Arithmetic Progression and Geometric Progression
Part 1: Arithmetic Progression (AP) Problems
Question 1: Find the 10th term of the AP: 2, 7, 12, ...
Solution:
Here, the first term $a = 2$, and the common difference $d = 7 - 2 = 5$.
The $n$th term of an AP is given by the formula $a_n = a + (n - 1)d$.
For the 10th term, substitute $n = 10$:
$a_{10} = 2 + (10 - 1)5 = 2 + 9(5) = 2 + 45 = 47$.
Therefore, the 10th term is 47.
Question 2: Which term of the AP: 21, 18, 15, ... is -81?
Solution:
Here, $a = 21$ and $d = 18 - 21 = -3$. Let the $n$th term be $a_n = -81$.
Using the formula $a_n = a + (n - 1)d$:
$-81 = 21 + (n - 1)(-3)$
$-81 - 21 = -3(n - 1)$
$-102 = -3(n - 1)$
$n - 1 = \frac{-102}{-3} = 34$
$n = 35$
Therefore, the 35th term is -81.
Question 3: Determine the AP whose 3rd term is 5 and the 7th term is 9.
Solution:
We are given $a_3 = 5$ and $a_7 = 9$.
Using $a_n = a + (n-1)d$, we get two equations:
$a + 2d = 5$ --- (Equation 1)
$a + 6d = 9$ --- (Equation 2)
Subtracting Equation 1 from Equation 2 gives:
$(a + 6d) - (a + 2d) = 9 - 5$
$4d = 4 \Rightarrow d = 1$
Substituting $d = 1$ into Equation 1:
$a + 2(1) = 5 \Rightarrow a = 3$
The required AP is $a, a+d, a+2d, \dots$ which is 3, 4, 5, 6, ...
Question 4: Find the sum of the first 22 terms of the AP: 8, 3, -2, ...
Solution:
Here, $a = 8$, $d = 3 - 8 = -5$, and $n = 22$.
The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
$S_{22} = \frac{22}{2}[2(8) + (22 - 1)(-5)]$
$S_{22} = 11[16 + 21(-5)]$
$S_{22} = 11[16 - 105]$
$S_{22} = 11[-89] = -979$
The sum of the first 22 terms is -979.
Question 5: If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Solution:
Here, $S_{14} = 1050$, $n = 14$, and $a = 10$.
Using $S_n = \frac{n}{2}[2a + (n - 1)d]$:
$1050 = \frac{14}{2}[2(10) + (14 - 1)d]$
$1050 = 7[20 + 13d]$
$\frac{1050}{7} = 20 + 13d \Rightarrow 150 = 20 + 13d$
$130 = 13d \Rightarrow d = 10$
Now, to find the 20th term ($a_{20}$):
$a_{20} = a + 19d = 10 + 19(10) = 10 + 190 = 200$.
The 20th term is 200.
Question 6: How many multiples of 4 lie between 10 and 250?
Solution:
The first multiple of 4 strictly after 10 is 12.
The last multiple of 4 before 250 is 248.
This forms an AP: 12, 16, 20, ..., 248 where $a = 12$, $d = 4$, and $a_n = 248$.
$a_n = a + (n - 1)d$
$248 = 12 + (n - 1)4$
$236 = 4(n - 1)$
$n - 1 = 59 \Rightarrow n = 60$
There are 60 multiples of 4 between 10 and 250.
Question 7: The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
We are given:
$a_4 + a_8 = 24 \Rightarrow (a + 3d) + (a + 7d) = 24 \Rightarrow 2a + 10d = 24 \Rightarrow a + 5d = 12$ --- (Eq 1)
$a_6 + a_{10} = 44 \Rightarrow (a + 5d) + (a + 9d) = 44 \Rightarrow 2a + 14d = 44 \Rightarrow a + 7d = 22$ --- (Eq 2)
Subtracting Eq 1 from Eq 2:
$(a + 7d) - (a + 5d) = 22 - 12$
$2d = 10 \Rightarrow d = 5$
Substitute $d = 5$ into Eq 1:
$a + 5(5) = 12 \Rightarrow a + 25 = 12 \Rightarrow a = -13$
The first three terms are $a, a+d, a+2d$ which equates to -13, -8, -3.
