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Find the first five terms of the sequence in which the $n^{th}$ term is given by (i) $t_n = 3n - 4$, (ii) $t_n = 2 - 5n$, and (iii) $t_n = n^2 - 2n + 3$ for $n \ge 1$.
Answer:
Find the $10^{th}$ and $15^{th}$ terms of the sequence $t_n = 5n - 3$ for $n \ge 1$.
Answer:
For the $10^{th}$ term, substitute $n = 10$: $t_{10} = 5(10) - 3 = 50 - 3 = \mathbf{47}$.
For the $15^{th}$ term, substitute $n = 15$: $t_{15} = 5(15) - 3 = 75 - 3 = \mathbf{72}$.
Determine whether 97 and 172 are terms of the sequence $t_n = 5n - 3$ for $n \ge 1$.
Answer:
Set $5n - 3 = 97 \implies 5n = 100 \implies n = 20$. Since 20 is a natural number, 97 is a term (the $20^{th}$ term).
Set $5n - 3 = 172 \implies 5n = 175 \implies n = 35$. Since 35 is a natural number, 172 is a term (the $35^{th}$ term).
Which term of the sequence $t_n = 5n - 3$ for $n \ge 1$ is 607?
Answer:
Set $5n - 3 = 607 \implies 5n = 610 \implies n = 122$.
607 is the $122^{nd}$ term of the sequence.
A sequence is given by the recursive rule $t_1 = -5, t_{n+1} = t_n + 3$ for $n \ge 1$. Find the first five terms of the sequence. Is 52 a term of this sequence? If so, which term is it?
Answer:
The first five terms are: -5, -2, 1, 4, 7.
The explicit rule for this AP is $t_n = -5 + (n-1)3 = 3n - 8$.
Set $3n - 8 = 52 \implies 3n = 60 \implies n = 20$.
Yes, 52 is a term of this sequence; it is the $20^{th}$ term.
Let $T_1 = 1, T_2 = 2, T_3 = 4$, and $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ for $n \ge 4$. Find $T_4, T_5, T_6, T_7$, and $T_8$.
Answer:
$T_4 = 1 + 2 + 4 = \mathbf{7}$
$T_5 = 2 + 4 + 7 = \mathbf{13}$
$T_6 = 4 + 7 + 13 = \mathbf{24}$
$T_7 = 7 + 13 + 24 = \mathbf{44}$
$T_8 = 13 + 24 + 44 = \mathbf{81}$
Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ...
Answer:
First term $a = 3$, common difference $d = 5$.
$10^{th}$ term: $t_{10} = a + 9d = 3 + 9(5) = 3 + 45 = \mathbf{48}$.
$26^{th}$ term: $t_{26} = a + 25d = 3 + 25(5) = 3 + 125 = \mathbf{128}$.
Which term of the AP: 21, 18, 15, ... is -81? Also, is 0 a term of this AP? Give reasons for your answer.
Answer:
First term $a = 21$, common difference $d = -3$.
Let $t_n = -81 \implies 21 + (n-1)(-3) = -81 \implies -3(n-1) = -102 \implies n-1 = 34 \implies n = 35$. So, -81 is the $35^{th}$ term.
Let $t_n = 0 \implies 21 + (n-1)(-3) = 0 \implies -3(n-1) = -21 \implies n-1 = 7 \implies n = 8$. Yes, 0 is a term; it is the $8^{th}$ term because $n$ is a positive integer.
Find the $n^{th}$ term of the AP: 11, 8, 5, 2... Write the recursive rule for this AP.
Answer:
First term $a = 11$, common difference $d = -3$.
$n^{th}$ term (explicit): $t_n = 11 + (n-1)(-3) = \mathbf{14 - 3n}$.
Recursive rule: $\mathbf{t_1 = 11, t_n = t_{n-1} - 3}$ for $n \ge 2$.
An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
Answer:
$t_3 = a + 2d = 12$
$t_{50} = a + 49d = 106$
Subtracting the first equation from the second gives $47d = 94 \implies d = 2$.
Substitute $d = 2$ back: $a + 4 = 12 \implies a = 8$.
The $29^{th}$ term is $t_{29} = a + 28d = 8 + 28(2) = 8 + 56 = \mathbf{64}$.
How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
Answer:
The 2-digit numbers divisible by 3 form an AP: 12, 15, 18, ..., 99.
Here, $a = 12, d = 3$. Let $t_n = 99 \implies 12 + (n-1)3 = 99 \implies 3(n-1) = 87 \implies n = 30$. There are 30 such numbers.
Sum = $\frac{n}{2}(a + l) = \frac{30}{2}(12 + 99) = 15 \times 111 = \mathbf{1665}$.
Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?
Answer:
This forms an AP with $a = 500,000$, $d = 20,000$, and $t_n = 700,000$.
$500,000 + (n-1)20,000 = 700,000 \implies 20,000(n-1) = 200,000 \implies n-1 = 10 \implies n = 11$.
His income reached ₹7,00,000 in the $11^{th}$ year, which means it happened after 10 years of receiving increments.
A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?
Answer:
The total number of marbles is the sum of the first 25 natural numbers.
$S_{25} = \frac{n(n+1)}{2} = \frac{25 \times 26}{2} = 25 \times 13 = \mathbf{325}$ marbles.
Find the 12th term of a GP with common ratio 2, whose 8th term is 192.
Answer:
$t_8 = ar^7 = 192$. Since $r = 2$, we have $a(2^7) = 192 \implies 128a = 192 \implies a = 1.5$.
$t_{12} = ar^{11} = 1.5 \times 2^{11} = 1.5 \times 2048 = \mathbf{3072}$.
(Alternatively, $t_{12} = t_8 \times r^4 = 192 \times 16 = 3072$).
Find the $10^{th}$ and $n^{th}$ terms of the GP: 5, 25, 125, ...
Answer:
First term $a = 5$, common ratio $r = 5$.
$10^{th}$ term: $t_{10} = ar^9 = 5(5^9) = \mathbf{5^{10}}$.
$n^{th}$ term: $t_n = ar^{n-1} = 5(5^{n-1}) = \mathbf{5^n}$.
A sequence is given by the recursive rule $t_1 = 2, t_{n+1} = 3t_n - 2$ for $n \ge 1$. Which term of the sequence is 730?
Answer:
Let's generate the sequence:
$t_1 = 2$
$t_2 = 3(2) - 2 = 4$
$t_3 = 3(4) - 2 = 10$
$t_4 = 3(10) - 2 = 28$
$t_5 = 3(28) - 2 = 82$
$t_6 = 3(82) - 2 = 244$
$t_7 = 3(244) - 2 = 730$
So, 730 is the $7^{th}$ term.
Which term of the GP: 2, 6, 18, ... is 4374? Write the explicit formula as well as the recursive formula for the $n^{th}$ term.
Answer:
Here $a = 2, r = 3$. Set $2(3^{n-1}) = 4374 \implies 3^{n-1} = 2187$. Since $3^7 = 2187$, $n-1 = 7 \implies n = 8$. It is the $8^{th}$ term.
Explicit formula: $\mathbf{t_n = 2 \times 3^{n-1}}$.
Recursive formula: $\mathbf{t_1 = 2, t_n = 3t_{n-1}}$ for $n \ge 2$.
A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues bouncing in this way.
(i) What height does the ball reach after the 5th bounce?
(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the $6^{th}$ time?
Answer:
(i) Height after 5th bounce = $80 \times (0.6)^5 = 80 \times 0.07776 = \mathbf{6.2208}$ metres.
(ii) The total distance includes the initial drop (80m) plus the up-and-down distance of the 5 bounces.
Total Distance = $80 + 2 \times 80(0.6^1 + 0.6^2 + 0.6^3 + 0.6^4 + 0.6^5)$.
The geometric series sum is $0.6 \frac{1 - 0.6^5}{1 - 0.6} = 0.6 \frac{1 - 0.07776}{0.4} = 1.5 \times 0.92224 = 1.38336$.
Total Distance = $80 + 160(1.38336) = 80 + 221.3376 = \mathbf{301.3376}$ metres.
Which term of the sequence $2, \sqrt{2}, 1, \dots$ is $1/128$?
Answer:
This is a GP with $a = 2$ and $r = \frac{1}{\sqrt{2}}$.
Set $t_n = 2 \left(\frac{1}{\sqrt{2}}\right)^{n-1} = \frac{1}{128}$.
$\left(\frac{1}{\sqrt{2}}\right)^{n-1} = \frac{1}{256}$.
Since $256 = 2^8 = (\sqrt{2})^{16}$, we have $\left(\frac{1}{\sqrt{2}}\right)^{n-1} = \left(\frac{1}{\sqrt{2}}\right)^{16}$.
Therefore, $n-1 = 16 \implies n = 17$. It is the $17^{th}$ term.
Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 is a square sheet of paper. To construct Stage 1, each side of the square is trisected... The centre square is then removed and the 8 smaller squares are retained...
(i) How many red squares are there in Stages 0 to 3?
(ii) Can you predict the number of red squares in Stages 4 and 5?
(iii) Can you find a rule for the number of red squares at the $n^{th}$ stage? Write the explicit formula as well as the recursive formula.
(iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the explicit and recursive formula for the area of the red region at the $n^{th}$ stage? What happens to this area as n increases?
Answer:
(i) Stage 0: 1, Stage 1: 8, Stage 2: 64, Stage 3: 512.
(ii) Stage 4: $8^4 = \mathbf{4096}$, Stage 5: $8^5 = \mathbf{32768}$.
(iii) Explicit formula: $\mathbf{t_n = 8^n}$. Recursive formula: $\mathbf{t_0 = 1, t_n = 8t_{n-1}}$.
(iv) In each stage, the area is multiplied by $\frac{8}{9}$ because 1 out of every 9 squares is removed. Explicit formula for area: $\mathbf{A_n = (\frac{8}{9})^n}$. Recursive formula: $\mathbf{A_0 = 1, A_n = \frac{8}{9} A_{n-1}}$. As n increases, the area gets closer and closer to 0.
Find the $31^{st}$ term of an AP whose $11^{th}$ term is 38 and $16^{th}$ term is 73.
Answer:
$a + 10d = 38$ and $a + 15d = 73$.
Subtracting gives $5d = 35 \implies d = 7$.
Substitute back: $a + 70 = 38 \implies a = -32$.
$31^{st}$ term = $a + 30d = -32 + 30(7) = -32 + 210 = \mathbf{178}$.
Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.
Answer:
$t_3 = a + 2d = 16$.
$t_7 - t_5 = (a + 6d) - (a + 4d) = 2d = 12 \implies d = 6$.
Substitute $d = 6$: $a + 12 = 16 \implies a = 4$.
The AP is 4, 10, 16, 22, 28, ...
How many three-digit numbers are divisible by 7?
Answer:
The first three-digit number divisible by 7 is 105. The last is 994.
This forms an AP with $a = 105, d = 7, t_n = 994$.
$105 + (n-1)7 = 994 \implies 7(n-1) = 889 \implies n-1 = 127 \implies n = \mathbf{128}$.
There are 128 three-digit numbers divisible by 7.
How many multiples of 4 lie between 10 and 250?
Answer:
The first multiple of 4 after 10 is 12. The last multiple before 250 is 248.
This forms an AP: 12, 16, 20, ..., 248. Here $a = 12, d = 4, t_n = 248$.
$12 + (n-1)4 = 248 \implies 4(n-1) = 236 \implies n-1 = 59 \implies n = \mathbf{60}$.
There are 60 multiples.
Find a GP for which the sum of the first two terms is 4 and the fifth term is 4 times the third term.
Answer:
Condition 1: $a + ar = 4 \implies a(1+r) = 4$.
Condition 2: $ar^4 = 4ar^2 \implies r^2 = 4 \implies r = \pm 2$.
Case 1 ($r = 2$): $a(1+2) = 4 \implies 3a = 4 \implies a = 4/3$. The GP is 4/3, 8/3, 16/3, 32/3...
Case 2 ($r = -2$): $a(1-2) = 4 \implies -a = 4 \implies a = -4$. The GP is -4, 8, -16, 32...
Find all possible ways of expressing 100 as the sum of consecutive natural numbers.
Answer:
The sum of $k$ consecutive integers starting from $a$ is $\frac{k(2a + k - 1)}{2} = 100 \implies k(2a + k - 1) = 200$.
The factors $k$ and $(2a + k - 1)$ must have opposite parity (one odd, one even). The odd factors of 200 are 1, 5, 25.
If odd factor is 5: $k=5, 2a+4=40 \implies 2a=36 \implies a=18$. The sequence is 18, 19, 20, 21, 22.
If odd factor is 25: $2a+k-1 = 25$ and $k=8 \implies 2a+7=25 \implies a=9$. The sequence is 9, 10, 11, 12, 13, 14, 15, 16.
(Other combinations yield non-natural starting numbers).
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the $2^{nd}$ hour, $4^{th}$ hour and $n^{th}$ hour?
Answer:
This forms a GP: Initial = 30. End of 1st hour = 60.
End of $2^{nd}$ hour: $30 \times 2^2 = \mathbf{120}$.
End of $4^{th}$ hour: $30 \times 2^4 = \mathbf{480}$.
End of $n^{th}$ hour: $\mathbf{30 \times 2^n}$.
The sum of the $4^{th}$ and $8^{th}$ terms of an AP is 24 and the sum of the $6^{th}$ and $10^{th}$ terms is 44. Find the first three terms of the AP.
Answer:
$t_4 + t_8 = 24 \implies (a+3d) + (a+7d) = 2a + 10d = 24 \implies a + 5d = 12$.
