Example 5 - Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Last updated at March 19, 2021 by Teachoo
Examples
Example 1
Example 2 Important
Example 3
Example 4 Important
Example 5 You are here
Example 6 Important
Example 7
Example 8 Important
Example 9
Example 10
Example 11
Example 12
Example 13 Important
Example 14 (i)
Example 14 (ii)
Example 15 Important
Example 16 Important Deleted for CBSE Board 2022 Exams
Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Serial order wise
Transcript
Example 5 Determine the AP whose 3rd term is 5 and the 7th term is 9. We know that an = a + (n – 1)d From (1) & (2) 5 – 2d = 9 – 6d 6d – 2d = 9 – 5 Given 3rd term is 5 a3 = a + (3 – 1)d 5 = a + 2d a = 5 – 2d Given 7th term is 9 a7 = a + (7 – 1)d 9 = a + 6d a = 9 – 6d 4d = 4 d = 4/4 d = 1 Putting d = 1 in (1) a = 5 – 2d a = 5 – 2(1) a = 5 – 2 a = 3 Hence , First term of an A.P = 3 Second term of AP = First term + Common Difference = 3 + 1 = 4 Third term of AP = Second term + Common Difference = 4 + 1 = 5 So the A.P is 3, 4, 5, …………
