Predicting What Comes Next: Sequences and Progressions | Exercise 8.2

Question 6

Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?

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Solution
Understand the Question
  • The annual salary increases by a constant amount each year, forming an Arithmetic Progression (AP).
  • First term (aa): Starting salary =₹5,00,000= \text{₹}5,00,000
  • Common difference (dd): Annual increment =₹20,000= \text{₹}20,000
  • nthn^{\text{th}} term (tnt_n): Target salary =₹7,00,000= \text{₹}7,00,000
  • We use the formula tn=a+(n−1)dt_n = a + (n - 1)d to find nn, which gives the year in which the target salary is reached, and calculate the number of elapsed years as n−1n - 1.

Step 1 · Set up the AP Equation

Given

a=₹5,00,000,d=₹20,000,tn=₹7,00,000a = \text{₹}5,00,000, \quad d = \text{₹}20,000, \quad t_n = \text{₹}7,00,000

Using the formula for the nthn^{\text{th}} term of an AP

tn=a+(n−1)dt_n = a + (n - 1)d ₹7,00,000=₹5,00,000+(n−1)₹20,000\text{₹}7,00,000 = \text{₹}5,00,000 + (n - 1)\text{₹}20,000

Step 2 · Solve for nn

₹7,00,000−₹5,00,000=(n−1)₹20,000₹2,00,000=(n−1)₹20,000₹2,00,000₹20,000=n−110=n−1n=10+1n=11\begin{aligned} \text{₹}7,00,000 - \text{₹}5,00,000 &= (n - 1)\text{₹}20,000 \\[0.6em] \text{₹}2,00,000 &= (n - 1)\text{₹}20,000 \\[0.6em] \dfrac{\text{₹}2,00,000}{\text{₹}20,000} &= n - 1 \\[0.6em] 10 &= n - 1 \\[0.6em] n &= 10 + 1 \\[0.6em] n &= 11 \end{aligned}

n=11n = 11 represents the 11th11^{\text{th}} year. Therefore, the income reached ₹7,00,000 after 1010 completed years.

Answer

10 years

Common Mistakes
  • Confusing Term Number with Elapsed Years: n=11n = 11 indicates the salary during the 11th11^{\text{th}} year. The question asks after how many years the salary reached ₹7,00,000, which requires n−1=10n - 1 = 10 years of increments.
  • Incorrect AP Formula: Using tn=a+ndt_n = a + nd instead of tn=a+(n−1)dt_n = a + (n - 1)d leads to an incorrect value of nn.

More questions in Exercise 8.2

Q1

Find the 10th10^{\text{th}} and 26th26^{\text{th}} terms of the AP: 3,8,13,18,…3, 8, 13, 18, \dots

Q2

Which term of the AP: 21,18,15,…21, 18, 15, \dots is −81-81? Also, is 00 a term of this AP? Give reasons for your answer.

Q3

Find the nthn^{\text{th}} term of the AP: 11,8,5,2,…11, 8, 5, 2, \dots. Write the recursive rule for this AP.

Q4

An AP consists of 50 terms in which the 3rd3^{\text{rd}} term is 12 and the last term is 106. Find the 29th29^{\text{th}} term.

(Hint: If aa is the first term and dd the common difference, then we arrive at the equations a+2d=12a + 2d = 12 and a+49d=106a + 49d = 106. Solve this pair of linear equations for aa and dd.)

Q5

How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?

Q6

Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?

Q7

A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?

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