Q1. Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Solution:
t_{11} = 38
a + (11-1)d = 38
a + 10d = 38 … (1)
t_{16} = 38
a + (16-1)d = 38
a + 15d = 73. … (2)
Subtracting (1) from (2):
5d = 35 \implies d = 7.
Substitute d:
a + 70 = 38 \implies a = -32.
t_{31} = a + 30d
t_{31} = -32 + 30(7)
t_{31} = -32 + 210
t_{31} = 178.
\boxed {t_{31} = 178}
Q2. Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.
Solution:
t_7 - t_5 = 12
(a+6d) - (a+4d) = 12
2d = 12 \implies d = 6.
t_3 = 16
a + 2d = 6
Substituting the value of d
a + 12 = 16 \implies a = 4.
Final Answer: The AP is: \boxed{ 4, 10, 16, 22, 28... }
Q3. How many three-digit numbers are divisible by 7?
Solution:
Smallest 3-digit multiple of 7 is 105.
Largest is 994.
Three digit numbers divisible by 7 form the following AP
AP: 105, 112, 119, ... 994
This is an AP with
a=105, d=7, t_n=994.
t_n = 994
a+(n-1)d = 994
105 + (n-1)7 = 994
(n-1)7 = 889
n-1 = 127
n = 128
Final Answer: 128, three digit numbers are divisible by 7.
Q4. How many multiples of 4 lie between 10 and 250?
Solution:
First multiple is 12.
Last multiple before 250 is 248.
Multiples of 4 between 10 and 250 form the following AP
AP: 12, 16, 20, ... 248
This is an AP with
a=12, d=4, t_n=248.
t_{n} = 248
a + (n-1)d = 248
12 + (n-1)4 = 248
4(n-1) = 236
n-1 = 59
n = 60
Final Answer: 60, multiples of 4 lie between 10 and 250.
Q5. Find a GP for which the sum of the first two terms is 4 and the fifth term is 4 times the third term.
Solution:
t_1 = a
t_2 = ar
t_5 = 4(t_3)
ar^4 = 4ar^2
r^2 = 4
r = 2 or -2 (assuming a \ne 0).
Case 1 (r = 2):
a + ar = 4
a(1+2) = 4
3a = 4
a = \frac{4}{3}.
GP is \boxed{ \frac{4}{3}, \frac{8}{3}, \frac{16}{3}... }
Case 2 (r = -2):
a + ar = 4
a(1-2) = 4
-a = 4
a = -4.
GP is \boxed{ -4, 8, -16... }
Q6. Find all possible ways of expressing 100 as the sum of consecutive natural numbers.
Solution:
Let the sum start at k and have m terms:
\frac{m}{2}(2k + m - 1) = 100 \implies m(2k + m - 1) = 200.
Since m and (2k + m - 1) must be factors of 200 with different parities (one odd, one even), we check odd factors of 200, which are 1, 5, 25.
If m = 5,
then 2k+4 = 40 \implies k=18.
Sequence: 18+19+20+21+22 = 100.
If m = 8,
then 2k+7 = 25 \implies k=9.
Sequence: 9+10+11+12+13+14+15+16 = 100.
(m=25 yields a negative starting number, so we ignore it).
There are exactly two ways.
Q7. The number of bacteria… doubles every hour. If there were 30 originally, how many at the end of the 2^{nd} hour, 4^{th} hour and n^{th} hour?
Solution:
Sequence: 30, 60, 120, 240 ...
This forms a GP:
a = 30 (at 0 hours),
r = 2.
End of 2nd hour:
30 \times 2^2 = 120.
End of 4th hour:
30 \times 2^4 = 480.
End of n^{th} hour:
30 \times 2^n
(Note: formula is ar^n because the original 30 is at n=0).
Q8. 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.
Solution:
t_{4} + t_{8} = 24
(a+3d) + (a+7d) = 24
2a + 10d = 24
a + 5d = 12 … (1)
t_{6} +_t{10} = 44
(a+5d)+ (a+9d) = 44
2a + 14d = 44
a + 7d = 22 … (2)
Subtracting: (1) from (2)
2d = 10 \implies d = 5.
Substituting d in eq. (1)
a + 25 = 12 \implies a = -13.
Final Answer: First three terms: \boxed{ -13, -8, -3}.
Q9. Find the smallest value of n such that the sum of the first n natural numbers is greater than 1,000.
Solution:
\frac{n(n+1)}{2} > 1000
n(n+1) > 2000
Since 44 \times 45 = 1980 and
45 \times 46 = 2070,
Final Answer: The smallest n is \boxed{45}.
Q10. Which term of the GP: 2, 8, 32, … is 131072? Write the explicit formula as well as the recursive formula.
Solution:
a = 2, r = 4.
