If the sum of the first 14 terms of an A.P. is 1050 and the first term is 10, then find the 20th term and the $n$th term.
Generated by claude-sonnet-4-6 · 2026-06-15 10:31 · grounding rag
Model Answer
Given: $S_{14} = 1050$, $a = 10$
Using $S_n = \dfrac{n}{2}[2a + (n-1)d]$:
$$1050 = \frac{14}{2}[2(10) + 13d]$$
$$1050 = 7[20 + 13d]$$
$$150 = 20 + 13d$$
$$d = \frac{130}{13} = 10$$
20th term:
$$a_{20} = a + 19d = 10 + 19(10) = 10 + 190 = \mathbf{200}$$
nth term:
$$a_n = a + (n-1)d = 10 + (n-1)(10) = 10 + 10n - 10 = \mathbf{10n}$$
Source: Chapter 5, Exercise 5.3 / Section 5.5
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Explanation
- First find $d$ using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$. Substituting $n = 14$, $S_{14} = 1050$, $a = 10$ gives $d = 10$.
- Then apply $a_n = a + (n-1)d$ for both $n = 20$ and general $n$.
- Examiners award marks for: correct formula used (1 mark), finding $d$ (1 mark), finding $a_{20}$ and $a_n$ (1 mark). Show all substitution steps clearly.