AP, GP & HP
Arithmetic Progression(AP):-
AP is a series of numbers where the difference between any two consecutive terms is constant. This constant difference is called the common difference.
Ex:-
General form: a, a + d, a + 2d, a + 3d, ……..till n term
Here a is the first term
& d is the common difference.
⟶ 1, 3, 5, 7, 9, …………. ⟹ a = 1 cd = 2
⟶ -4, -3, -2, -1, 0, 1, 2, 3, ………… ⟹ a = -4 cd = 1
⟶ 2, 5, 8, 11, 14, 17, …….. ⟹ a = 2 cd = 3
● Sum of first n term of AP
● Sum of terms equidistant from both end is always the same and this sum is equal to the sum of the first term and the last term of the AP.
● If the same quantity be added to, or subtracted from, all the terms of an AP, the resulting terms will also form an AP, but with ther same common difference as before.
● If all terms of an AP be multiplied or divided by the same quantity, the resulting terms will also form an AP, but with a new common difference, which will be the multiplication or division of the old common difference as the case may be
● If a₁, a₂, a₃, a₄, …………, aₙ are in AP then
arithmetic mean = \(\frac{{{a_1} + {a_2} + {a_3} + ………. + {a_n}}}{n}\)
● Arithmetic mean of two numbers a & b = \(\frac{{a + b}}{2}\)
● The arithmetic mean (AM) of the AP which has odd number of terms is the middle most term of the AP itself.
● If number of terms ‘n’ is odd then middle term is \({\left( {\frac{{n + 1}}{2}} \right)^{th}}\) term
● If number of terms ‘n’ is even then there are two middle term \({\left( {\frac{n}{2}} \right)^{th}}\) & \({\left( {\frac{n}{2} + 1} \right)^{th}}\) term
● Arithmetic mean of an AP which has even number of terms is \(\frac{{sum\;of\;two\;middle\;terms}}{2}\) or \(\frac{{sum\;of\;terms\;equidistant\;from\;both\;ends}}{2}\)
●⟶ If we are to assume terms of an AP then assume them as follows:
● 3 terms: a – d, a, a + d
● 4 terms: a-3d, a – d, a + d, a + 3d
● 5 terms: a – 2d, a – d, a, a + d, a + 2d
● 6 terms: a – 5d, a – 3d, a – d, a + d, a + 3d, a + 5d
Geometric Progression(GP)
GP:-
A sequence of numbers where each term after the first is found by multiplying the previous term by a constant called the common ratio.
General form: a, ar, ar², ar³, ar⁴, ………….
First term = a
common ratio = r
Ex:-
⟶ 2, 4, 8, 16, 32, ……….
⟶ 3, 12, 48, …………
⟶ 3, 9, 27, 81, …………..
Sum of Infinite Geometric Progression:-
● If all the terms of a GP be multiplied or divided by the same quantity, the resulting term will also form a GP with the same common ratio as before.
● If a₁, a₂, a₃, …………, aₙ are in GP then
Geometric mean = \({\left( {{a_1}.{a_2}.{a_3}………..{a_n}} \right)^{\frac{1}{n}}}\)
= \(\sqrt[n]{{\left( {{a_1}.{a_2}.{a_3}………..{a_n}} \right)}}\)
● If a & b are in GP then
Geometric mean = \(\sqrt {a.b} \)
● In a finite GP, the product of terms equidistant from both the ends is the same and equal to the product of the first and last term.
● A GP can not have any ‘0’ term, as multiplying by 0 would break the constant ratio rule.
● Reciprocals of all the terms in a GP also form a GP with the reciprocal of the original common ratio.
● If each term of a GP is raised to the same non-zero power, the resulting sequence is also a GP, with the common ratio raised to the same power.