Let ar^{n-1} = 131072
2(4^{n-1}) = 131072
4^{n-1} = 65536.
4^{n-1} = 4^8 ,
n-1 = 8 \implies n = 9
131072 is the \boxed{ \text{9th term} } .
Explicit Rule:
t_n = 2 \times 4^{n-1}.
Recursive:
t_1 = 2, t_n = 4t_{n-1} for n \ge 2.
11. 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.
Solution:
Let the terms be \frac{a}{r}, a, ar.
Product = \frac{a}{r} \times a \times ar = a^3 = -1 \implies a = -1.
Sum = -\frac{1}{r} - 1 - r = \frac{13}{12} \implies r + \frac{1}{r} = -\frac{25}{12}.
Multiplying by 12r gives
12r^2 + 25r + 12 = 0
(4r+3)(3r+4) = 0
So r = -\frac{3}{4} or r = -\frac{4}{3}.
Final Answer: In either case, the terms are: \boxed{ \frac{4}{3}, -1, \frac{3}{4} }.
Q12. 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.
Solution:
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 they are equal, x, y, z form a GP with a common ratio of r^6.
Q13. 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.
Solution:
t_1 = a
t_2 = ar
t_3 = ar^2
t_1 + t_2 + t_3 = 26
a + ar + ar^2 = 26
a(1 + r + r^2) = 26 … (1)
{t_1}^2 + {t_2}^2 + {t_3}^2 = 364
a^2 + (ar)^2 + (ar^2)^2 = 364
a^2(1 + r^2 + r^4) = 364 … (2)
Dividing square of (1) by (2)
\frac{[a(1+r+r^2)]^2}{a^2(1+r^2+r^4)} = \frac{26^2}{364}
\frac{1+r+r^2}{1-r+r^2} = \frac{676}{364}
\frac{1+r+r^2}{1-r+r^2} = \frac{13}{7}.
Cross-multiplying:
7 + 7r + 7r^2 = 13 - 13r + 13r^2
6r^2 - 20r + 6 = 0
3r^2 - 10r + 3 = 0
3r^2 -9r -r +3 = 0
3r (r-3) - 1(r-3) = 0
(3r-1)(r-1) = 0
r = 3 or r = \frac{1}{3}.
Substitute r=3 into sum:
a(1+3+9) = 26 \implies a=2.
Terms are: \boxed{ 2, 6, 18}
Substitute r=\frac{1}{3} into sum:
a(1+\frac{1}{3}+\frac{1}{9})= 26 \implies a= 18
Terms are: \boxed{ 18, 6, 2}.
Q14. Suppose P_1=1, P_2=2 and for n>2, P_n = P_1 + P_2 + ... + P_{n-1} + 1. Find P_1 to P_8. Can you find a simpler recursive formula for P_n ? Can you give an explicit formula?
Solution:
P_3 = P_1 + P_2 + 1
P_3 = 1+2+1 = 4.
P_4 = P_1 + P_2 + P_3 + 1
P_4 = 1+2+4+1 = 8.
P_5 = P_1 + P_2 + P_3 +P_4 + 1
P_5 = 1+2+4+8+1 = 16
P_6 = P_1 + P_2 + P_3 +P_4 + P_5 +1
P_6 = 1+2+4+8+16+1 = 32
P_7 = P_1 + P_2 + P_3 +P_4 + P_5 + P_6 +1
P_7 = 1+2+4+8+16+32+1 = 64
P_8 = P_1 + P_2 + P_3 +P_4 + P_5 + P_6 + P_7 +1
P_8 = 1+2+4+8+16+32+64+1 = 128
Simpler recursive:
Notice each term is double the previous term.
P_n = 2P_{n-1} for n \ge 3.
Explicit:
P_n = 2^{n-1} for n \ge 2, and P_1=1.
Q15. Suppose W_1=1, W_2=2 and for n>2, W_n = W_1 + W_2 + ... + W_{n-2} + 2. Find W_1 to W_8. Do you recognise this sequence?
Solution:
W_3 = W_1 + 2 = 1+2 = 3.
W_4 = W_1 + W_2 + 2 = 1+2+2 = 5.
W_5 = W_1 + W_2 + W_3 + 2 = 1+2+3+2 = 8.
W_6 = W_1 + W_2 + W_3 + W_4 + 2 = 1+2+3+5+2 = 13.
W_7 = W_1 + W_2 + W_3 + W_4 + W_5+ 2 = 1+2+3+5+8+2 = 21.
W_8 = W_1 + W_2 + W_3 + W_4 + W_5 + W_6 + 2 = 1+2+3+5+8+13+2 = 34.
Sequence is: \boxed{ 1, 2, 3, 5, 8, 13, 21, 34}.
This is the famous Virahānka-Fibonacci sequence, where each term is the sum of the two preceding terms!