| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Find sum to infinity |
| Difficulty | Standard +0.3 This is a straightforward application of standard geometric series formulas (S_∞ = a/(1-r) and S_n). Part (i) requires solving simultaneous equations from given conditions, which is routine algebraic manipulation. Part (ii) involves substituting into the sum formula and solving an inequality, which is slightly more involved but still standard C2 content with no novel insight required. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a + ar = 34\) or \(\frac{a(1-r^2)}{(1-r)} = 34\) or \(\frac{a(r^2-1)}{(r-1)} = 34\) | B1 | Correct equation connecting \(a\), \(r\) and 34 |
| \(\frac{a}{1-r} = 162\) | B1 | Correct equation connecting \(a\), \(r\) and 162 |
| Eliminate \(a\): \((1+r)(1-r) = \frac{17}{81}\) or \(1-r^2 = \frac{34}{162}\) | aM1 | Eliminates \(a\) correctly; not a cubic |
| \(r^2 = \frac{64}{81}\) and \(r = \frac{8}{9}\) only | aA1 | Must have positive value only; accept 0.8 recurring or 8/9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(r = \frac{8}{9}\) \((0 < r < 1)\) into correct formula | bM1 | |
| \(a = 18\) | bA1 | Must be 18 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Eliminate \(r\): \(\frac{34-a}{a} = 1 - \frac{a}{162}\) | bM1 | Eliminates \(r\) correctly; equivalent to \(a^2 - 324a + 5508 = 0\) |
| \(a = 18\) (rejects 306) | bA1 | Correct value; award only after 306 rejected |
| Substitute \(a=18\) to give \(r =\) | aM1 | |
| \(r = \frac{8}{9}\) | aA1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{42(1-\left(\frac{6}{7}\right)^n)}{1-\frac{6}{7}} > 290\) | M1 | Allow \(n\) or \(n-1\) and any inequality symbol |
| \(\left(\frac{6}{7}\right)^n < \left(\frac{4}{294}\right)\) or equivalent e.g. \(\left(\frac{7}{6}\right)^n > \left(\frac{294}{4}\right)\) | A1 | Must be power \(n\) (not \(n-1\)) |
| \(n > \frac{\log\left(\frac{4}{294}\right)}{\log\left(\frac{6}{7}\right)}\) or \(\log_{\frac{6}{7}}\left(\frac{4}{294}\right)\) | M1 | Uses logs correctly; must be log of positive quantity |
| \(n > 27.9\), so \(n = 28\) | A1 |
# Question 5:
## Part (i)(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $a + ar = 34$ or $\frac{a(1-r^2)}{(1-r)} = 34$ or $\frac{a(r^2-1)}{(r-1)} = 34$ | B1 | Correct equation connecting $a$, $r$ and 34 |
| $\frac{a}{1-r} = 162$ | B1 | Correct equation connecting $a$, $r$ and 162 |
| Eliminate $a$: $(1+r)(1-r) = \frac{17}{81}$ or $1-r^2 = \frac{34}{162}$ | aM1 | Eliminates $a$ correctly; not a cubic |
| $r^2 = \frac{64}{81}$ and $r = \frac{8}{9}$ only | aA1 | Must have positive value only; accept 0.8 recurring or 8/9 |
## Part (i)(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $r = \frac{8}{9}$ $(0 < r < 1)$ into correct formula | bM1 | |
| $a = 18$ | bA1 | Must be 18 |
## Way 2 Part (b) first:
| Answer | Mark | Guidance |
|--------|------|----------|
| Eliminate $r$: $\frac{34-a}{a} = 1 - \frac{a}{162}$ | bM1 | Eliminates $r$ correctly; equivalent to $a^2 - 324a + 5508 = 0$ |
| $a = 18$ (rejects 306) | bA1 | Correct value; award only after 306 rejected |
| Substitute $a=18$ to give $r =$ | aM1 | |
| $r = \frac{8}{9}$ | aA1 | |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{42(1-\left(\frac{6}{7}\right)^n)}{1-\frac{6}{7}} > 290$ | M1 | Allow $n$ or $n-1$ and any inequality symbol |
| $\left(\frac{6}{7}\right)^n < \left(\frac{4}{294}\right)$ or equivalent e.g. $\left(\frac{7}{6}\right)^n > \left(\frac{294}{4}\right)$ | A1 | Must be power $n$ (not $n-1$) |
| $n > \frac{\log\left(\frac{4}{294}\right)}{\log\left(\frac{6}{7}\right)}$ or $\log_{\frac{6}{7}}\left(\frac{4}{294}\right)$ | M1 | Uses logs correctly; must be log of positive quantity |
| $n > 27.9$, so $n = 28$ | A1 | |
---\begin{enumerate}
\item (i) All the terms of a geometric series are positive. The sum of the first two terms is 34 and the sum to infinity is 162
\end{enumerate}
Find\\
(a) the common ratio,\\
(b) the first term.\\
(ii) A different geometric series has a first term of 42 and a common ratio of $\frac { 6 } { 7 }$.
Find the smallest value of $n$ for which the sum of the first $n$ terms of the series exceeds 290\\
\hfill \mbox{\textit{Edexcel C2 2015 Q5 [10]}}