An arithmetic series has first term $\frac{3}{2}$ and fourth term ${{u}_{4}}$.
A geometric series also has first term $\frac{3}{2}$ and fourth term ${{u}_{4}}$.
Given that the common ratio of the geometric series is non-negative and the sum of its first $3$ terms is $\frac{21}{2}$, find the sum of the first $10$ odd-numbered terms of the arithmetic series.[5]

An arithmetic series has first term $\frac{3}{2}$ and fourth term ${{u}_{4}}$.

A geometric series also has first term $\frac{3}{2}$ and fourth term ${{u}_{4}}$.

Given that the common ratio of the geometric series is non-negative and the sum of its first $3$ terms is $\frac{21}{2}$, find the sum of the first $10$ odd-numbered terms of the arithmetic series.

Geometric Progression

APGP

International Baccalaureate Higher Level Mathematics

IB HL Math

2020 RI P2 Q1

GP: $S_3 = \frac{a(r^3 - 1)}{r - 1}$

$\frac{21}{2} = \frac{\frac{3}{2}(r^3 - 1)}{r - 1}$

$7(r - 1) = r^3 - 1$

$7r - 7 = r^3 - 1$

$0 = r^3 - 7r + 6$

From GC, $r = 1$ or $2$ or $-3$

(reject $r = 1$ as $r \neq 1$, reject $r = -3$ as $r \geq 0$)

$\therefore r = 2$

AP: $u_4 = \frac{3}{2} + 3d = \frac{3}{2}(2)^3 = 12$

$d = \frac{7}{2}$

The odd-numbered terms $T_1, T_3, T_5, T_7, T_9, \ldots$ form a new AP with first term $= \frac{3}{2}$, common difference $= 2d = 2\left(\frac{7}{2}\right) = 7$

Required Sum, $S_{10} = \frac{10}{2}\left[2\left(\frac{3}{2}\right) + (10-1)7\right]$

$= 5[3 + 63]$

$= 330$

2020 RI P2 Q1

2020 RI P2 Q1

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