Moderate -0.8 This is a straightforward application of standard geometric series formulas with all values given. Part (a) uses S_∞ = a/(1-r) to find a, part (b) applies the nth term formula, and part (c) uses the sum formula. All steps are direct substitution into memorized formulas with no problem-solving or conceptual challenges required.
A1: accept awrt 12466 (even following use of 0.9) Correct answer implies M1A1 even with no method shown. Accept correct equivalents such as mixed or improper fractions
(2)
(c)
\(S = \frac{a(1-r^{12})}{1-r}\) or lists and adds their first twelve terms with their \(a\)
M1
\(S = \frac{"19000"(1-(-0.9)^{12})}{1-(-0.9)}\) or \(S = 10000(1-(-0.9)^{12})\)
A1ft
A1ft: Correct unsimplified with their \(a\) and \(r = +0.9\) or –0.9 or for listing method as follows: \(19000 + -17100 + 15390 + -13851 + 12465.9 + -11219.31 + 10097.379 + -9087.6411 + 8178.87699 + -7360.989291 + 6624.890362 + -5962.401326 = \) (Do not follow through for listing method)
\(= 7176\) only
A1cso
A1cso: 7176 only
(3)
[7]
Notes:
(a) M1: Correct use of formula for sum to infinity as above, or states correct formula and makes small slip such as replacing \(r\) with 0.9 instead of –0.9. A1: Correct answer.
(b) M1: Correct use of formula with \(n–1 = 4\), allow 0.9 instead of –0.9 here. Condone invisible brackets. A1: accept awrt 12466 (even following use of 0.9) Correct answer implies M1A1 even with no method shown. Accept correct equivalents such as mixed or improper fractions.
(c) M1: Correct use of formula with power 12 (or adds 12 terms) with their \(a\) (not 10000) and \(r = +0.9\) or –0.9.
A1ft: Correct unsimplified with their \(a\) and \(r = +0.9\) or –0.9 or for listing method as follows: \(19000 + -17100 + 15390 + -13851 + 12465.9 + -11219.31 + 10097.379 + -9087.6411 + 8178.87699 + -7360.989291 + 6624.890362 + -5962.401326 = \) (Do not follow through for listing method)
A1cso: 7176 only
Special case: \(S = \frac{a(1-r^n)}{1-r}\) so \(S = \frac{"19000"(1+(0.9)^{12})}{1+(0.9)}\) is M1A0A0.
Whereas \(S = \frac{"19000"(1-(-0.9)^{12})}{1-(-0.9)}\) on its own with no formula quoted is M0A0A0.
\(S = \frac{"19000"(1+(-0.9)^{12})}{1+(0.9)}\) should have M1 (bod) then final two A marks depend on whether answer is correct so if this is followed by 7176 the A1A1 should be awarded. If it is followed by 12824 then A0A0 is implied.
| **Part** | **Answer/Working** | **Marks** | **Notes** |
|---|---|---|---|
| (a) | $10000 = \frac{a}{1-(-0.9)}$ | M1 | M1: Correct use of formula for sum to infinity as above, or states correct formula and makes small slip such replacing $r$ with 0.9 instead of –0.9 |
| | $a = 19000$ | A1 | A1: Correct answer |
| | | **(2)** | |
| (b) | Use $ar^4$ | M1 | M1: Correct use of formula with $n - 1 = 4$, allow 0.9 instead of –0.9 here. Condone invisible brackets. |
| | $19000 \times (-0.9)^4 = 12465.9$ (accept awrt 12466) | A1 | A1: accept awrt 12466 (even following use of 0.9) Correct answer implies M1A1 even with no method shown. Accept correct equivalents such as mixed or improper fractions |
| | | **(2)** | |
| (c) | $S = \frac{a(1-r^{12})}{1-r}$ or lists and adds their first twelve terms with their $a$ | M1 | M1: Correct use of formula with power 12 (or adds 12 terms) with their $a$ (not 10000) and $r = +0.9$ or –0.9 |
| | $S = \frac{"19000"(1-(-0.9)^{12})}{1-(-0.9)}$ or $S = 10000(1-(-0.9)^{12})$ | A1ft | A1ft: Correct unsimplified with their $a$ and $r = +0.9$ or –0.9 or for listing method as follows: $19000 + -17100 + 15390 + -13851 + 12465.9 + -11219.31 + 10097.379 + -9087.6411 + 8178.87699 + -7360.989291 + 6624.890362 + -5962.401326 = $ (Do not follow through for listing method) |
| | $= 7176$ only | A1cso | A1cso: 7176 only |
| | | **(3)** | |
| | | **[7]** | |
**Notes:**
(a) M1: Correct use of formula for sum to infinity as above, or states correct formula and makes small slip such as replacing $r$ with 0.9 instead of –0.9. A1: Correct answer.
(b) M1: Correct use of formula with $n–1 = 4$, allow 0.9 instead of –0.9 here. Condone invisible brackets. A1: accept awrt 12466 (even following use of 0.9) Correct answer implies M1A1 even with no method shown. Accept correct equivalents such as mixed or improper fractions.
(c) M1: Correct use of formula with power 12 (or adds 12 terms) with their $a$ (not 10000) and $r = +0.9$ or –0.9.
A1ft: Correct unsimplified with their $a$ and $r = +0.9$ or –0.9 or for listing method as follows: $19000 + -17100 + 15390 + -13851 + 12465.9 + -11219.31 + 10097.379 + -9087.6411 + 8178.87699 + -7360.989291 + 6624.890362 + -5962.401326 = $ (Do not follow through for listing method)
A1cso: 7176 only
**Special case:** $S = \frac{a(1-r^n)}{1-r}$ so $S = \frac{"19000"(1+(0.9)^{12})}{1+(0.9)}$ is M1A0A0.
Whereas $S = \frac{"19000"(1-(-0.9)^{12})}{1-(-0.9)}$ on its own with no formula quoted is M0A0A0.
$S = \frac{"19000"(1+(-0.9)^{12})}{1+(0.9)}$ should have M1 (bod) then final two A marks depend on whether answer is correct so if this is followed by 7176 the A1A1 should be awarded. If it is followed by 12824 then A0A0 is implied.
---
\begin{enumerate}
\item A geometric series with common ratio $r = - 0.9$ has sum to infinity 10000 For this series,\\
(a) find the first term,\\
(b) find the fifth term,\\
(c) find the sum of the first twelve terms, giving this answer to the nearest integer.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2018 Q6 [7]}}