● If three terms a, b, c are in GP then
b² = a.c
● Assuming terms of a GP
3 terms: \(\frac{a}{r},\;a,\;ar\)
4 terms: \(\frac{a}{{{r^2}}},\;\frac{a}{r},\;ar,\;a{r^2}\)
5 terms: \(\frac{a}{{{r^2}}},\;\frac{a}{r},\;a,\;ar,\;a{r^2}\)
6 terms: \(\frac{a}{{{r^5}}},\;\frac{a}{{{r^3}}},\;\frac{a}{r},\;ar,\;a{r^3},\;a{r^5}\)
● square of any term in a GP is equal to the product of terms just before it and just after it
aₙ² = aₙ₋₁ × aₙ₊₁
Harmonic Progression
HP:-
An HP is a sequence of real numbers where each term is the reciprocal of a term in an AP, Provided the AP does not contain zero.
i.e. reciprocal of terms of an HP will form an AP.
\(\frac{1}{{{a_1}}},\;\frac{1}{{{a_2}}},\;\frac{1}{{{a_3}}},\;\frac{1}{{{a_4}}},\;…………..\)
where a₁, a₂, a₃, a₄, ………. will form an AP
● ⟶ \(\frac{1}{2},\;\frac{1}{4},\;\frac{1}{6},\;\frac{1}{8},\;…………..\) is an HP
● ⟶ \(\frac{1}{6},\;\frac{1}{9},\;\frac{1}{12},\;\frac{1}{15},\;…………..\) is an HP
● Harmonic mean of n terms of an HP
= \(\frac{n}{{\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \frac{1}{{{a_3}}} + ………. + \frac{1}{{{a_n}}}}}\)
● If two terms a & b are in HP
then Harmonic mean = \(\frac{2}{{\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}}}}\) = \(\frac{{2ab}}{{a + b}}\)
● If three terms a, b & c are in HP then
Harmonic mean = \(\frac{{3ab}}{{ab + bc + ca}}\)
● Relation between AP, GP and HP
For any two number
Arithmetic mean ⟶ AM
Geometric mean ⟶ GM
Harmonic mean ⟶ HM
Example
(Q) A series is given below:
6, 9, 12, 15, ………….
(i) Find its 20ᵗʰ term
(ii) Find the sum of 20 terms
Sol:-
Given series is AP with first term a = 6 & common difference d 9 – 6 = 3
(i) ∴ T₂₀ = 6 + (20 – 1).3 = 6 + 57 = 63 Answer
(ii) & S₂₀ = \(\frac{{20}}{2}[6 + 63]\) = 690 Answer
(Q). Find the sum of 10 terms of following AP
-9, -7, -5, -3, …………..
Sol:-
S₁₀ = \(\frac{{10}}{2}[2 \times ( – 9) + (10 – 1).(2)]\)
= 5[-18 + 18]
= 0 Answer
(Q). If T₁ + T₅ +T₁₀ + T₁₅ + T₂₀ +T₂₄ = 555, WHERE T₁ is first term of AP. Find the sum of its 24 terms.
Sol:-
T₁ = a
T₅ = a + 4d
T₁₀ = a + 9d
T₁₅ = a + 14d
T₂₀ = a + 19d
T₂₄ = a + 23d
Sum = 6a + 69d = 555
⟹ 2a + 23d = 185 ……………… (i)
Now sum of first 24 terms
S₂₄ = \(\frac{{24}}{2}\)[a + (24 – 1).d]
= 12[2a + 23d]
put value from equation (i)
S₂₄ = 12[185]
= 2220 Answer
(Q). Find the sum of series till 100 terms:
1, 5, 7, 6, 13, 7, 19, 8, 25, 9, ………..
Sol:-
Method(1):
The above series has two separate AP’s:
Iˢᵗ AP:- 1, 7, 13, 19, 25, ………. till 50 terms
2ⁿᵈ AP:- 5, 6, 7, 8, 9, ………. till 50 terms
Sum of Iˢᵗ AP = \(\frac{{50}}{2}\)[2×1 + (50 – 1)×6]
= 7400
Sum of 2ⁿᵈ AP = \(\frac{{50}}{2}\)[2×5 + (50 – 1)×1]
= 1475
∴ Sum of given series = Sum of Iˢᵗ AP + Sum of 2ⁿᵈ AP
= 8875 Answer
Method(2):
By observation we find that starting from first term sum of two consecutive terms of the given series are in AP.
