Verified step-by-step solutions for your class notes.
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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