Progressions (A.P., G.P., H.P.) Questions and Answers

Answer: D

She received 2046 apples.

The apples form a geometric progression 2, 4, 8, ... with first term 2 and common ratio 2, for 10 terms:

\[ \begin{aligned} S_{10} &= \frac{2(2^{10} - 1)}{2 - 1} \\ &= 2(1024 - 1) = 2046 \end{aligned} \]

1024 is the number given on the tenth day alone (\(2^{10}\)), not the total; 2060 and 1760 do not come from the series.

Answer: D

The sum t20 to t30 has 11 terms and equals 0, so the middle term t25 is 0, which makes (A) true. Terms t10, t20, t30 and t40 are equally spaced, so they form an A.P. (B), and for n1 + n2 = 50 the terms are placed symmetrically about t25, so their sum is 0 (C).

Since t25 = a + 24d = 0, a = -24d, so if d is positive then a is negative. Hence (D) is false.

If the sum of 11 consecutive terms of an arithmetic progression is zero, the middle term is zero, so t25 = a + 24d = 0 and a = -24d. For positive d this makes a negative, not positive, which is why statement D is false. Statement C follows from the symmetry about t25.

Answer: D

Since \(t_{7}-t_{4}=3d=20\), the common difference is \(d=\tfrac{20}{3}\), which fixes d but not the first term, so (A) is true.

The difference between the 10th and 1st terms is 9d = 60, so (B) is true, and the A.M. of the 12th and 20th terms is the 16th term, so (C) is true. The A.M. of the 5th and 30th terms is the \(\tfrac{5+30}{2}=17.5\)th term, not the 17th, so (D) is false.

Since d = 20/3, the 10th term minus the first is 9 x 20/3 = 60, which agrees with statement (B).

Answer: A

The sum is 324.

The series adds consecutive odd numbers from 1 to 35. First count the terms, then use the fact that the sum of the first n odd numbers is n squared.

\[ \begin{aligned} n &= \frac{35 + 1}{2} = 18 \\ S &= n^{2} = 18^{2} = 324 \end{aligned} \]

The same result comes from the arithmetic series formula, number of terms times the average of the first and last terms:

\[ S = \frac{18}{2}(1 + 35) = 9 \times 36 = 324 \]

The other options are not the square of the number of terms; 423 simply reverses the digits of the correct answer.

Answer: A

The clock strikes 156 times in 24 hours.

In each 12-hour cycle it strikes 1 + 2 + ... + 12 times, and a day has two such cycles.

\[ \begin{aligned} 1 + 2 + \cdots + 12 &= \frac{12 \times 13}{2} = 78 \\ 2 \times 78 &= 156 \end{aligned} \]

144 is \(12 \times 12\), which treats every hour as 12 strokes, and 164 and 168 match no counting rule. The key step is remembering that the 1-to-12 cycle happens twice a day.

Answer: A

The sum is 676.

The odd numbers from 1 to 51 are 1, 3, 5, ..., 51. There are \(\frac{51 + 1}{2} = 26\) of them, and the sum of the first \(n\) odd numbers is \(n^{2}\):

\[ 1 + 3 + 5 + \cdots + 51 = 26^{2} = 676 \]

625 is \(25^{2}\), the sum up to 49, which leaves out 51. 576 is \(24^{2}\), and 696 does not come from the formula.

The formula works because adding consecutive odd numbers builds perfect squares: 1, 1 + 3 = 4, 1 + 3 + 5 = 9, and so on.

Answer: A

The series has 25 terms.

201, 208, 215, ... is an arithmetic progression with first term 201 and common difference 7. If 369 is the \(n\)th term:

\[ \begin{aligned} T_n &= a + (n-1)d \\ 369 &= 201 + (n-1) \times 7 \\ n - 1 &= 168 \div 7 = 24 \\ n &= 25 \end{aligned} \]

24 is the number of gaps between terms, the answer you get by forgetting to add 1 for the first term. 26 and 23 are off-by-one slips in the other direction.

Answer: D

The sum is 441.

The odd numbers from 1 to 41 are 1, 3, ..., 41: there are 21 of them, and the sum of the first n odd numbers is \(n^{2}\).

\[ 1 + 3 + \dots + 41 = 21^{2} = 441 \]

As a check, the average of the first and last terms is 21, and 21 x 21 = 441. 400 (\(20^{2}\)) would miss the last term; 372, 398 and 505 fit no count of odd numbers.

Answer: A

In this A.P., the first term \(a = 2\) and the common difference \(d = 7 - 2 = 5\). The formula for the nth term is \(a_n = a + (n-1)d\). For the 10th term (n=10), we have \(a_{10} = 2 + (10-1) \times 5 = 2 + 9 \times 5 = 2 + 45 = 47\).

Answer: B

Divide any term by the one before it.

\[ \frac{6}{3} = \frac{12}{6} = \frac{24}{12} = 2 \]