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

Important Instructions
81.

In an A.P, \(t_{7} - t_{4} = 20\). Which one of the following is false?

Exam Year(s)

TNPSC Group 1 - 2025, TNPSC - 2025,

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).

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82.

1 + 3 + 5 + ........ 33 + 35 =

Exam Year(s)

Tripura Public Service Commission - 2015,

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.

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83.

A clock strikes once at 1 O'clock, twice at 2 O'clock, thrice at 3 O'clock and so on. How many times in total will it strike in 24 hours ?

Exam Year(s)

Tripura Public Service Commission - 2014,

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.

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84.

The sum of all odd numbers from 1 to 51 is

Exam Year(s)

Tripura Public Service Commission - 2019,

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.

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85.

How many terms are there in the series 201, 208, 215, ..... 369 ?

Exam Year(s)

Tripura Public Service Commission - 2019,

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.

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Tags: A.P. G.P. H.P. Progressions Questions and Answers || Arithmetic Progression MCQ Questions || Geometric Progression GK Questions || Harmonic Progression MCQ || Quantitative Aptitude Progressions GK