AP Word Problems on Loans and Production | Class 10 Maths

Quick answer: In AP word problems, a quantity that increases or decreases by the same amount each time forms an arithmetic progression. Identify the first term $a$, the common difference $d$ and what is asked: a single value is the $n^{\text{th}}$ term, $a_n = a + (n - 1)d$, and a total is the sum, $S_n = \frac{n}{2}[2a + (n - 1)d]$.

Many real-life quantities change by a fixed amount each time. A borrower may increase each monthly loan payment by the same sum, a factory may raise its production by the same number of units every year, and a runner may add the same distance to each day's run.

When a quantity increases or decreases by a constant amount ($d$), the values form an Arithmetic Progression (AP).

In CBSE Class 10 Mathematics, Chapter 5 (Arithmetic Progressions), word problems on loan repayment, savings and production that increases uniformly are common board exam questions. The key skill is turning the words into $a$, $d$, $n$, $a_n$ and $S_n$.

What You Will Learn

• How to tell whether a real-life situation forms an AP
• The two key formulas: the $n^{\text{th}}$ term $a_n = a + (n - 1)d$, and the sum $S_n = \frac{n}{2}[2a + (n - 1)d] = \frac{n}{2}(a + l)$
• Problem 1: loan repayment with instalments that increase every month
• Problem 2: the NCERT TV set production problem
• How to set out your working to score full marks

1. Turning Words into AP Terms

Tip: Ask yourself whether the question wants one value (use $a_n$) or a total (use $S_n$).

2. Problem 1: Loan Repayment with Increasing Instalments

Question

A person repays a loan of ₹$118000$ by paying ₹$1000$ in the first month and then increasing the payment by ₹$100$ every month. How many months will it take to clear the loan, and what is the last instalment?

Solution

Step 1: Write the AP values.

• First instalment: $a = 1000$
• Monthly increase: $d = 100$
• Total loan: $S_n = 118000$
• To find: the number of months $n$ and the last instalment $a_n$

Step 2: Use the sum formula.

$S_n = \frac{n}{2}[2a + (n - 1)d]$

$118000 = \frac{n}{2}[2(1000) + (n - 1) \times 100]$

Multiplying both sides by $2$:

$236000 = n(2000 + 100n - 100) = n(100n + 1900) = 100n^2 + 1900n$

Dividing by $100$:

$n^2 + 19n - 2360 = 0$

Step 3: Factorise. We need two numbers whose product is $-2360$ and whose sum is $19$. These are $59$ and $-40$, since $59 \times (-40) = -2360$ and $59 - 40 = 19$.

$n^2 + 59n - 40n - 2360 = 0$

$n(n + 59) - 40(n + 59) = 0$

$(n - 40)(n + 59) = 0$

Step 4: Choose the valid root. So $n = 40$ or $n = -59$. The number of months cannot be negative, so $n = 40$.

Step 5: Find the last instalment.

$a_{40} = a + (40 - 1)d = 1000 + 39 \times 100 = 4900$

Check: $S_{40} = \frac{40}{2}(1000 + 4900) = 20 \times 5900 = 118000$. ✓

Answer: The loan will be cleared in 40 months (3 years and 4 months), and the last instalment is ₹4900.

3. Problem 2: TV Set Production (NCERT)

Question

A manufacturer of TV sets produced $600$ sets in the third year and $700$ sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find:

(i) the production in the 1st year

(ii) the production in the 10th year

(iii) the total production in the first 7 years

Solution

Step 1: Form two equations. Let the production in the first year be $a$ and the yearly increase be $d$.

$a_3 = a + 2d = 600 \qquad \text{...(1)}$

$a_7 = a + 6d = 700 \qquad \text{...(2)}$

Step 2: Find $d$. Subtracting (1) from (2):

$4d = 100 \implies d = 25$

So production increases by $25$ sets every year.

Step 3: Part (i), production in the 1st year. From (1):

$a + 2(25) = 600 \implies a = 550$

Answer (i): $550$ TV sets.

Step 4: Part (ii), production in the 10th year.

$a_{10} = a + 9d = 550 + 9 \times 25 = 550 + 225 = 775$

Answer (ii): $775$ TV sets.

Step 5: Part (iii), total production in the first 7 years. Using $S_n = \frac{n}{2}(a + l)$ with $l = a_7 = 700$:

$S_7 = \frac{7}{2}(550 + 700) = \frac{7}{2} \times 1250 = 4375$

Answer (iii): $4375$ TV sets.

4. Summary Table

5. Exam Tips and Common Mistakes

Exam Tip: $n$ is a number of terms, so it must be a positive whole number. Always reject a negative or fractional value and say why.

Exam Tip: When the first and last terms are known, $S_n = \frac{n}{2}(a + l)$ is the quickest formula.

Exam Tip: Check your answer by substituting back. In Problem 1, $S_{40} = 118000$ confirms that $n = 40$ is correct.

Common Mistake: Mixing up $a_n$ and $S_n$. $a_n$ is the payment in one month, while $S_n$ is the total paid over $n$ months.

Frequently Asked Questions

How do I know if a word problem is an AP?

If a quantity increases or decreases by the same amount every time (every month, year or day), it forms an AP.

When should I use aₙ and when Sₙ?

Use $a_n = a + (n - 1)d$ for a single value, such as the payment in the 10th month. Use $S_n$ for a total, such as the total amount paid.

How many months will it take to repay ₹118000 if the first instalment is ₹1000 and each increases by ₹100?

It will take $40$ months, and the last instalment will be ₹$4900$.

What was the TV production in the 1st year if 600 sets were made in the 3rd year and 700 in the 7th year?

The yearly increase is $25$ sets, so the production in the first year was $550$ sets.

Why do we reject the negative value of n?

$n$ is the number of terms (months or years), so it cannot be negative.