The ratio of the 11th term to the 18th term of an AP is \(2:3\). Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.
\(a_5:a_{21}=1:3\), \(S_5:S_{21}=5:49\)
Step 1: General term of an AP
The (n^{th}) term of an AP is given by: \(a_n = a + (n-1)d\), where (a) is the first term and (d) is the common difference.
Step 2: Write the 11th and 18th terms
\(a_{11} = a + (11-1)d = a + 10d\)
\(a_{18} = a + (18-1)d = a + 17d\)
Step 3: Use the given ratio
We are told that \(a_{11}:a_{18} = 2:3\).
This means: \(\dfrac{a+10d}{a+17d} = \dfrac{2}{3}\).
Step 4: Solve for relation between (a) and (d)
Cross multiply: \(3(a + 10d) = 2(a + 17d)\)
\(3a + 30d = 2a + 34d\)
\(3a - 2a = 34d - 30d\)
\(a = 4d\)
Step 5: Find the 5th and 21st terms
\(a_5 = a + 4d = 4d + 4d = 8d\)
\(a_{21} = a + 20d = 4d + 20d = 24d\)
So, ratio \(a_5:a_{21} = 8d:24d = 1:3\).
Step 6: Formula for sum of (n) terms of an AP
\(S_n = \dfrac{n}{2}[2a + (n-1)d]\)
Step 7: Find sum of first 5 terms
\(S_5 = \dfrac{5}{2}[2a + 4d]\)
Substitute (a = 4d): \(S_5 = \dfrac{5}{2}[2(4d) + 4d]\)
\(= \dfrac{5}{2}(8d + 4d) = \dfrac{5}{2}(12d) = 30d\)
Step 8: Find sum of first 21 terms
\(S_{21} = \dfrac{21}{2}[2a + 20d]\)
Substitute (a = 4d): \(S_{21} = \dfrac{21}{2}[2(4d) + 20d]\)
\(= \dfrac{21}{2}(8d + 20d) = \dfrac{21}{2}(28d) = 294d\)
Step 9: Ratio of sums
\(S_5:S_{21} = 30d:294d = 30:294 = 5:49\)
Final Answer: \(a_5:a_{21} = 1:3\), \(S_5:S_{21} = 5:49\)