State the values of u₁ and d for this sequence.

Use an appropriate formula to show that the sum of the natural numbers from 1 to n is given by \(\frac{1}{2}n (n +1)\).

Calculate the sum of the natural numbers from 1 to 200.

The sum of the first n terms of G₁ is 29 524. Find n.

A second geometric progression G₂ has the form \(1,\frac{1}{3},\frac{1}{9},\frac{1}{{27}}...\)

Calculate the sum of the first 10 terms of G₂.

Explain why the sum of the first 1000 terms of G₂ will give the same answer as the sum of the first 10 terms, when corrected to three significant figures.

Using your results from parts (a) to (c), or otherwise, calculate the sum of the first 10 terms of the sequence \(2,3\frac{1}{3},9\frac{1}{9},27\frac{1}{{27}}...\)

Give your answer correct to one decimal place.

\(u_1 = d = 1\).     (A1)(A1)

[2 marks]

Sum is \(\frac{1}{2}n(2{u_1} + d(n - 1))\) or \(\frac{1}{2}n({u_1} + {u_n})\)     (M1)

Award (M1) for either sum formula seen, even without substitution.

So sum is \(\frac{1}{2}n(2 + (n - 1)) = \frac{1}{2}n(n + 1)\)     (A1)(AG)

Award (A1) for substitution of \({u_1} = 1 = d\) or \({u_1} = 1\) and \({u_n} = n\) with simplification where appropriate. \(\frac{1}{2}n(n + 1)\) must be seen to award this (A1).

[2 marks]

\(\frac{1}{2}(200)(201) = 20 100\)     (M1)(A1)(G2)

(M1) is for correct formula with correct numerical input. Original sum formula with u, d and n can be used.

[2 marks]

\(\frac{{1 - {3^n}}}{{1 - 3}} = 29524\)     (M1)(A1)

(M1) for correctly substituted formula on one side, (A1) for = 29524 on the other side.

n = 10.     (A1)(G2)

Trial and error is a valid method. Award (M1) for at least \(\frac{{1 - {3^{10}}}}{{1 - 3}}\) seen and then (A1) for = 29524, (A1) for \(n = 10\). For only unproductive trials with \(n \ne 10\), award (M1) and then (A1) if the evaluation is correct.

[3 marks]

Common ratio is \(\frac{1}{3}\), (0.333 (3sf) or 0.3)     (A1)

Accept ‘divide by 3’.

[1 mark]

\(\frac{{1 - {{\left( {\frac{1}{3}} \right)}^{10}}}}{{1 - \frac{1}{3}}}\)     (M1)

= 1.50 (3sf)     (A1)(ft)(G1)

1.5 and \(\frac{3}{2}\) receive (A0)(AP) if AP not yet used Incorrect formula seen in (a) or incorrect value in (b) can follow through to (c). Can award (M1) for \(1 + \left( {\frac{1}{3}} \right) + \left( {\frac{1}{9}} \right) + ......\)

[2 marks]

Both \({\left( {\frac{1}{3}} \right)^{10}}\) and \({\left( {\frac{1}{3}} \right)^{1000}}\) (or those numbers divided by 2/3) are 0 when corrected to 3sf, so they make no difference to the final answer.     (R1)

Accept any valid explanation but please note: statements which only convey the idea of convergence are not enough for (R1). The reason must show recognition that the convergence is adequately fast (though this might be expressed in a much less technical manner).

[1 mark]

The sequence given is \(G_1 + G_2\)     (M1)

The sum is 29 524 + 1.50     (A1)(ft)

= 29 525.5     (A1)(ft)(G2)

The (M1) is implied if the sum of the two numbers is seen. Award (G1) for 29 500 with no working. (M1) can be awarded for
\(2 + 3\frac{1}{3} + ...\) Award final (A1) only for answer given correct to 1dp.

[3 marks]

(i) Identification of u₁ and d was fine. In (b) many candidates failed to recognise the need for a general proof and simply gave an example substitution. Part (c) was well done.

(i) Identification of u₁ and d was fine. In (b) many candidates failed to recognise the need for a general proof and simply gave an example substitution. Part (c) was well done.

(i) Identification of u₁ and d was fine. In (b) many candidates failed to recognise the need for a general proof and simply gave an example substitution. Part (c) was well done.

(ii) Too many candidates here failed to swap to the GP formulae. Those who did know what was happening here often performed the calculations well and got decent marks. The explanations in (d) were often unsatisfactory but some allowance was made for the language difficulties encountered by candidates writing in a 2^nd or higher language. The last part (e) of the question, intended as a high-grade discriminator performed that task very well.

(ii) Too many candidates here failed to swap to the GP formulae. Those who did know what was happening here often performed the calculations well and got decent marks. The explanations in (d) were often unsatisfactory but some allowance was made for the language difficulties encountered by candidates writing in a 2^nd or higher language. The last part (e) of the question, intended as a high-grade discriminator performed that task very well.

(ii) Too many candidates here failed to swap to the GP formulae. Those who did know what was happening here often performed the calculations well and got decent marks. The explanations in (d) were often unsatisfactory but some allowance was made for the language difficulties encountered by candidates writing in a 2^nd or higher language. The last part (e) of the question, intended as a high-grade discriminator performed that task very well.

(ii) Too many candidates here failed to swap to the GP formulae. Those who did know what was happening here often performed the calculations well and got decent marks. The explanations in (d) were often unsatisfactory but some allowance was made for the language difficulties encountered by candidates writing in a 2^nd or higher language. The last part (e) of the question, intended as a high-grade discriminator performed that task very well.

(ii) Too many candidates here failed to swap to the GP formulae. Those who did know what was happening here often performed the calculations well and got decent marks. The explanations in (d) were often unsatisfactory but some allowance was made for the language difficulties encountered by candidates writing in a 2^nd or higher language. The last part (e) of the question, intended as a high-grade discriminator performed that task very well.