A person saves Rs. 1000 more than he did the previous year. If he saves Rs. 2000 in the first year, in how many years will he save Rs. 170000?
17 years
Calculating Savings Years with Arithmetic Progression
The problem describes a situation where a person's savings increase by a fixed amount each year compared to the previous year. This pattern of saving forms an Arithmetic Progression (AP).
Here's how we can break down the problem:
- The amount saved in the first year is the first term of the AP, denoted as \(a_1\).
- The amount by which the saving increases each year is the common difference of the AP, denoted as \(d\).
- The total amount the person wants to save is the sum of the AP over a certain number of years, denoted as \(S_n\).
- We need to find the number of years, which is \(n\).
Given Information:
- First year's saving, \(a_1 = \text{Rs. } 2000\)
- Yearly increase in saving, \(d = \text{Rs. } 1000\)
- Total target saving, \(S_n = \text{Rs. } 170000\)
The formula for the sum of the first \(n\) terms of an Arithmetic Progression is:
\(S_n = \frac{n}{2}[2a_1 + (n-1)d]\)
Now, let's substitute the given values into the formula:
\(170000 = \frac{n}{2}[2(2000) + (n-1)1000]\)
Let's solve this equation for \(n\):
\(170000 = \frac{n}{2}[4000 + 1000n - 1000]\)
\(170000 = \frac{n}{2}[3000 + 1000n]\)
To eliminate the fraction, multiply both sides by 2:
\(170000 \times 2 = n[3000 + 1000n]\)
\(340000 = 3000n + 1000n^2\)
Rearrange the terms to form a standard quadratic equation (\(an^2 + bn + c = 0\)):
\(1000n^2 + 3000n - 340000 = 0\)
We can simplify the equation by dividing all terms by 1000:
\(n^2 + 3n - 340 = 0\)
Now, we need to solve this quadratic equation for \(n\). We can factor the quadratic expression. We look for two numbers that multiply to -340 and add up to 3. The numbers are 20 and -17.
\(n^2 + 20n - 17n - 340 = 0\)
Factor by grouping:
\(n(n + 20) - 17(n + 20) = 0\)
Factor out the common term \((n + 20)\):
\((n - 17)(n + 20) = 0\)
This gives two possible solutions for \(n\):
- \(n - 17 = 0 \implies n = 17\)
- \(n + 20 = 0 \implies n = -20\)
Since \(n\) represents the number of years, it must be a positive value. Therefore, the valid solution is \(n = 17\).
The person will save Rs. 170000 in 17 years.
Revision Table: Savings Calculation AP
| Concept | Description | Value in Problem |
|---|---|---|
| Arithmetic Progression (AP) | A sequence where the difference between consecutive terms is constant. | Applicable to yearly savings growth. |
| First Term (\(a_1\)) | The first value in the sequence. | Rs. 2000 (First year saving). |
| Common Difference (\(d\)) | The constant difference between consecutive terms. | Rs. 1000 (Yearly increase). |
| Number of Terms (\(n\)) | The number of elements in the sequence (years). | What we need to find. |
| Sum of \(n\) terms (\(S_n\)) | The total value after \(n\) terms. | Rs. 170000 (Total target saving). |
| AP Sum Formula | \(S_n = \frac{n}{2}[2a_1 + (n-1)d]\) | Used to solve for \(n\). |
| Quadratic Equation | An equation of the form \(ax^2 + bx + c = 0\). | Resulting equation from the AP sum formula. |
Additional Information: Arithmetic Progression Concepts
An arithmetic progression is a fundamental concept in sequences and series. Understanding its properties is key to solving problems like this savings calculation.
- Terms of an AP: The terms are \(a_1, a_1+d, a_1+2d, a_1+3d, \dots, a_1+(n-1)d\). The \(n\)-th term is given by \(a_n = a_1 + (n-1)d\).
- Sum Formula Derivation: The sum formula \(S_n = \frac{n}{2}(a_1 + a_n)\) is another way to express the sum, which comes from adding the first and last terms, the second and second-to-last terms, and so on, which always sum to \(a_1 + a_n\). There are \(n/2\) such pairs. Substituting \(a_n = a_1 + (n-1)d\) into this formula gives \(S_n = \frac{n}{2}(a_1 + a_1 + (n-1)d) = \frac{n}{2}[2a_1 + (n-1)d]\).
- Applications: APs are used in various real-world scenarios, including calculating simple interest, loan payments, growth patterns, and sequences in nature or physics problems where a quantity changes by a constant amount.
- Solving Quadratic Equations: When the sum of an AP leads to a quadratic equation in \(n\), we must solve it, often by factoring, completing the square, or using the quadratic formula: \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Remember that in real-world problems, \(n\) must be a positive integer.
Important Questions from Arithmetic Progression
If the arithmetic mean of a, b, c is \(\rm \frac M 3\) and \(\rm \frac{1}{a} + \frac{1}{b} = -\frac{1}{c} \) , then the arithmetic mean of a 2, b 2, c 2 is
How many two-digit numbers are divisible by 3 ?
A car starts with a speed of 60 km/h with its speed increasing every one hour by 5 km/h. In how many hours will it cover 435 kms?
How many natural numbers lie between 3 and 200 which are divisible by 7?
The arithmetic mean of 16 student's scores is 320. Find the sum of these scores.
