Example 4 - Chapter 9 Class 11 Sequences and Series (Term 1)
Last updated at Oct. 7, 2021 by Teachoo
Examples
Example 1 (i)
Example 1 (ii)
Example 2
Example 3 Important
Example 4 Deleted for CBSE Board 2023 Exams You are here
Example 5 Deleted for CBSE Board 2023 Exams
Example 6 Important Deleted for CBSE Board 2023 Exams
Example 7 Deleted for CBSE Board 2023 Exams
Example 8 Important
Example 9
Example 10 Important
Example 11
Example 12 Important
Example 13
Example 14 Important
Example 15 Important
Example 16
Example 17 Important
Example 18 Important
Example 19 Important Deleted for CBSE Board 2023 Exams
Example 20 Deleted for CBSE Board 2023 Exams
Example 21 Important
Example 22
Example 23
Example 24 Important
Chapter 9 Class 11 Sequences and Series
Serial order wise
- Ex 9.1
- Ex 9.2
- Ex 9.3
- Ex 9.4
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Examples
- Miscellaneous
Transcript
Example 4, In an A.P. if mth term is n and the nth term is m, where m n, find the pth term. We know that an = a + (n 1) d i.e. nth term = a + (n 1) d Thus, mth term = am = a + (m 1) d It is given that mth term is n a + (m 1) d = n Also, it is given that nth term is m a + (n 1) d = m First we find common difference, Subtracting (2) from (1) [a + (m 1) d] [a + (n 1) d] = n m a + (m 1)d a (n 1)d = n m a a + (m 1)d (n 1)d = n m (m 1)d (n 1)d = n m md d nd + d = n m md nd = n m d(m n) = n m d = ( )/( ) d = (( ) )/( ) 1 d = 1 Now we have to calculate a Putting d = 1 in (2) a + (n 1) d = m a + (n 1) (-1) = m a n + 1 = m a = m + n 1 For pth term, we use the formula, an = a + (n 1)d putting n = p, d = -1 and a = m + n 1 ap = (m + n 1) + ( p 1) ( 1) = m + n 1 + ( p + 1) = m + n 1 p + 1 = m + n p Thus, pth term = m + n p
