Geometric Sequence worksheet

1. Term 9 (aₙ): 0.05

1. Each term is the previous one multiplied by r, so the nth term is the first term times r to the power n − 1

   $\mathrm{aₙ}$$=\mathrm{a₁}\times r^{n-1}$

2. Put in your first term, ratio and position

   $\mathrm{a₉}$$=3\times {0.6}^8$$=0.05038848$

3. The sum of the first n terms (multiply the sum by r, subtract, and divide by 1 − r)

   $\mathrm{Sₙ}$$=\frac{\mathrm{a₁}\times \left(1-r^n\right)}{1-r}$

4. Substitute the values

   $\mathrm{S₉}$$=\frac{3\times \left(1-{0.6}^9\right)}{1-0.6}$$=7.42441728$

5. Because |r| < 1, the terms shrink toward 0 and rⁿ vanishes as n grows, so the infinite sum settles on one value

   $\mathrm{S\infty }$$=\frac{\mathrm{a₁}}{1-r}$$=\frac{3}{1-0.6}$$=7.5$

2. Term 5 (aₙ): 0.05

1. Each term is the previous one multiplied by r, so the nth term is the first term times r to the power n − 1

   $\mathrm{aₙ}$$=\mathrm{a₁}\times r^{n-1}$

2. Put in your first term, ratio and position

   $\mathrm{a₅}$$=2\times {0.4}^4$$=0.0512$

3. The sum of the first n terms (multiply the sum by r, subtract, and divide by 1 − r)

   $\mathrm{Sₙ}$$=\frac{\mathrm{a₁}\times \left(1-r^n\right)}{1-r}$

4. Substitute the values

   $\mathrm{S₅}$$=\frac{2\times \left(1-{0.4}^5\right)}{1-0.4}$$=3.2992$

5. Because |r| < 1, the terms shrink toward 0 and rⁿ vanishes as n grows, so the infinite sum settles on one value

   $\mathrm{S\infty }$$=\frac{\mathrm{a₁}}{1-r}$$=\frac{2}{1-0.4}$$=3.3333333333$

3. Term 7 (aₙ): 0.01

1. Each term is the previous one multiplied by r, so the nth term is the first term times r to the power n − 1

   $\mathrm{aₙ}$$=\mathrm{a₁}\times r^{n-1}$

2. Put in your first term, ratio and position

   $\mathrm{a₇}$$=3\times {0.4}^6$$=0.012288$

3. The sum of the first n terms (multiply the sum by r, subtract, and divide by 1 − r)

   $\mathrm{Sₙ}$$=\frac{\mathrm{a₁}\times \left(1-r^n\right)}{1-r}$

4. Substitute the values

   $\mathrm{S₇}$$=\frac{3\times \left(1-{0.4}^7\right)}{1-0.4}$$=4.991808$

5. Because |r| < 1, the terms shrink toward 0 and rⁿ vanishes as n grows, so the infinite sum settles on one value

   $\mathrm{S\infty }$$=\frac{\mathrm{a₁}}{1-r}$$=\frac{3}{1-0.4}$$=5$

4. Term 11 (aₙ): 2.93e-3

1. Each term is the previous one multiplied by r, so the nth term is the first term times r to the power n − 1

   $\mathrm{aₙ}$$=\mathrm{a₁}\times r^{n-1}$

2. Put in your first term, ratio and position

   $\mathrm{a₁₁}$$=3\times {0.5}^{10}$$=0.0029296875$

3. The sum of the first n terms (multiply the sum by r, subtract, and divide by 1 − r)

   $\mathrm{Sₙ}$$=\frac{\mathrm{a₁}\times \left(1-r^n\right)}{1-r}$

4. Substitute the values

   $\mathrm{S₁₁}$$=\frac{3\times \left(1-{0.5}^{11}\right)}{1-0.5}$$=5.9970703125$

5. Because |r| < 1, the terms shrink toward 0 and rⁿ vanishes as n grows, so the infinite sum settles on one value

   $\mathrm{S\infty }$$=\frac{\mathrm{a₁}}{1-r}$$=\frac{3}{1-0.5}$$=6$

5. Term 9 (aₙ): 0.07

1. Each term is the previous one multiplied by r, so the nth term is the first term times r to the power n − 1

   $\mathrm{aₙ}$$=\mathrm{a₁}\times r^{n-1}$

2. Put in your first term, ratio and position

   $\mathrm{a₉}$$=4\times {0.6}^8$$=0.06718464$

3. The sum of the first n terms (multiply the sum by r, subtract, and divide by 1 − r)

   $\mathrm{Sₙ}$$=\frac{\mathrm{a₁}\times \left(1-r^n\right)}{1-r}$

4. Substitute the values

   $\mathrm{S₉}$$=\frac{4\times \left(1-{0.6}^9\right)}{1-0.6}$$=9.89922304$

5. Because |r| < 1, the terms shrink toward 0 and rⁿ vanishes as n grows, so the infinite sum settles on one value

   $\mathrm{S\infty }$$=\frac{\mathrm{a₁}}{1-r}$$=\frac{4}{1-0.6}$$=10$