Question 8: Find the sum of the first 40 positive integers divisible by 6.
Solution:
The positive integers divisible by 6 form an AP: 6, 12, 18, 24, ...
Here, $a = 6$, $d = 6$, and $n = 40$.
Using $S_n = \frac{n}{2}[2a + (n - 1)d]$:
$S_{40} = \frac{40}{2}[2(6) + (40 - 1)6]$
$S_{40} = 20[12 + 39(6)]$
$S_{40} = 20[12 + 234] = 20[246]$
$S_{40} = 4920$
The sum is 4920.
Question 9: Find the middle term of the AP: 6, 13, 20, ..., 216.
Solution:
First, find the total number of terms ($n$). Here, $a = 6$, $d = 7$, and $a_n = 216$.
$a_n = a + (n - 1)d$
$216 = 6 + (n - 1)7$
$210 = 7(n - 1) \Rightarrow n - 1 = 30 \Rightarrow n = 31$
Since $n$ is odd, there is a single middle term given by the $\frac{n + 1}{2}$th term.
Middle term position = $\frac{31 + 1}{2} = 16$. We need to find $a_{16}$.
$a_{16} = a + 15d = 6 + 15(7) = 6 + 105 = 111$.
The middle term is 111.
Question 10: In an AP, given $a = 5$, $d = 3$, $a_n = 50$, find $n$ and $S_n$.
Solution:
First, find $n$ using the $n$th term formula:
$a_n = a + (n - 1)d$
$50 = 5 + (n - 1)3$
$45 = 3(n - 1) \Rightarrow n - 1 = 15 \Rightarrow n = 16$
Next, find $S_{16}$ using $S_n = \frac{n}{2}(a + a_n)$:
$S_{16} = \frac{16}{2}(5 + 50)$
$S_{16} = 8(55) = 440$
Therefore, $n = 16$ and $S_{16} = 440$.
Part 2: Geometric Progression (GP) Problems
Question 1: Find the 7th term of the GP: 3, 6, 12, ...
Solution:
Here, the first term $a = 3$, and the common ratio $r = \frac{6}{3} = 2$.
The $n$th term of a GP is given by $a_n = a r^{n-1}$.
For the 7th term ($n = 7$):
$a_7 = 3(2)^{7-1} = 3(2^6)$
$a_7 = 3(64) = 192$.
The 7th term is 192.
Question 2: Which term of the GP: 2, 8, 32, ... is 131072?
Solution:
Here, $a = 2$, $r = \frac{8}{2} = 4$, and $a_n = 131072$.
Using $a_n = a r^{n-1}$:
$131072 = 2(4)^{n-1}$
$\frac{131072}{2} = 4^{n-1} \Rightarrow 65536 = 4^{n-1}$
We know that $4^8 = 65536$, therefore:
$4^8 = 4^{n-1} \Rightarrow n - 1 = 8 \Rightarrow n = 9$.
It is the 9th term.
Question 3: Determine the GP whose 3rd term is 24 and 6th term is 192.
Solution:
Given $a_3 = 24 \Rightarrow a r^2 = 24$ --- (Eq 1)
And $a_6 = 192 \Rightarrow a r^5 = 192$ --- (Eq 2)
Dividing Eq 2 by Eq 1:
$\frac{a r^5}{a r^2} = \frac{192}{24}$
$r^3 = 8 \Rightarrow r = 2$
Substituting $r = 2$ into Eq 1:
$a(2^2) = 24 \Rightarrow 4a = 24 \Rightarrow a = 6$
The GP is $a, ar, ar^2, \dots$ which is 6, 12, 24, 48, ...
Question 4: Find the sum of the first 5 terms of the GP: 1, 3, 9, ...
Solution:
Here, $a = 1$, $r = 3$ (where $r > 1$), and $n = 5$.
The sum formula for $r > 1$ is $S_n = \frac{a(r^n - 1)}{r - 1}$.
$S_5 = \frac{1(3^5 - 1)}{3 - 1}$
$S_5 = \frac{243 - 1}{2} = \frac{242}{2} = 121$.
The sum of the first 5 terms is 121.