∴ required sum = \(\frac{{50}}{2}\)[2×6 + (50 – 1)×7]
= 8875 Answer
(Q). Find the value of :
1 – 2 – 3 -+ 2 – 3 – 4 + 3 – 4 – …………… till 100 terms
Sol:-
Above series has following three AP’s
Iˢᵗ AP : 1 + 2 + 3 + 4 + ……….. till 34 terms
2ⁿᵈ AP : -2 – 3 – 4- 5 – ……….. till 33 terms
3ʳᵈ AP : -3 – 4 – 5 – ……………. till 33 terms
finding the number of terms in each series:
If we group 3-3 terms then total 33 groups will be formed and remaining 1 element will be the Iˢᵗ element of 34ᵗʰ group, which is of Iˢᵗ AP.
Hence Iˢᵗ AP has 34 terms while 2ⁿᵈ & 3ʳᵈ AP’s have 33-33 terms
Sum of Iˢᵗ AP = \(\frac{{34}}{2}\)[2×1 + (34 – 1)×1]
= 595
Sum of 2ⁿᵈ AP = \(\frac{{33}}{2}\)[2×(-2) + (33 – 1)(-1)]
= – 594
Sum of 3ʳᵈ AP = \(\frac{{33}}{2}\)[2×(-3) + (33 – 1)(-1)]
= – 627
∴ Sum of given series = 595 – 594 – 627
= – 626 Answer
(Q). If the sum of first 13 terms of an AP is equal to the sum o first 27 terms of that AP. Find the sum of first 40 terms of that AP.
Sol:-
S₁₃ = S₂₇
\(\frac{{13}}{2}\)[2a + 12d] = \(\frac{{27}}{2}\)[2a + 26d]
14a + 273d = 0
2a + 39d = 0
Now S₄₀ = \(\frac{{40}}{2}\)[2a + 39d]
= 20[0]
= 0 Answer
(Q). A number 20 is divided into 4 parts that are in AP such that product of the first and fourth is in the ratio of 2:3 to the product of second sand third. Find the second largest part.
Sol:-
Let 4 number in AP are:
a – 3d, a – d, a + d, a + 3d
∴ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20
4a = 20
a = 5
& \(\frac{{(a – 3d)(a + 3d)}}{{(a – d)(a + d)}} = \frac{2}{3}\)
\(\frac{{{a^2} – 9{d^2}}}{{{a^2} – {d^2}}} = \frac{2}{3}\)
Put value of a
\(\frac{{25 – 9{d^2}}}{{25 – {d^2}}} = \frac{2}{3}\)
25d² = 25
d² = 1
∴ Four parts are : a – 3d, a – d, a + d, a + 3d
2, 4, 6, 8
∴ Second largest part = 6 Answer
(Q). Find the total number of terms between 500 and 1000 which are divisible by 6.
Sol:-
We are to find number of terms in following AP:
504, 510, 516, 522, …………………, 996
Number of terms = \(\frac{{996 – 504}}{6} + 1\)
= 83 Answer
(Q). Find the arithmetic mean of 4 and 9.
Sol:- AM = \(\sqrt {4 \times 9} \) = 6 Answer
(Q). Given a series:
2, 6, 18, 54, ……………… till 20 terms.
(i) 20ᵗʰ term
(ii) Sum of 20 terms of series.
Sol:-
Given series is GP with a = 2 r = \(\frac{6}{2}\) = 3
(i). T₂₀ = a.r²⁰⁻¹ = 2.13¹⁹ Answer
(ii). S₂₀ = \(\frac{{a({r^n} – 1)}}{{r – 1}} = \frac{{2({3^{20}} – 1)}}{{3 – 1}}\) = 3²⁰ – 1 Answer
(Q). Given a series:
2, 4, 8, 16…………….
