Practice Questions for Algebra - Progressions

1. PE 2 - Progressions | Algebra - Progressions

Simplify p^1/3 × p^1/9 × p^1/27 × … infinite terms

A.
p
B.
p^2/3
C.
p²
D.
p^1/2

Answer: Option D

Explanation :

Given, p^1/3 × p^1/9 × p^1/27 × … infinite terms

= p^{(1/3 × 1/9 × 1/27 … infinite terms)}

Now, 1/3 × 1/9 × 1/27 … infinite terms = $\frac{1 / 3}{1 - 1 / 3}$ =1/2

∴ Required sum = p^1/2

Hence, option (d).

2. PE 2 - Progressions | Algebra - Progressions

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 = $\frac{6 \pm \sqrt{6^{2} - 4 \times 1 \times 1}}{2}$ = 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).

3. PE 2 - Progressions | Algebra - Progressions

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) = ${[\frac{4}{5} (2 + d)]}^{2}$

⇒ 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).

4. PE 2 - Progressions | Algebra - Progressions

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 13^th 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’.

T_n = a + (n – 1)d and

S_2n-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) × T_n

∴ $\frac{S_{2 n - 1}}{{S^{'}}_{2 n - 1}} = \frac{T_{n}}{{T^{'}}_{n}}$

Substituting n = 13

⇒ $\frac{T_{13}}{{T^{'}}_{13}}$ = $\frac{S_{2 \times 13 - 1}}{{S^{'}}_{2 \times 13 - 1}}$ = $\frac{S_{25}}{{S^{'}}_{25}}$ = $\frac{5 \times 25 + 5}{7 \times 25 - 5}$ = $\frac{130}{170}$ = $\frac{13}{17}$

Hence, option (b).

5. PE 2 - Progressions | Algebra - Progressions

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

⇒ $\frac{a (r^{10} - 1)}{r - 1}$ = 2046

Case 1: r = +2
⇒ $\frac{a (1024 - 1)}{2 - 1}$ = 2046
⇒ a = 2

Case 2: r = -2
⇒ $\frac{a (1024 - 1)}{- 2 - 1}$ = 2046
⇒ a = -6

∴ a = -6 or 2

Hence, option (c).

6. PE 2 - Progressions | Algebra - Progressions

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, $\frac{S_{\infty}}{T_{2}} = \frac{16}{3}$

⇒ $\frac{a / (1 - r)}{a r} = \frac{16}{3}$

⇒ $\frac{1}{r (1 - r)} = \frac{16}{3}$

⇒ 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).

7. PE 2 - Progressions | Algebra - Progressions

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

8. PE 2 - Progressions | Algebra - Progressions

Find the sum of first 11 terms of the series: $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + . . .$

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:

T_n = $\frac{2^{n} - 1}{2^{n}}$ = 1 - $\frac{1}{2^{n}}$

Hence,
T₁ = 1 – 1/2
T₂ = 1 – 1/2²
T₃ = 1 – 1/2³
…
T₁₁ = 1 – 1/2¹¹

S₁₁ = 20 – $(\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + . . . \frac{1}{2^{11}})$
= 20 - $\frac{\frac{1}{2} [{(\frac{1}{2})}^{11} - 1]}{\frac{1}{2} - 1}$
= 20 + 1/2¹¹ – 1
= 19 + 1/2¹¹

Hence, option (b).

9. PE 2 - Progressions | Algebra - Progressions

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 = $\frac{x}{1 - x^{2}}$ - $\frac{x^{2}}{1 - x^{2}}$

⇒ a = $\frac{x - x^{2}}{1 - x^{2}}$

⇒ a = $\frac{x (1 - x)}{(1 + x) (1 - x)}$

⇒ a = $\frac{x}{(1 + x)}$

⇒ a + ax = x

⇒ a = x(1 - a)

⇒ x = a/(1 - a)

Hence, option (a).

10. PE 2 - Progressions | Algebra - Progressions

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).

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PE 2 - Progressions | Algebra - Progressions

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