1. **Find the 8th term of the GP 55, 11, 11/5, ...** The problem is to find the 8th term of a geometric progression (GP) where the first three terms are 55, 11, and 11/5. The formula for the $n^{th}$ term of a GP is: $$a_n = a \times r^{n-1}$$ where $a$ is the first term and $r$ is the common ratio. Step 1: Find the common ratio $r$. $$r = \frac{11}{55} = \frac{1}{5}$$ Step 2: Use the formula to find the 8th term. $$a_8 = 55 \times \left(\frac{1}{5}\right)^{7}$$ Step 3: Calculate the power. $$\left(\frac{1}{5}\right)^7 = \frac{1}{5^7} = \frac{1}{78125}$$ Step 4: Multiply. $$a_8 = 55 \times \frac{1}{78125} = \frac{55}{78125}$$ 2. **A geometric progression is given by 6, 12, 24, ..., 6144. Find: (i) the number of terms in the GP (ii) the sum of all the terms.** Step 1: Identify $a=6$, $r=\frac{12}{6}=2$, last term $a_n=6144$. Step 2: Use the $n^{th}$ term formula: $$a_n = a \times r^{n-1}$$ $$6144 = 6 \times 2^{n-1}$$ Divide both sides by 6: $$\frac{6144}{6} = 2^{n-1}$$ $$1024 = 2^{n-1}$$ Step 3: Express 1024 as a power of 2: $$1024 = 2^{10}$$ Step 4: Equate exponents: $$2^{n-1} = 2^{10} \Rightarrow n-1=10 \Rightarrow n=11$$ Step 5: Find the sum of $n$ terms: $$S_n = a \times \frac{r^n - 1}{r - 1}$$ $$S_{11} = 6 \times \frac{2^{11} - 1}{2 - 1} = 6 \times (2048 - 1) = 6 \times 2047 = 12282$$ 3. **The 2nd and the 6th terms of a GP are 1 and 16/625 respectively. Find: (i) the first term and the common ratio (ii) the general (n^th) term (iii) the first five terms of the GP (iv) the sum of the first 10 terms.** Step 1: Let first term be $a$ and common ratio $r$. Given: $$a_2 = a r = 1$$ $$a_6 = a r^5 = \frac{16}{625}$$ Step 2: Divide $a_6$ by $a_2$: $$\frac{a r^5}{a r} = r^4 = \frac{16/625}{1} = \frac{16}{625}$$ Step 3: Solve for $r$: $$r^4 = \frac{16}{625} = \left(\frac{2}{5}\right)^4$$ $$r = \frac{2}{5}$$ Step 4: Find $a$: $$a r = 1 \Rightarrow a = \frac{1}{r} = \frac{1}{2/5} = \frac{5}{2}$$ Step 5: General term: $$a_n = a r^{n-1} = \frac{5}{2} \times \left(\frac{2}{5}\right)^{n-1}$$ Step 6: First five terms: $$a_1 = \frac{5}{2}$$ $$a_2 = 1$$ $$a_3 = \frac{5}{2} \times \left(\frac{2}{5}\right)^2 = \frac{5}{2} \times \frac{4}{25} = \frac{2}{5}$$ $$a_4 = \frac{5}{2} \times \left(\frac{2}{5}\right)^3 = \frac{5}{2} \times \frac{8}{125} = \frac{4}{25}$$ $$a_5 = \frac{5}{2} \times \left(\frac{2}{5}\right)^4 = \frac{5}{2} \times \frac{16}{625} = \frac{8}{125}$$ Step 7: Sum of first 10 terms: $$S_{10} = a \times \frac{r^{10} - 1}{r - 1} = \frac{5}{2} \times \frac{\left(\frac{2}{5}\right)^{10} - 1}{\frac{2}{5} - 1}$$ Calculate denominator: $$\frac{2}{5} - 1 = \frac{2}{5} - \frac{5}{5} = -\frac{3}{5}$$ Calculate numerator: $$\left(\frac{2}{5}\right)^{10} = \frac{2^{10}}{5^{10}} = \frac{1024}{9765625}$$ So: $$S_{10} = \frac{5}{2} \times \frac{\frac{1024}{9765625} - 1}{-\frac{3}{5}} = \frac{5}{2} \times \frac{\frac{1024 - 9765625}{9765625}}{-\frac{3}{5}} = \frac{5}{2} \times \frac{-9764601/9765625}{-3/5}$$ Step 8: Simplify: $$= \frac{5}{2} \times \frac{-9764601}{9765625} \times \frac{5}{-3} = \frac{5}{2} \times \frac{5}{3} \times \frac{9764601}{9765625} = \frac{25}{6} \times \frac{9764601}{9765625} \approx 4.1667$$ 4. **The $n^{th}$ term of a GP is given by $a_n = \frac{4}{3^{1-n}}$. Find $S_n$, $S_6$ and $S_{10}$.** Step 1: Rewrite $a_n$: $$a_n = \frac{4}{3^{1-n}} = 4 \times 3^{n-1}$$ Step 2: Identify $a=4$, $r=3$. Step 3: Sum of $n$ terms: $$S_n = a \times \frac{r^n - 1}{r - 1} = 4 \times \frac{3^n - 1}{3 - 1} = 2 \times (3^n - 1)$$ Step 4: Calculate $S_6$: $$S_6 = 2 \times (3^6 - 1) = 2 \times (729 - 1) = 2 \times 728 = 1456$$ Step 5: Calculate $S_{10}$: $$S_{10} = 2 \times (3^{10} - 1) = 2 \times (59049 - 1) = 2 \times 59048 = 118096$$ 5. **Find the sum of all the integers between 10 and 50,000 which have the common ratio 3 and make a GP e.g. $3^3, 3^4, 3^5, ..., 3^9$.** Step 1: Identify terms as powers of 3 between 10 and 50000. Step 2: Powers of 3: $$3^3 = 27, 3^4 = 81, 3^5 = 243, 3^6 = 729, 3^7 = 2187, 3^8 = 6561, 3^9 = 19683, 3^{10} = 59049$$ Step 3: Terms between 10 and 50000 are $3^3$ to $3^9$. Step 4: Sum these terms: $$S = 3^3 + 3^4 + 3^5 + 3^6 + 3^7 + 3^8 + 3^9$$ Step 5: Use sum formula for GP: $$a = 3^3 = 27, r=3, n=7$$ $$S_7 = a \times \frac{r^n - 1}{r - 1} = 27 \times \frac{3^7 - 1}{3 - 1} = 27 \times \frac{2187 - 1}{2} = 27 \times \frac{2186}{2} = 27 \times 1093 = 29511$$ 6. **Calculate how many terms will make the sum equal to 16384 whose $n^{th}$ term is given by $2^{n-1}$.** Step 1: Identify $a=1$ (since $2^{0} = 1$), $r=2$, sum $S_n=16384$. Step 2: Sum formula: $$S_n = a \times \frac{r^n - 1}{r - 1} = \frac{2^n - 1}{2 - 1} = 2^n - 1$$ Step 3: Set equal to 16384: $$2^n - 1 = 16384$$ $$2^n = 16385$$ Step 4: Since $2^{14} = 16384$, $2^{14} = 16384 < 16385 < 2^{15} = 32768$, so $n$ is not an integer. Step 5: Check if sum can be exactly 16384: $$S_{14} = 2^{14} - 1 = 16383$$ $$S_{15} = 2^{15} - 1 = 32767$$ No integer $n$ satisfies $S_n=16384$ exactly. Step 6: If the problem means sum closest to 16384, then $n=14$ gives sum 16383. --- Final answers: 1. $a_8 = \frac{55}{78125}$ 2. (i) $n=11$ (ii) $S_{11} = 12282$ 3. (i) $a=\frac{5}{2}$, $r=\frac{2}{5}$ (ii) $a_n=\frac{5}{2} \times \left(\frac{2}{5}\right)^{n-1}$ (iii) $\frac{5}{2}, 1, \frac{2}{5}, \frac{4}{25}, \frac{8}{125}$ (iv) $S_{10} \approx 4.1667$ 4. $S_n=2(3^n -1)$, $S_6=1456$, $S_{10}=118096$ 5. Sum = 29511 6. No integer $n$ satisfies $S_n=16384$ exactly; closest is $n=14$ with $S_{14}=16383$.