(i) Find 10th term
(ii) Sum of first 10 terms
Sol:-
(i). T₁₀ = 2.2¹⁰⁻¹ = 2¹⁰ = 1024 Answer
(ii). S₁₀ = \(\frac{{2({2^{10 – 1}} – 1)}}{{2 – 1}} = 2({2^9} – 1)\) = 2(2⁹ – 1) = 1022 Answer
(Q). Find the sum of series:
16, 8, 4, 2, 1, ……………….. ∞
Sol:-
a = 16 r = \(\frac{8}{{16}}\) = \(\frac{1}{2}\)
S∞ = \(\frac{a}{{1 – r}} = \frac{{16}}{{1 – \frac{1}{2}}}\) = 32 Answer
(Q). Find the remainder when \({36^{(120 + 40 + 8 + 1\frac{3}{8} + ……..\infty )}}\)
Sol:-
120 + 40 + 8 + \(1\frac{3}{8}\) + ……….. ∞ ⟹ This is an infinite GP where a = 120 & r = \(\frac{1}{5}\)
∴ S = \(\frac{{120}}{{1 – \frac{1}{5}}}\) = 150
(Q). Find the Geometric mean of 45, 12, 50, 30.
Sol:-
45 = 5×3²
12 = 2²×3
50 = 5²×2
30 = 2×3×5
∴ 45×12×50×30 = 2⁴×3⁴×5⁴
∴ AM = \({\left( {{2^4} \times {3^4} \times {5^4}} \right)^{\frac{1}{4}}}\) = 2×3×5 = 30 Answer
(Q). What will be the sum of n term of the series:
9 + 99 + 999 + 9999 + ………….. till n terms.
Sol:-
(Q). Find the sum of series:
7 + 77 + 777 + 7777 + …………….. till n terms.
Sol:-
7 + 77 + 777 + 7777 + ……………… till n terms
= 7(1 + 11 + 111 + 1111 + …………….. till n terms)
= \(\frac{7}{9}\)(9 + 99 + 999 + 9999 + ………….. till n terms)
= \(\frac{7}{9}\)[\(\frac{{10}}{9}\)(10ⁿ – 1) – n]
= \(\frac{{70}}{{81}}({10^n} – 1) – \frac{7}{9}n\) Answer
(Q). A ball is thrown from a height of 500m on the ground. The ball bounce \(\frac{4}{5}\) times of the every last bounce then find the total distance covered by the ball when it stops.
Sol:-
● This will form an infinite GP whose terms are (a + b), (b + c), (c + d), (d + e), (e + f), ………….. ∞
Here,
a = 500 m
b = 500 × \(\frac{4}{5}\) = 400
c = 400 × \(\frac{4}{5}\) = 320
d = 300 × \(\frac{4}{5}\) = 256
∴ series will be (500 + 400), (400 + 320), (320 + 256), ……………. ∞
⟹ 900, 720, 576, ……………… ∞
∴ Sum of this infinite GP = total distance covered by ball
= \(\frac{{900}}{{1 – \frac{{720}}{{900}}}}\)
= 4500 m Answer
(Q). The side of a square is 18 cm. Infinite squares are made by joining the mid-points of the square. Calculate the area of all the squares made.
Sol:-
● This will form an infinite GP whose terms are area of squares
Here, First term a = 18 × 18 = 324
= \(9\sqrt 2 \times 9\sqrt 2 \) = 162
∴ common ratio = \(\frac{{16}}{{324}}\) = 0.5
∴ Sum of areas of all squares = Sum of this infinite GP
= \(\frac{{324}}{{1 – 0.5}}\) = 648 cm² Answer
⟶ Area of square made by joining the mid-points of a square is half of the actual square.
(Q). The sides of a right-angle triangle are 3, 4 and 5 cm. respectively. A new right angle triangle is made by joining the mid-points of all the sides. This process continues for infinite then calculate the area of all the triangles so made.
Sol:-
● This will form an infinite GP whose terms are area of the triangles.
