NCERT Class XI Mathematics - Sequences and Series - Solutions
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Question : 30
Total: 106
Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m.
Solution:
Let the sequence be 1, A 1 , A 2 A m
Then 31 is (m + 2)th term, a = 1, let d be the common difference
⇒ 31 = a + (m + 2 – 1) d
⇒ 31 = 1 + (m + 1) d
⇒ (m + 1)d = 30 ⇒ d =
We are given,
⇒ 9 + 63d = 155 – 10d ⇒ 73d = 146 ⇒ d = 2
∴ d =
Then 31 is (m + 2)th term, a = 1, let d be the common difference
⇒ 31 = a + (m + 2 – 1) d
⇒ 31 = 1 + (m + 1) d
⇒ (m + 1)d = 30 ⇒ d =
We are given,
⇒ 9 + 63d = 155 – 10d ⇒ 73d = 146 ⇒ d = 2
∴ d =
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