An arithmetic sequence has $u_{3} = 11$ and $u_{8} = 26$. Find the first term and the sum of the first $15$ terms.
Question HKC78-AIHL-SEQ-01
[Maximum mark: 5]
An arithmetic sequence has $u_{3} = 11$ and $u_{8} = 26$. Find the first term and the sum of the first $15$ terms.
Mark Scheme
$u_{3} = a + 2d = 11$ and $u_{8} = a + 7d = 26$. Subtract: $5d = 15$, so $d = 3$ and $a = 5$. [3]
$S_{15} = dfrac{15}{2}big(2cdot 5 + 14cdot 3big) = dfrac{15}{2} cdot 52 = 390$. [2]
Total: 5 marks
A geometric series has first term $250$ and common ratio $0.8$. Find the smallest number of terms whose sum exceeds $1000$.
Question HKC78-AIHL-SEQ-02
[Maximum mark: 6]
A geometric series has first term $250$ and common ratio $0.8$. Find the smallest number of terms whose sum exceeds $1000$.
Mark Scheme
$S_{n} = 250 cdot dfrac{1 – 0.8^{n}}{1 – 0.8} = 1250big(1 – 0.8^{n}big)$. [2]
$1250big(1 – 0.8^{n}big) > 1000$ gives $0.8^{n} < 0.2$. [1]
$n > dfrac{ln 0.2}{ln 0.8} approx 7.213$, because $ln 0.8 < 0$ reverses the inequality when dividing. So $n ge 8$. [2]
Check: $S_{7} = 1250(1 – 0.8^{7}) approx 987.86$ and $S_{8} approx 1040.28$. The smallest number of terms is $8$. [1]
Total: 6 marks
