HSBC Series and Progression Questions and Answers Quiz-1

Explanation For the first A.P. a + 2D₁ = b i.e. D₁= [b – a]/2.
For the second A.P. a + 4D₂ = b i.e. D₂ = [b – a]/4.
Now the value of A_n+1will be a + (n + 1 – 1) * [b – a]/2.
Similarly the value of a_2n+1 will be a + (2n + 1 – 1) * [b – a]/4. Taking the ratio of these two, we get

option: 3

Explanation: Let the first three numbers be a /r, a, ar.
This also implies the 4th number will be (ar + 6).
Out of these the first and the fourth numbers are same i.e. a/r = ar + 6 (ii).
It can also be said that ar = a + 6
⇒ ar – a = 6 ⇒ a(r – 1) = 6 or a = 6/(r – 1) (iii).
Put the value of a from (iii) in (ii), we get the quadratic equation as 2r²– r – 1 = 0.
Solving it, we get the value of r as – 1/2 or 1/2.
Putting r as – 1/2, we get the value of a is – 4 and putting r as 1/2 we get the value of a as – 12.
The first term of the GP becomes 8 and – 24 respectively.
Similarly, the third term of the GP becomes 2 and – 6 respectively.
The two possible GP series are 8, -4, 2 and –24, - 12, - 6.
The last terms of these 2 series become 8 and 0 respectively.
See in the second possible series, the first and the fourth numbers are not same, as specified in the question. Thus the only possible for numbers are 8, - 4, 2, 8, thus first option is the answer.

Explanation: The middle three terms of this Arithmetic Progression represent the eleventh, twelth & thirteenth term.
The average of these three terms will represent the twelth term i.e. the 12th term will be 720/3 = 240. The average of the last three terms will be the twenty second term i.e. 1320/3 = 440
The difference of twenty second term and the twelth term will give us ten times the difference i.e. 440 – 240 = 200/10 = 20 will be the difference. If the twelth term is 240, the 18th term will be 240 + 6d i.e. 240 + 6 * 20 = 360 will be the 18thterm. If you want to apply the Arithmetic Progression formula to solve, you can also do it like that. Therefore, 2 nd option is the correct answer.

Explanation: The sum of all terms of this Arithmetic Progression is (n/2) (a + l) = 750. This gives us n = 15 terms. The 15th term of this Arithmetic Progression is (a + 14d) = 85. Substituting for a, we get d = 5. Therefore, the 6th term of this Arithmetic Progression is (a + 5d) = 40.

Explanation: The sum of all terms is (n/2) (a + l) = 190. This gives us n = 10.
The 10th term is (a + 9d) = 28. Substituting for a, we get d = 2.

Explanation: Assuming that the numbers are (a – d), a, (a + d) and their sum is 45, we get the middle number as 15. Now, the product (a – d) (a + d) = 200. Solving, we get d = 5. Therefore, the numbers are 10, 15 and 20.

Explanation: The sum of six terms is 3 * (2a + 5 * 2) = 48. Solving this, we get a = 3.
The 4th term is (a + 3d) = 9.

Explanation: Assuming that the numbers in the Arithmetic Progression are (a – d), a, (a + d) and their sum is 72, we get the middle number as 24. Now, the product (a – d) (a + d) = 495. Solving, we get d = 9. Therefore, the numbers are 15, 24 and 33.

Explanation: The sum of all terms is (n/2)(a + l) = 150. This gives us n = 15. The 15th term is (a + 14d) = 17. Substituting for a, we get d = 1. Therefore, the 5th term is (a + 4d) = 7.

Explanation: The 5th and 9th terms are (a + 4d) and (a + 8d). Equating these terms with their values and solving as simultaneous equations, we get a = 3/2 and d = 1/3. So the 12th term is (a + 11d) = (3/2) + (11/3) = 31/6.