Area of first triangle = \(\frac{1}{2}\)×4×3 = 6 cm²
Area of second triangle = \(\frac{1}{2}\)×2×\(\frac{2}{3}\) = \(\frac{1}{4}\)×6 cm²
∴ a = 6
r = \(\frac{1}{4}\)
S∞ = \(\frac{6}{{1 – \frac{1}{4}}}\) = 8
∴ required sum = 8 cm² Answer
⟶ The figure made by joining the mid-point of a right angle triangle is the right angle triangle whose area is the \({\frac{1}{4}^{th}}\) of the original one.
(Q). Find the sum of n terms of series:
12 + 104 + 1006 + …………….. till n terms.
Sol:-
The series can be re-written as:
10 + 2 + 100 + 4 + 1000 + 6 + ……………. till n terms
This is mixture of two series: one AP and second GP.
Iˢᵗ series:
a = 10
r =10
Sₙ = \(\frac{{10({{10}^n} – 1)}}{{10 – 1}} = \frac{{10}}{9}({10^n} – 1)\)
2ⁿᵈ series:
2 + 4 + 6 + 8 + …………… till n terms
= sum of first n even natural number
= n(n + 1)
∴ Sum of given series = \(\frac{{10}}{9}\)(10ⁿ – 1) + n(n + 1) Answer
(Q). Two number a and b are such that their AM is \(1\frac{1}{{12}}\) times more than their GM. Find the ratio between the numbers.
Sol:-
\(\frac{{\frac{{a + b}}{2}}}{{\sqrt {ab} }} = \frac{{17}}{{15}}\)
\(\frac{{a + b}}{{2\sqrt {ab} }} = \frac{{17}}{{15}}\)
Apply componendo-dividendo:
\(\frac{{a + b + 2\sqrt {ab} }}{{a + b – 2\sqrt {ab} }} = \frac{{17 + 15}}{{17 – 15}}\)
\({\frac{{(\sqrt a + \sqrt b )}}{{{{(\sqrt a + \sqrt b )}^2}}}^2} = 16\)
\(\frac{{\sqrt a + \sqrt b }}{{\sqrt a – \sqrt b }} = 4\)
\(\sqrt a + \sqrt b = 4\sqrt a – 4\sqrt b \)
\(5\sqrt b = 3\sqrt a \)
\(\frac{a}{b} = \frac{{25}}{9}\) Answer
(Q). Find the 5ᵗʰ and 8ᵗʰ term of Harmonic progression
6, 4, 3, …………….
Sol:-
6, 4, 3, ………….. is HP
∴ \(\frac{1}{6},\frac{1}{4},\frac{1}{3},\) ………….. are in AP
a = \(\frac{1}{6}\)
d = \(\frac{1}{4} – \frac{1}{6} = \frac{1}{{12}}\)
T₅ = \(\frac{1}{6}\) + (5 – 1).\(\frac{1}{{12}}\) = \(\frac{1}{{2}}\)
and T₈ = \(\frac{1}{6}\) + (8 – 1).\(\frac{1}{{12}}\) = \(\frac{3}{{4}}\)
∴ 5ᵗʰ term of HP = \(\frac{1}{{{T_5}}}\) = 2
& 8ᵗʰ term of HP = \(\frac{1}{{{T_8}}}\) = \(\frac{4}{3}\) Answer
(Q). Find the 16ᵗʰ term of an HP whose 6ᵗʰ and 11ᵗʰ term are 10 and 18 respectively.
Sol:-
if this HP is written in term of AP
then 6ᵗʰ term of AP = \(\frac{1}{{10}}\)
a + 5d = \(\frac{1}{{10}}\) ……………… (i)
& 11ᵗʰ term of AP = \(\frac{1}{{18}}\)
a + 10d = \(\frac{1}{{18}}\) ……………….. (ii)
From equation (i) & (ii)
d = \(\frac{{ – 2}}{{225}}\)
a = \(\frac{13}{{90}}\)
∴ 16ᵗʰ term of this AP = a + 15d
= \(\frac{{13}}{{90}} + 15\left( {\frac{{ – 2}}{{225}}} \right)\)
= \(\frac{1}{{90}}\)
∴ 16ᵗʰ term of HP = \(\frac{1}{{{{16}^{th}}term\;of\;AP}}\)
= 90 Answer