Question 5: The 4th term of a GP is the square of its second term, and the first term is -3. Determine its 7th term.
Solution:
Given $a = -3$ and $a_4 = (a_2)^2$.
$a r^3 = (a r)^2$
$a r^3 = a^2 r^2$
Dividing both sides by $ar^2$ (assuming $r \neq 0$):
$r = a$
Since $a = -3$, we have $r = -3$.
We need to find the 7th term ($a_7$):
$a_7 = a r^6 = -3(-3)^6 = -3(729) = -2187$.
The 7th term is -2187.
Question 6: Find the sum to infinity of the GP: 1, 1/2, 1/4, ...
Solution:
Here, $a = 1$ and $r = \frac{1/2}{1} = \frac{1}{2}$.
Since $|r| < 1$, the sum to infinity formula is $S_\infty = \frac{a}{1 - r}$.
$S_\infty = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$.
The sum to infinity is 2.
Question 7: How many terms of the GP 3, 3/2, 3/4, ... are needed to give the sum 3069/512?
Solution:
Here, $a = 3$, $r = \frac{1}{2}$, and $S_n = \frac{3069}{512}$.
Using $S_n = \frac{a(1 - r^n)}{1 - r}$ (since $r < 1$):
$\frac{3069}{512} = \frac{3(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}}$
$\frac{3069}{512} = \frac{3(1 - (\frac{1}{2})^n)}{\frac{1}{2}}$
$\frac{3069}{512} = 6(1 - (\frac{1}{2})^n)$
Dividing by 6:
$\frac{3069}{3072} = 1 - (\frac{1}{2})^n$
$(\frac{1}{2})^n = 1 - \frac{3069}{3072} = \frac{3}{3072} = \frac{1}{1024}$
We know that $2^{10} = 1024$, so $(\frac{1}{2})^n = (\frac{1}{2})^{10} \Rightarrow n = 10$.
10 terms are needed.
Question 8: If the 3rd, 8th and 13th terms of a GP are $x$, $y$ and $z$, respectively. Prove that $y^2 = xz$.
Solution:
Let the first term of the GP be $a$ and common ratio be $r$.
Then, $x = a_3 = a r^2$
$y = a_8 = a r^7$
$z = a_{13} = a r^{12}$
Now, evaluate $y^2$:
$y^2 = (a r^7)^2 = a^2 r^{14}$
Next, evaluate $xz$:
$xz = (a r^2)(a r^{12}) = a^2 r^{2+12} = a^2 r^{14}$
Since both $y^2$ and $xz$ equal $a^2 r^{14}$, it is proven that $y^2 = xz$.
Question 9: Find the sum of the sequence 7, 77, 777, 7777, ... to $n$ terms.
Solution:
Let $S_n = 7 + 77 + 777 + \dots \text{ to } n \text{ terms}$
Take 7 as a common factor:
$S_n = 7(1 + 11 + 111 + \dots \text{ to } n \text{ terms})$
Multiply and divide by 9:
$S_n = \frac{7}{9}(9 + 99 + 999 + \dots \text{ to } n \text{ terms})$
Rewrite each term using powers of 10:
$S_n = \frac{7}{9}[(10 - 1) + (10^2 - 1) + (10^3 - 1) + \dots + (10^n - 1)]$
Group the powers of 10 and the 1s:
$S_n = \frac{7}{9}[(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots n \text{ times})]$
The first part is a GP with $a = 10$, $r = 10$. Applying the sum formula:
$S_n = \frac{7}{9} \left[ \frac{10(10^n - 1)}{10 - 1} - n \right] = \frac{7}{9} \left[ \frac{10(10^n - 1)}{9} - n \right]$.
Question 10: A person has 2 parents, 4 grandparents, 8 great-grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own.
Solution:
The number of ancestors in each generation forms a GP: 2, 4, 8, ...
Here, $a = 2$, $r = 2$, and we need the sum for $n = 10$ generations.
Using the formula $S_n = \frac{a(r^n - 1)}{r - 1}$:
$S_{10} = \frac{2(2^{10} - 1)}{2 - 1}$
$S_{10} = 2(1024 - 1)$
$S_{10} = 2(1023) = 2046$.
The total number of ancestors is 2046.