Show that a1, a2, . . ., an, . . . form an AP where an is defined as below : (i) an = 3 + 4n (ii) an = 9 – 5n Also find the sum of the first 15 terms in each case.
Step-by-Step Solution
Key Concept: A sequence is an arithmetic progression (AP) iff the difference between successive terms is constant, i.e., a_{n+1} - a_n = d (independent of n). Once AP is established, the sum of first n terms is given by \(S_n = \frac{n}{2}\bigl(a_1 + a_n\bigr) = \frac{n}{2}\bigl[2a_1 + (n-1)d\bigr]\).
1. Check for AP
- For (i) \(a_n = 3 + 4n\):
\[a_{n+1} - a_n = (3 + 4(n+1)) - (3 + 4n) = 4\]
The difference is constant (\(d = 4\)), hence \(\{a_n\}\) is an AP.
- For (ii) \(a_n = 9 - 5n\):
\[a_{n+1} - a_n = (9 - 5(n+1)) - (9 - 5n) = -5\]
The difference is constant (\(d = -5\)), so \(\{a_n\}\) is also an AP.
2. Find the first term \(a_1\) and the 15th term \(a_{15}\)
- (i) \(a_1 = 3 + 4\times1 = 7\)
\(a_{15} = 3 + 4\times15 = 3 + 60 = 63\)
- (ii) \(a_1 = 9 - 5\times1 = 4\)
\(a_{15} = 9 - 5\times15 = 9 - 75 = -66\)
3. Use the sum formula for AP
\[S_{15} = \frac{15}{2}\bigl(a_1 + a_{15}\bigr)\]
- (i) \[S_{15} = \frac{15}{2}(7 + 63) = \frac{15}{2}\times70 = 15\times35 = 525\]
- (ii) \[S_{15} = \frac{15}{2}(4 + (-66)) = \frac{15}{2}\times(-62) = 15\times(-31) = -465\]
4. Result
- The sequences defined by (i) and (ii) are arithmetic progressions with common differences \(4\) and \(-5\) respectively.
- Sum of the first 15 terms: \(S_{15}=525\) for (i) and \(S_{15}=-465\) for (ii).
Correct Answer: Both sequences are APs. For (i) common difference d = 4, a₁ = 7, a₁₅ = 63, and S₁₅ = 525. For (ii) common difference d = –5, a₁ = 4, a₁₅ = –66, and S₁₅ = –465.