Q1. The first and third terms of an A.P. {A1} are A1 = a and A3 = b, and a1 = a and a5 = b respectively be the first and fifth terms of another A.P. {a1}. Find the ratio of An+1 and a2n+1.
A. 2
B. 4
C. 1
D. b/a
Sol : Option C 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 An+1will be a + (n + 1 – 1) * [b – a]/2.
Similarly the value of a2n+1 will be a + (2n + 1 – 1) * [b – a]/4. Taking the ratio of these two, we get:
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Thus 3rd option is the answer.
Q2. There is a set of four numbers p, q, r and s respectively in such a manner that first three are in G.P. and the last three are in A.P. with a difference of 6. If the first and the fourth numbers are the same, find the value of p.
Sol : Option A 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.
Q3. An Arithmetic Progression has 23 terms, the sum of the middle three terms of this arithmetic progression is 720, and the sum of the last three terms of this Arithmetic Progression is 1320. What is the 18th term of this Arithmetic Progression?
A. 240
B. 360
C. 340
D. 440
Sol : Option B 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.
Q4. The first term of an Arithmetic Progression is 15 and the last term is 85. If the sum of all terms is 750, what is the 6th term?
A. 30
B. 40
C. 45
D. 55
Sol : Option B 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.
5. The first term of an AP is 10 and the last term is 28. If the sum of all terms is 190, what is the common difference?
A. 5
B. 3
C. 1
D. 2
Sol : Option D 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.
Q.6. . The sum of three numbers in an Arithmetic Progression is 45 and their product is 3000. What are the three numbers?
A. 5, 15, 25
B. 12, 15, 18
C. 10, 15, 20
D. -10, -15, -20
Sol : Option C 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.
Q.7. The sum of the first six terms of an AP is 48 and the common difference is 2. What is the 4th term?
A. 6
B. 7
C. 9
D. 10
Sol : Option C 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.
Q8. The sum of three numbers in Arithmetic Progression is 72 and their product is 11880. What are the numbers?
Sol : Option D 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.
Q9. The first term of an AP is 3 and the last term is 17. If the sum of all terms is 150, what is the 5th term?
A. 5
B. 7
C. 9
D. 11
Sol : Option B 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.
Q10. The 5th term of an AP is 17/6 and the 9th term is 25/6. What is the 12th term?
A. 29/6
B. 31/6
C. 33/6
D. 34/6
Sol : Option B 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.