Easy -1.2 This is a straightforward application of the arithmetic series formula requiring only substitution of n=20 into the given nth term formula and using S_n = n/2(first term + last term). It's a single-step calculation with no problem-solving or conceptual challenge, making it easier than average but not trivial since students must recognize which formula to use.
B1 for \(a = 11\) and B1 for \(d = 5\) or \(20^{\text{th}}\) term \(= 106\) and M1 for \(20/2[\text{their}(a) + \text{their}(106)]\) or \(20/2[2\text{their}(a)+(20-1)\times\text{their}(d)]\) OR M1 for \(6\times20\) and M2 for \(5\left(\frac{20}{2}[20+1]\right)\) o.e.
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1170 | B1 for $a = 11$ and B1 for $d = 5$ or $20^{\text{th}}$ term $= 106$ and M1 for $20/2[\text{their}(a) + \text{their}(106)]$ or $20/2[2\text{their}(a)+(20-1)\times\text{their}(d)]$ OR M1 for $6\times20$ and M2 for $5\left(\frac{20}{2}[20+1]\right)$ o.e. | 4