All questions of Progressions (Sequences & Series) for Interview Preparation Exam

39=a+7d, 59=a+11d, on solving. 4d=20,so d=5,so 59=a+(11×5), 59=a+55,so a=4.

Can't we find with t

Solution:

To solve this problem, we can use the formula for the sum of an arithmetic progression:

Sn = n/2[2a + (n-1)d]

where Sn is the sum of the first n terms, a is the first term, d is the common difference, and n is the number of terms.

We know that the man saves 4 more each year than he did the year before, so the common difference is d = 4.

We also know that he saves `20 in the first year, so the first term is a = 20.

We want to find out after how many years his savings will be more than `1000 altogether, so we want to find the value of n that satisfies the inequality Sn > 1000.

Substituting the values that we know into the formula, we get:

Sn = n/2[2a + (n-1)d]
Sn = n/2[2(20) + (n-1)(4)]
Sn = n/2[40 + 4n - 4]
Sn = n/2[36 + 4n]
Sn = 18n + 2n^2

Now we can set up the inequality:

Sn > 1000
18n + 2n^2 > 1000
2n^2 + 18n - 1000 > 0

We can solve this quadratic inequality by factoring or using the quadratic formula, but we can also estimate the value of n by looking at the options given.

Option A is 19 years, which means that the man's savings would be:

S19 = 19/2[2(20) + (19-1)(4)]
S19 = 19/2[40 + 72]
S19 = 19/2[112]
S19 = 1064

This is more than `1000, so option A is the correct answer.

We can also check the other options:

Option B: S20 = 20/2[2(20) + (20-1)(4)] = 1120
Option C: S21 = 21/2[2(20) + (21-1)(4)] = 1204
Option D: S18 = 18/2[2(20) + (18-1)(4)] = 988

Therefore, the correct answer is option A, 19 years.

Let the side of the outermost square be $s$. Then its diagonal is $s\sqrt{2}=6$, so $s=3\sqrt{2}$.

The first figure formed by joining the midpoints of adjacent sides is a square whose side length is $\frac{s}{\sqrt{2}}=\frac{3\sqrt{2}}{2}$. Its area is $\left(\frac{3\sqrt{2}}{2}\right)^2=\frac{27}{4}$.

The second figure formed by joining the midpoints of adjacent sides of the first figure is a square whose side length is $\frac{s}{2}$. Its area is $\left(\frac{s}{2}\right)^2=\frac{18}{4}=4.5$.

The third figure formed by joining the midpoints of adjacent sides of the second figure is a square whose side length is $\frac{s}{2\sqrt{2}}=\frac{3}{2}$. Its area is $\left(\frac{3}{2}\right)^2=\frac{9}{4}$.

This process continues indefinitely, with each successive figure being a square whose side length is half the previous square's side length. The sum of the areas of all these squares is therefore:

$$\frac{27}{4}+4.5+\frac{9}{4}+\frac{9}{16}+\frac{1}{4}+\frac{1}{64}+\cdots$$

This is an infinite geometric series with first term $\frac{27}{4}$ and common ratio $\frac{1}{8}$. Therefore, the sum is:

$$\frac{\frac{27}{4}}{1-\frac{1}{8}}=\frac{\frac{27}{4}}{\frac{7}{8}}=\frac{27}{4}\cdot\frac{8}{7}=\boxed{12}$$

Given, 4th term = 1/3 and 10th term = 243.

Let the first term be 'a' and the common ratio be 'r'.

We know that the nth term of a GP is given by ar^(n-1).

So, we can write:

ar^3 = 1/3 ...(1)

ar^9 = 243 ...(2)

Dividing equation (2) by equation (1), we get:

r^6 = 729

r = 3

Substituting this value of r in equation (1), we get:

a(3^3) = 1/3

a = 1/27

Therefore, the first term of the GP is 1/27 and the second term can be found using the formula:

ar = (1/27)*3 = 1/9

Hence, the correct answer is option (c) 1/27.