$t_6 + t_{10} = 44 \implies (a+5d) + (a+9d) = 2a + 14d = 44 \implies a + 7d = 22$.
Subtracting gives $2d = 10 \implies d = 5$.
Substitute back: $a + 25 = 12 \implies a = -13$.
The first three terms are -13, -8, -3.
Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.
Answer:
We need $\frac{n(n+1)}{2} > 1000 \implies n(n+1) > 2000$.
We know $44 \times 45 = 1980$ and $45 \times 46 = 2070$.
The smallest value of $n$ is 45.
Which term of the GP: 2, 8, 32, ... is 131072? Write the explicit formula as well as the recursive formula for the $n^{th}$ term.
Answer:
Here $a=2, r=4$. Set $2(4^{n-1}) = 131072 \implies 4^{n-1} = 65536 = 4^8$.
$n-1 = 8 \implies n=9$. It is the $9^{th}$ term.
Explicit formula: $\mathbf{t_n = 2 \times 4^{n-1}}$.
Recursive formula: $\mathbf{t_1 = 2, t_n = 4t_{n-1}}$ for $n \ge 2$.
The sum of the first three terms of a GP is 13/12 and their product is -1. Find the common ratio and the terms.
Answer:
Let terms be $a/r, a, ar$. Product = $a^3 = -1 \implies a = -1$.
Sum = $-1/r - 1 - r = 13/12 \implies \frac{-(r^2 + r + 1)}{r} = \frac{13}{12} \implies 12r^2 + 25r + 12 = 0$.
Factorizing: $(3r+4)(4r+3) = 0 \implies r = -4/3$ or $r = -3/4$.
The terms are 3/4, -1, 4/3 (or in reverse order).
If the $4^{th}$, $10^{th}$ and $16^{th}$ terms of a GP are x, y and z respectively, prove that x, y, z are in GP.
Answer:
Let the GP be $A, AR, AR^2 \dots$
$x = AR^3$, $y = AR^9$, $z = AR^{15}$.
Check if $y^2 = xz$: $y^2 = (AR^9)^2 = A^2R^{18}$.
$xz = (AR^3)(AR^{15}) = A^2R^{18}$.
Since $y^2 = xz$, the terms $x, y, z$ are in GP.
The sum of the first three terms of a geometric progression is 26, and the sum of their squares is 364. Find the terms of the GP.
Answer:
$a(1+r+r^2) = 26$ and $a^2(1+r^2+r^4) = 364$.
We know $1+r^2+r^4 = (1+r+r^2)(1-r+r^2)$.
Divide the squares sum by the square of the first sum:
$\frac{a^2(1+r+r^2)(1-r+r^2)}{a^2(1+r+r^2)^2} = \frac{364}{26^2} \implies \frac{1-r+r^2}{1+r+r^2} = \frac{364}{676} = \frac{7}{13}$.
Cross-multiplying gives $13 - 13r + 13r^2 = 7 + 7r + 7r^2 \implies 6r^2 - 20r + 6 = 0 \implies 3r^2 - 10r + 3 = 0$.
Roots are $r = 3$ or $r = 1/3$.
If $r = 3$, $a(1+3+9) = 26 \implies 13a = 26 \implies a = 2$.
The terms are 2, 6, 18.
Suppose $P_1 = 1$, $P_2 = 2$ and for $n > 2$, $P_n = P_1 + P_2 + \dots + P_{n-1} + 1$. Find the values of $P_1, P_2, \dots, P_8$. Can you find a simpler recursive formula for $P_n$? Can you give an explicit formula?
Answer:
$P_1 = 1, P_2 = 2$
$P_3 = 1 + 2 + 1 = 4$
$P_4 = 1 + 2 + 4 + 1 = 8$
$P_5 = 16, P_6 = 32, P_7 = 64, P_8 = 128$.
Simpler recursive formula: $\mathbf{P_n = 2P_{n-1}}$ for $n \ge 3$.
Explicit formula: $\mathbf{P_n = 2^{n-1}}$.
Suppose $W_1 = 1, W_2 = 2$ and for $n > 2$, $W_n = W_1 + W_2 + \dots + W_{n-2} + 2$. Find the values of $W_1, W_2, \dots, W_8$. Do you recognise this sequence?
Answer:
$W_1 = 1, W_2 = 2$
$W_3 = W_1 + 2 = 1 + 2 = 3$
$W_4 = W_1 + W_2 + 2 = 1 + 2 + 2 = 5$
$W_5 = 1 + 2 + 3 + 2 = 8$
$W_6 = 1 + 2 + 3 + 5 + 2 = 13$
$W_7 = 21, W_8 = 34$.
This is the famous Virahānka-Fibonacci sequence.
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