Example 11
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Example 11, In a G.P., the 3rd term is 24 and the 6th term is 192. Find the 10th term. We know that an = arn – 1 where an = nth term of GP n is the number of terms a is the first term r is the common ratio Here, 3rd term is 24 i.e. a3 = 24 Putting an = 24, n = 3 in an formula 24 = a 𝑟^(3−1) 24 = ar2 ar2 = 24 Similarly, Given 6th term is 192 i.e. a6 = 192 Putting an = 192, n = 6 in an formula 192 = a 𝑟^(6−1) 192 = ar5 ar5 = 192 Now, our equations are ar2 = 24 …(1) & ar5 = 192 …(2) Dividing (2) by (1), ar5/ar2 = 192/24 𝑟5/𝑟2 = 8 𝑟5/𝑟2 = 8 r5 – 2 = 8 r3 = 8 r3 = 2 × 2 × 2 r3 = (2)3 r = 2. putting r in (1) ar2 = 24 a(2)2 = 24 a (4) = 24 a = 24/4 a = 6 Thus, a = 6 & r = 2 Now we need to find 10th term, we use the formula an = arn – 1 Putting n = 10 , a = 6, r = 2 a10 = ar10 – 1 = (6)(2)10 – 1 = 6 (2)9 = 3072 Hence, 10th term of GP is 3072
Examples
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Example 6 Important
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Example 9
Example 10
Example 11 You are here
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Example 14 Important
Example 15 Important
Example 16
Example 17 Important
Example 18 Important
Example 19 Important
Example 20
Example 21 Important
Example 22
Example 23 Important
Example 24 Important
Chapter 9 Class 11 Sequences and Series
Serial order wise
About the Author
CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .