Ex 8.2, 6

Transcript

Ex 8.2, 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? Now, Salary in 1st year = ₹ 5,00,000 Salary in 2nd year = 5,00,000 + 20,000 = ₹ 5,20,000 Salary in 3rd year = 5,20,000 + 20,000 = ₹ 5,40,000 So, our sequence is 500000, 520000, 540000, …. We notice that difference is the same ∴ It is an AP Now, First term = a = 5,00,000 Common difference = d = 20,000 We need to find after how many years did his income reach ₹ 7,00,000 Thus, an = 7,00,000 We need to find n We know that an = a + (n – 1) d Putting an = 7,00,000, a = 5,00,000, d = 20,000 in formula 7,00,000 = 5,00,000 + (n – 1) × 20,000 7,00,000 – 5,00,000 = (n – 1) × 20,000 2,00,000 = (n – 1) × 20,000 (n – 1) × 20,000 = 2,00,000 (n – 1) = (𝟐,𝟎𝟎,𝟎𝟎𝟎)/(𝟐𝟎,𝟎𝟎𝟎) n – 1 = 10 n = 10 + 1 n = 11 Thus, in 11th year his salary was ₹ 7,00,000 We need to find After how many years did his income reach ₹ 7,00,000 Thus, it took 10 years of receiving increments to reach that amount.