing remainder 3. Total remainder = 98 x 3 = 294, and 294 = 6 x 49, so the sum is exactly divisible by 6. The remainder is 0. The official UPSC key confirms option (a).

Correct answer: B - 8th January, 2023

Savings on day n = 2n - 1 (Rs. 1, 3, 5, ...), so total savings after n days = 1 + 3 + 5 + ... + (2n - 1) = n^2, always a perfect square. For n^2 to also be a perfect cube, every prime exponent in n^2 must be divisible by 3, which means n itself must be a perfect cube. The smallest such n greater than 1 is n = 8: total = 64 = 8^2 = 4^3, both a perfect square and a perfect cube. Hence 8th January, 2023.

Correct answer: D - 225

The multipliers between consecutive terms rise by 0.5 each step: 20 x 0.5 = 10, 10 x 1 = 10, 10 x 1.5 = 15, 15 x 2 = 30, 30 x 2.5 = 75, so X = 75 x 3 = 225. Option (d) 225 is correct. 105 would need a multiplier of 1.4, 120 a multiplier of 1.6 and 150 a multiplier of 2, none of which continues the +0.5 pattern of the series.

Correct answer: D - 64/729th part of the population.

Each year one-third migrates, so two-thirds of the population remains at the end of each year. Starting from the whole (1), after each year multiply by 2/3. After 6 years the leftover fraction = (2/3)^6 = 2^6 / 3^6 = 64 / 729. Hence (d) 64/729. Option (c) 32/729 would be (2/3)^? mismatched (it has 2^5 in the numerator but 3^6 in the denominator, which is inconsistent), and the other options use the wrong powers entirely.

Correct answer: B - 252

The nth term is n^2 (n + 1), equivalently n^3 + n^2: 1 x 2 = 2, 4 x 3 = 12, 9 x 4 = 36, 16 x 5 = 80, 25 x 6 = 150, and the sixth term is 36 x 7 = 252, option (b). The pattern can also be seen from differences: first differences 10, 24, 44, 70 have second differences 14, 20, 26 (increasing by 6), so the next first difference is 70 + 32 = 102 and X = 150 + 102 = 252. Options (a), (c) and (d) do not fit either the formula or the constant third difference.