Question

In an AP of five terms, the sum of the last four terms is 38, and the product of the
second and the fifth terms is 70. Find the terms of the progression.
3. a) Find the 12 th term of the GP, if the second term is 75 and the 5 th term is -3/5.
b) Find the 45 th term of the GP, if the 20 th term is 1200 and the 30 th term is also 1200.

Answer 1

Progression is simply a group of numbers that follows the same pattern.

• The terms of the progression are: $\mathbf{2, 5, 8, 11,\ and\ 14}$
• The 12th term is: $\mathbf{ -\frac{3}{390625}}$
• The 45th term is $\mathbf{1200}$

1. Arithmetic Progression

From the question, we have:

$\mathbf{T_2 + T_3 + T_4 + T_5 = 38 }$

$\mathbf{T_2 \times T_5 = 70}$

The nth term of an AP is:

$\mathbf{T_n = a + (n - 1)d}$

So, the above equations become:

$\mathbf{T_2 + T_3 + T_4 + T_5 = 38 }$

$\mathbf{a + d + a + 2d + a + 3d + a + 4d = 38}$

$\mathbf{4a + 10d = 38}$

Divide through by 2

$\mathbf{2a + 5d = 19}$

$\mathbf{T_2 \times T_5 = 70}$

$\mathbf{(a + d) \times (a + 4d) = 70}$

Expand

$\mathbf{a^2 + 4ad + ad + 4d^2 = 70}$

$\mathbf{a^2 + 5ad + 4d^2 = 70}$

So, the two equations are:

$\mathbf{2a + 5d = 19}$

$\mathbf{a^2 + 5ad + 4d^2 = 70}$

Make a, the subject in $\mathbf{2a + 5d = 19}$

$\mathbf{a = \frac{19 - 5d}{2}}$

Substitute $\mathbf{a = \frac{19 - 5d}{2}}$ in $\mathbf{a^2 + 5ad + 4d^2 = 70}$

$\mathbf{(\frac{19 - 5d}{2})^2 + 5 \times (\frac{19 - 5d}{2}) \times d + 4d^2 = 70}$

$\mathbf{\frac{361 - 190d + 25d^2}{4} + \frac{95 - 25d}{2} \times d + 4d^2 = 70}$

$\mathbf{\frac{361 - 190d + 25d^2}{4} + \frac{95d - 25d^2}{2} + 4d^2 = 70}$

Take LCM

$\mathbf{\frac{361 - 190d + 25d^2 + 2(95d - 25d^2) + 4 \times 4d^2}{4} = 70}$

$\mathbf{\frac{361 - 190d + 25d^2 + 190d - 50d^2 + 16d^2}{4} = 70}$

Multiply through by 4

$\mathbf{361 - 190d + 25d^2 + 190d - 50d^2 + 16d^2 = 280}$

Evaluate like terms

$\mathbf{361 - 9d^2 = 280}$

Collect like terms

$\mathbf{ - 9d^2 = 280 - 361}$

$\mathbf{ - 9d^2 = -81}$

Divide both sides by -9

$\mathbf{ d^2 = 9}$

Take square roots of both sides

$\mathbf{d = 3}$

Recall that:

$\mathbf{a = \frac{19 - 5d}{2}}$

$\mathbf{a = \frac{19 - 5 \times 3}{2}}$

$\mathbf{a = \frac{19 - 15}{2}}$

$\mathbf{a = \frac{4}{2}}$

$\mathbf{a = 2}$

The terms of the progression is calculated using:

$\mathbf{T_n = a + (n - 1)d}$

So, we have:

$\mathbf{T_1 = 2}$

$\mathbf{T_2 = 2 + 3 = 5}$

$\mathbf{T_3 = 2 + 2 \times 3 = 8}$

$\mathbf{T_4 = 2 + 3 \times 3 = 11}$

$\mathbf{T_5 = 2 + 4 \times 3 = 14}$

2. Geometric Progression

The given parameters are:

$\mathbf{T_{2} = 75}$

$\mathbf{T_{5} = -\frac 35}$

The nth term of a GP is:

$\mathbf{T_{n} = ar^{n-1}}$

So, we have:

$\mathbf{T_{2} = ar^{2-1}}$

$\mathbf{T_{2} = ar}$

Substitute $\mathbf{T_{2} = 75}$

$\mathbf{ ar = 75}$

Similarly

$\mathbf{T_{5} = ar^{5-1}}$

$\mathbf{T_{5} = ar^4}$

Substitute $\mathbf{T_{5} = -\frac 35}$

$\mathbf{ar^4 = -\frac{3}{5}}$

Divide $\mathbf{ar^4 = -\frac{3}{5}}$ by $\mathbf{ ar = 75}$

$\mathbf{\frac{ar^4}{ar} = \frac{-3/5}{75}}$

$\mathbf{r^3 = -\frac{3}{375}}$

Take cube roots of both sides

$\mathbf{r = -\sqrt[3]{\frac{3}{375}}}$

$\mathbf{r = -\sqrt[3]{\frac{1}{125}}}$

$\mathbf{r = -\frac{1}{5}}$

Recall that:

$\mathbf{ ar = 75}$

$\mathbf{a = \frac{75}{r}}$

Substitute $\mathbf{r = -\frac{1}{5}}$

$\mathbf{a = \frac{75}{-1/5}}$

$\mathbf{a = -375}$

Recall that:

$\mathbf{T_{n} = ar^{n-1}}$

So, the 12th term is:

$\mathbf{T_{12} = ar^{11}}$

Substitute $\mathbf{a = -375}$ and $\mathbf{r = -\frac{1}{5}}$

$\mathbf{T_{12} = -375 \times (-1/5)^{11}}$

$\mathbf{T_{12} = -\frac{3}{390625}}$

3. Geometric Progression

The given parameters are:

$\mathbf{T_{20} = 1200}$

$\mathbf{T_{30} = 1200}$

The nth term of a GP is:

$\mathbf{T_{n} = ar^{n-1}}$

So, we have:

$\mathbf{T_{20} = ar^{19}}$

Substitute $\mathbf{T_{20} = 1200}$

$\mathbf{ar^{19} = 1200}$

Similarly

$\mathbf{T_{30} = ar^{29}}$

Substitute $\mathbf{T_{30} = 1200}$

$\mathbf{ar^{29} = 1200}$

Divide $\mathbf{ar^{29} = 1200}$ by $\mathbf{ar^{19} = 1200}$

$\mathbf{\frac{ar^{29}}{ar^{19}} = \frac{1200}{1200}}$

$\mathbf{r^{10} = 1}$

Take tenth roots of both sides

$\mathbf{r = 1}$

Recall that:

$\mathbf{ar^{29} = 1200}$

$\mathbf{a = \frac{1200}{r^{29}}}$

Substitute $\mathbf{r = 1}$

$\mathbf{a = \frac{1200}{1^{29}}}$

$\mathbf{a = 1200}$

Recall that:

$\mathbf{T_{n} = ar^{n-1}}$

So, the 45th term is:

$\mathbf{T_{45} = ar^{44}}$

Substitute $\mathbf{a = 1200}$ and $\mathbf{r = 1}$

$\mathbf{T_{45} = 1200 \times 1^{44}}$

$\mathbf{T_{45} = 1200}$

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