PE 2 - Progressions | Algebra - Progressions
Simplify p1/3 × p1/9 × p1/27 × … infinite terms
- A.
p
- B.
p2/3
- C.
p2
- D.
p1/2
Answer: Option D
Explanation :
Given, p1/3 × p1/9 × p1/27 × … infinite terms
= p(1/3 × 1/9 × 1/27 … infinite terms)
Now, 1/3 × 1/9 × 1/27 … infinite terms ==1/2
∴ Required sum = p1/2
Hence, option (d).
Workspace:
Three numbers are in an increasing geometric progression. If the second number is multiplied with 3, then the numbers would be in arithmetic progression. Find the value of the common ratio.
- A.
3 - 2√2
- B.
3 + 2√2
- C.
Both (a) and (b)
- D.
None of these
Answer: Option C
Explanation :
Let the first term be ‘a’ and common ratio be ‘r’.
∴ The three terms in GP are a, ar, ar2.
Also given, a, 3ar and ar2 are in Arithmetic Progression.
∴ a + ar2 = 2 × 3ar
⇒ r2 – 6r + 1 = 0
⇒ r = = 3 ± 2√2
Since the terms are in an increasing geometric progression hence r should be greater than 1.
⇒ r = 3 + 2√2
Hence, option (b).
Workspace:
Three numbers are in an arithmetic progression. If the third number is multiplied with 4/5, then the numbers would be in geometric progression. If the first term of the arithmetic progression is 2, find the common difference of the arithmetic progression.
- A.
-3/4
- B.
1
- C.
3
- D.
Cannot be determined
Answer: Option D
Explanation :
Given first term = 2
Let the common difference of the AP be ‘d’
∴ Terms in AP are 2, 2 + d, 2 + 2d
Also, 2, 4/5 × (2 + d) and 2 + 2d are in GP
⇒ 2 × (2 + 2d) =
⇒ 25(1 + d) = 4(2 + d)2
⇒ 25 + 25d = 16 + 16d + 4d2
⇒ 4d2 -9d – 9 = 0
⇒ (4d + 3)(d - 3) = 0
⇒ d = -3/4 or 3.
Hence, option (d).
Workspace:
If the ratio of the sums up to n terms of the two arithmetic progressions is (5n + 5) : (7n − 5), find the ratio of the 13th terms of the two progressions.
- A.
68/87
- B.
13/17
- C.
5/7
- D.
Cannot be determined
Answer: Option B
Explanation :
For any arithmetic Progression whose first term is ‘a’ and common difference is ‘d’.
Tn = a + (n – 1)d and
S2n-1 = (2n - 1)/2 × (2a + (2n – 1 - 1)d)
= (2n - 1)/2 × (2a + 2(n - 1)d)
= (2n - 1) × (a + (n - 1)d)
= (2n - 1) × Tn
∴
Substituting n = 13
⇒ = = = = =
Hence, option (b).
Workspace:
The sum of first 10 terms of a geometric progression is 2046. Find the first term if the tenth term is four times the eight term.
- A.
2
- B.
-6
- C.
2 or -6
- D.
1 or -3
Answer: Option C
Explanation :
Let the first term of the GP be ‘a’ and the common ratio is ‘r’.
Given, T10 = 4 × T8
⇒ ar9 = 4 × ar7
⇒ r = ±2
Now, S10 = 2046
⇒ = 2046
Case 1: r = +2
⇒ = 2046
⇒ a = 2
Case 2: r = -2
⇒ = 2046
⇒ a = -6
∴ a = -6 or 2
Hence, option (c).
Workspace:
If the sum of infinite terms and the second term of a geometric progression is 16 : 3, then find the common ratio?
- A.
1/4
- B.
2/3
- C.
1/3
- D.
3/5
Answer: Option A
Explanation :
Let the first term of the GP be ‘a’ and the common ratio be ‘r’.
Given,
⇒
⇒
⇒ 16r2 - 16r + 3 = 0
⇒ r = ¼ or 3/4
Since only ¼ is there in one of the options and there is no option as ’Cannot be determined’, we will mark the correct answer as option (a).
Hence, option (a).
Workspace:
There are three numbers in geometric progression. Their product is 1728 and the sum of the products of the numbers taken two at a time is 504. Find the largest of the three numbers.
- A.
24
- B.
12
- C.
36
- D.
None of these
Answer: Option A
Explanation :
Let the three terms in GP be a/r, a and ar.
∴ a/r × a × ar = 1728
⇒ a3 = 1728
⇒ a = 12
Also,
a/r × a + a × ar + ar × a/r = 504
⇒ a2/r + a2 + a2r = 504
⇒ 2/r + 2 + 2r = 7 [∵ a = 12]
⇒ 2r2 – 5r + 2 = 0
⇒ (2r - 1)(r - 2) = 0
⇒ r = ½ or 2
Case 1: r = ½
Largest number will be a/r = 24.
Case 1: r = ½
Largest number will be ar = 24.
In any case largest number will be 24.
Hence, option (a).
Workspace:
Find the sum of first 11 terms of the series:
- A.
19 + 1/212
- B.
19 + 1/211
- C.
20 + 1/212
- D.
20 + 1/211
Answer: Option B
Explanation :
In this special series, nth term can be written as:
Tn = = 1 -
Hence,
T1 = 1 – 1/2
T2 = 1 – 1/22
T3 = 1 – 1/23
…
T11 = 1 – 1/211
S11 = 20 –
= 20 -
= 20 + 1/211 – 1
= 19 + 1/211
Hence, option (b).
Workspace:
If a = x – x2 + x3 – x4 + x5 – x6 + … infinite terms and |x| < 1. Find x in terms of a.
- A.
x = a/(1 – a)
- B.
x = 1/a
- C.
x = a/(1 + a)
- D.
Cannot be determined
Answer: Option A
Explanation :
Given, a = (x + x3 + x5 + …) – (x2 + x4 + x6 + …)
⇒ a = -
⇒ a =
⇒ a =
⇒ a =
⇒ a + ax = x
⇒ a = x(1 - a)
⇒ x = a/(1 - a)
Hence, option (a).
Workspace:
Five stones are kept, on the path connecting A and B, at intervals of 5 m, with the first stone at A itself. Vishal started from A and began to move all the stones to B by carrying one stone at a time. If the distance AB is 100 m find the minimum distance he has to travel (in km).
Answer: 0.8
Explanation :
1st Stone: Vishal will pick first stone from A and go till B. Distance travelled = 100 m. Vishal is now at B
2nd Stone: Vishal goes till 2nd stone and back to B. Distance travelled = 2 × 95 = 190 m. Vishal is now at B again.
Similarly, distance travelled for 3rd, 4th and 5th stones will be 2 × 90, 2 × 85 and 2 × 80 respectively.
∴ Total distance travelled = 100 + 2 × 95 + 2 × 90 + 2 × 85 + 2 × 80 = 800 m = 0.8 km
Hence, 0.8.
Workspace:
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