Question 7 - A-Level Maths

CAIE P1 2023 November — Question 7 7 marks

Exam Board	CAIE
Module	P1 (Pure Mathematics 1)
Year	2023
Session	November
Marks	7
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 geometric progression problem requiring students to set up two simultaneous equations (S₂ = 15 and S∞ = 125/7) and solve for the first term and common ratio. While it involves algebraic manipulation and understanding of GP formulas, it's a standard textbook exercise with no novel insight required, making it slightly easier than average.
Spec	1.04i Geometric sequences: nth term and finite series sum 1.04j Sum to infinity: convergent geometric series \|r\|<1

7 The sum of the first two terms of a geometric progression is 15 and the sum to infinity is \(\frac { 125 } { 7 }\). The common ratio of the progression is negative. Find the third term of the progression.

Show mark scheme Show mark scheme source

Question 7:

Answer	Marks	Guidance
Answer	Mark	Guidance
\(a(1+r) = 15\)	B1	Accept \(\frac{a(1-r^2)}{1-r} = 15\) for first B1
\(\frac{a}{1-r} = \frac{125}{7}\)	B1
\(\frac{125}{7}(1-r)(1+r) = 15\)	M1	Eliminate \(a\)
\(1 - r^2 = \frac{105}{125}\)	M1
\(r^2 = \frac{4}{25}\) leading to \(r = -\frac{2}{5}\)	A1	Condone \(\frac{2}{5}\) or \(\pm\frac{2}{5}\)
\(a = \frac{125}{7} \times \frac{7}{5} = 25\)	A1	Ignore 2nd answer
3rd term \(= 25 \times \frac{4}{25} = 4\)	A1	CAO

Alternative Method:

Answer	Marks	Guidance
Answer	Mark	Guidance
\(a(1+r) = 15\)	B1
\(\frac{a}{1-r} = \frac{125}{7}\)	B1
\(7(15-15r) = (125-125r)(1-r^2)\)	M1
\(125r^3 - 125r^2 - 20r + 20 = 0\)	M1
\(r = \frac{-2}{5}\) \([1, \frac{2}{5}]\)	A1	Condone extra answer of \(r = 1\)
\(a = 25\)	A1	Ignore 2nd answer
3rd term \(= 4\)	A1	CAO

Alternative Method (eliminate \(r\)):

Answer	Marks	Guidance
Answer	Mark	Guidance
\(a(1+r) = 15\)	B1
\(\frac{a}{1-r} = \frac{125}{7}\)	B1
\(\frac{a}{1-\left(\frac{15}{a}-1\right)} = \frac{125}{7}\)	M1	Eliminate \(r\)
\(7a^2 - 250a + 1875 [= 0]\)	M1
\(a = 25\), \(\left[\frac{75}{7}\right]\)	A1	Condone extra answer of \(r = \left[\frac{75}{7}\right]\)
\(r = \frac{-2}{5}\)	A1	Ignore 2nd answer
3rd term \(= 25 \times \frac{4}{25} = 4\)	A1	CAO

## Question 7:

| Answer | Mark | Guidance |
|--------|------|----------|
| $a(1+r) = 15$ | B1 | Accept $\frac{a(1-r^2)}{1-r} = 15$ for first B1 |
| $\frac{a}{1-r} = \frac{125}{7}$ | B1 | |
| $\frac{125}{7}(1-r)(1+r) = 15$ | M1 | Eliminate $a$ |
| $1 - r^2 = \frac{105}{125}$ | M1 | |
| $r^2 = \frac{4}{25}$ leading to $r = -\frac{2}{5}$ | A1 | Condone $\frac{2}{5}$ or $\pm\frac{2}{5}$ |
| $a = \frac{125}{7} \times \frac{7}{5} = 25$ | A1 | Ignore 2nd answer |
| 3rd term $= 25 \times \frac{4}{25} = 4$ | A1 | CAO |

**Alternative Method:**

| Answer | Mark | Guidance |
|--------|------|----------|
| $a(1+r) = 15$ | B1 | |
| $\frac{a}{1-r} = \frac{125}{7}$ | B1 | |
| $7(15-15r) = (125-125r)(1-r^2)$ | M1 | |
| $125r^3 - 125r^2 - 20r + 20 = 0$ | M1 | |
| $r = \frac{-2}{5}$ $[1, \frac{2}{5}]$ | A1 | Condone extra answer of $r = 1$ |
| $a = 25$ | A1 | Ignore 2nd answer |
| 3rd term $= 4$ | A1 | CAO |

**Alternative Method (eliminate $r$):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $a(1+r) = 15$ | B1 | |
| $\frac{a}{1-r} = \frac{125}{7}$ | B1 | |
| $\frac{a}{1-\left(\frac{15}{a}-1\right)} = \frac{125}{7}$ | M1 | Eliminate $r$ |
| $7a^2 - 250a + 1875 [= 0]$ | M1 | |
| $a = 25$, $\left[\frac{75}{7}\right]$ | A1 | Condone extra answer of $r = \left[\frac{75}{7}\right]$ |
| $r = \frac{-2}{5}$ | A1 | Ignore 2nd answer |
| 3rd term $= 25 \times \frac{4}{25} = 4$ | A1 | CAO |

---

Show LaTeX source

7 The sum of the first two terms of a geometric progression is 15 and the sum to infinity is $\frac { 125 } { 7 }$. The common ratio of the progression is negative.

Find the third term of the progression.\\

\hfill \mbox{\textit{CAIE P1 2023 Q7 [7]}}

This paper (10 questions)

View full paper

Q1 4 Q2 4 Q3 3 Q4 6 Q5 6 Q6 7 Q7 7 Q8 9 Q9 9 Q10 11