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Ex 9.2
Ex 9.2, 2 Deleted for CBSE Board 2023 Exams
Ex 9.2, 3 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 4 Deleted for CBSE Board 2023 Exams
Ex 9.2, 5 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 6 Deleted for CBSE Board 2023 Exams
Ex 9.2, 7 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 8 Deleted for CBSE Board 2023 Exams
Ex 9.2, 9 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 10 Deleted for CBSE Board 2023 Exams
Ex 9.2, 11 Important Deleted for CBSE Board 2023 Exams
Ex 9.2, 12 Deleted for CBSE Board 2023 Exams
Ex 9.2, 13 Deleted for CBSE Board 2023 Exams
Ex 9.2, 14 Important
Ex 9.2, 15 Important
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Ex 9.2, 17 Deleted for CBSE Board 2023 Exams
Ex 9.2, 18 Important Deleted for CBSE Board 2023 Exams
Ex 9.2
Last updated at March 30, 2023 by Teachoo
Ex 9.2,16 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. We know that to insert n numbers between a & b common difference (d) = (𝑏 − 𝑎)/(𝑛 + 1) Here, We have to insert m numbers between 1 and 31 So , b = 31 , a = 1 & number of terms to be inserted = n = m Therefore, d = (31 − 1)/(𝑚 + 1) = 30/(𝑚 + 1 ) Now, a = 1 , d = 30/(𝑚 + 1 ), b = 31 We need to find 7th and (m – 1)th numbers inserted Now it is given that ratio of (7^𝑡ℎ 𝑛𝑢𝑚𝑏𝑒𝑟)/((𝑚 − 1)^𝑡ℎ 𝑛𝑢𝑚𝑏𝑒𝑟) = 5/9 (1 + 7𝑑)/(1 + (𝑚 − 1)𝑑) = 5/9 (1 + 7𝑑)/(1 + (𝑚 − 1)𝑑) = 5/9 (1 + 7d)9 = 5[1 + (m – 1)d] 9 + 63d = 5 + 5d(m – 1) 9 + 63d = 5 + 5dm – 5d 9 – 5 + 63d + 5d = 5dm 4 + 63d = 5dm Putting d = 30/(𝑚 + 1) 4 + 63(30/(𝑚 + 1)) = 5(30/(𝑚 + 1))m (4(𝑚 + 1) + 63 × 30)/(𝑚 + 1 ) = (5 × 30 × 𝑚)/(𝑚 + 1) 4(m + 1) + 2040 = 150m 4m + 4 + 2040 = 150m 2044 = 150m – 4m 2044 = 150m – 4m 2044 = 146m m = 2044/146 m = 14 Hence m = 14