Approach:
Let the first term of the AP be a and the common difference be d.
Given, Sum of all the terms = 105
Also, greatest term = 6 times the least term
Therefore, the greatest term = a + (n-1)d = 6a
where n is the number of terms in the AP.

Calculation:
1. Sum of all terms in an AP:
Sum of n terms in an AP can be given by the formula:
Sum = (n/2)[2a + (n-1)d]
Here, Sum = 105
105 = (n/2)[2a + (n-1)d]

2. Greatest term is 6 times the least term:
a + (n-1)d = 6a
5a = (n-1)d

3. Substitute the value of d in equation 1:
105 = (n/2)[2a + 5a(n-1)/(n-1)]
105 = (n/2)[(7a-5a+5a(n-1))/(n-1)]
105 = (n/2)[(2a+5a(n-1))/(n-1)]

4. Simplify the equation:
210 = n[2a + 5a(n-1)]
Divide both sides by a:
210/a = n(2 + 5n - 5)
42/a = n(5n-3)

5. Check for values of a:
We need to find the lowest value of a.
From the above equation, we can see that a must be a factor of 42.
Therefore, the possible values of a are 1, 2, 3, 6, 7, 14, 21, 42.

6. Substitute values of a to find n:
For each value of a, we can find the corresponding value of n.
If the value of n is a positive integer, then that value of a is valid.
The values of a and n are as follows:
a = 1, n = 15
a = 2, n = 6
a = 3, n = 3.6 (not valid)
a = 6, n = 2.4 (not valid)
a = 7, n = 2.1 (not valid)
a = 14, n = 1.5 (not valid)
a = 21, n = 1.2 (not valid)
a = 42, n = 0.75 (not valid)

7. Find the common difference:
From the equation, 5a = (n-1)d, we get the common difference as:
d = 5a/(n-1)

8. Check for valid values of d:
If the value of d is positive, then the corresponding value of a is valid.
The values of a, n, and d are as follows:
a = 1, n = 15, d = 1/2 (valid)
a = 2, n = 6, d = 2/5 (not valid)
a = 3, n = 3.6, d = 3 (not valid)
a = 6, n = 2.4, d = 6 (not valid)
a = 7, n = 2.1, d

To solve this problem, we can use the formulas for the sum of a geometric progression (GP) and the sum of the squares of a GP.

Let's assume that the three numbers in the GP are "a/r", "a", and "ar", where "a" is the first term and "r" is the common ratio.

The sum of the three numbers is given as 14:
a/r + a + ar = 14 ----(1)

The sum of their squares is given as 84:
(a/r)^2 + a^2 + (ar)^2 = 84 ----(2)

Now, let's simplify equations (1) and (2) to solve for "a" and "r".

Simplifying equation (1):
a/r + a + ar = 14
a(1/r + 1 + r) = 14
a(r^2 + r + 1)/r = 14
a(r^2 + r + 1) = 14r ----(3)

Simplifying equation (2):
(a/r)^2 + a^2 + (ar)^2 = 84
a^2/r^2 + a^2 + a^2r^2 = 84
a^2(1/r^2 + 1 + r^2) = 84
a^2(r^2 + 1 + r^4)/r^2 = 84
a^2(r^4 + r^2 + 1) = 84r^2 ----(4)

Now, let's substitute the value of "a" from equation (3) into equation (4).

(a(r^2 + r + 1))^2 = 84r^2
(a^2(r^2 + r + 1)^2) = 84r^2
[(14r)/(r^2 + r + 1)]^2(r^2 + r + 1)^2 = 84r^2
(14r)^2 = 84r^2
196r^2 = 84r^2
112r^2 = 0
r^2 = 0

Since the common ratio "r" cannot be zero, this means that there is no valid solution for this problem. Therefore, the given information is inconsistent and there is no largest number that satisfies both conditions.

Hence, the correct answer would be considered as "None of the above" or "Not possible to determine".