6. Term 9 (aₙ): 2.62e-3

1. Each term is the previous one multiplied by r, so the nth term is the first term times r to the power n − 1

   $\mathrm{aₙ}$$=\mathrm{a₁}\times r^{n-1}$

2. Put in your first term, ratio and position

   $\mathrm{a₉}$$=4\times {0.4}^8$$=0.00262144$

3. The sum of the first n terms (multiply the sum by r, subtract, and divide by 1 − r)

   $\mathrm{Sₙ}$$=\frac{\mathrm{a₁}\times \left(1-r^n\right)}{1-r}$

4. Substitute the values

   $\mathrm{S₉}$$=\frac{4\times \left(1-{0.4}^9\right)}{1-0.4}$$=6.66491904$

5. Because |r| < 1, the terms shrink toward 0 and rⁿ vanishes as n grows, so the infinite sum settles on one value

   $\mathrm{S\infty }$$=\frac{\mathrm{a₁}}{1-r}$$=\frac{4}{1-0.4}$$=6.6666666667$

7. Term 8 (aₙ): 0.03

1. Each term is the previous one multiplied by r, so the nth term is the first term times r to the power n − 1

   $\mathrm{aₙ}$$=\mathrm{a₁}\times r^{n-1}$

2. Put in your first term, ratio and position

   $\mathrm{a₈}$$=4\times {0.5}^7$$=0.03125$

3. The sum of the first n terms (multiply the sum by r, subtract, and divide by 1 − r)

   $\mathrm{Sₙ}$$=\frac{\mathrm{a₁}\times \left(1-r^n\right)}{1-r}$

4. Substitute the values

   $\mathrm{S₈}$$=\frac{4\times \left(1-{0.5}^8\right)}{1-0.5}$$=7.96875$

5. Because |r| < 1, the terms shrink toward 0 and rⁿ vanishes as n grows, so the infinite sum settles on one value

   $\mathrm{S\infty }$$=\frac{\mathrm{a₁}}{1-r}$$=\frac{4}{1-0.5}$$=8$

8. Term 7 (aₙ): 0.03

1. Each term is the previous one multiplied by r, so the nth term is the first term times r to the power n − 1

   $\mathrm{aₙ}$$=\mathrm{a₁}\times r^{n-1}$

2. Put in your first term, ratio and position

   $\mathrm{a₇}$$=2\times {0.5}^6$$=0.03125$

3. The sum of the first n terms (multiply the sum by r, subtract, and divide by 1 − r)

   $\mathrm{Sₙ}$$=\frac{\mathrm{a₁}\times \left(1-r^n\right)}{1-r}$

4. Substitute the values

   $\mathrm{S₇}$$=\frac{2\times \left(1-{0.5}^7\right)}{1-0.5}$$=3.96875$

5. Because |r| < 1, the terms shrink toward 0 and rⁿ vanishes as n grows, so the infinite sum settles on one value

   $\mathrm{S\infty }$$=\frac{\mathrm{a₁}}{1-r}$$=\frac{2}{1-0.5}$$=4$

9. Term 7 (aₙ): 0.05

1. Each term is the previous one multiplied by r, so the nth term is the first term times r to the power n − 1

   $\mathrm{aₙ}$$=\mathrm{a₁}\times r^{n-1}$

2. Put in your first term, ratio and position

   $\mathrm{a₇}$$=3\times {0.5}^6$$=0.046875$

3. The sum of the first n terms (multiply the sum by r, subtract, and divide by 1 − r)

   $\mathrm{Sₙ}$$=\frac{\mathrm{a₁}\times \left(1-r^n\right)}{1-r}$

4. Substitute the values

   $\mathrm{S₇}$$=\frac{3\times \left(1-{0.5}^7\right)}{1-0.5}$$=5.953125$

5. Because |r| < 1, the terms shrink toward 0 and rⁿ vanishes as n grows, so the infinite sum settles on one value

   $\mathrm{S\infty }$$=\frac{\mathrm{a₁}}{1-r}$$=\frac{3}{1-0.5}$$=6$

10. Term 10 (aₙ): 7.86e-4

1. Each term is the previous one multiplied by r, so the nth term is the first term times r to the power n − 1

   $\mathrm{aₙ}$$=\mathrm{a₁}\times r^{n-1}$

2. Put in your first term, ratio and position

   $\mathrm{a₁₀}$$=3\times {0.4}^9$$=0.000786432$

3. The sum of the first n terms (multiply the sum by r, subtract, and divide by 1 − r)

   $\mathrm{Sₙ}$$=\frac{\mathrm{a₁}\times \left(1-r^n\right)}{1-r}$

4. Substitute the values

   $\mathrm{S₁₀}$$=\frac{3\times \left(1-{0.4}^{10}\right)}{1-0.4}$$=4.999475712$

5. Because |r| < 1, the terms shrink toward 0 and rⁿ vanishes as n grows, so the infinite sum settles on one value

   $\mathrm{S\infty }$$=\frac{\mathrm{a₁}}{1-r}$$=\frac{3}{1-0.4}$$=5$