Progressions  Vikas Saini

Arithmetic Progression
Basic formulae :
T_{n} = a + (n1)d.
S_{n}= (n/2) [2a+(n1)d].
S_{n } S_{n1 }= d.
Here T_{n }= nth term.
S_{n }= Sum of n terms.
a = first term.
d = difference.
Q. Sum of 6 terms of an AP is 6. 3^{rd} term of AP is 0.
Find the 6^{th} term of it.
Solution :
Suppose 3^{rd} term of AP is a.
2^{nd} & 1^{st} term are ad & a2d respectively.
4^{th}, 5^{th} & 6^{th} terms are a+d,a+2d,a+3d respectively.
So (ad)+(a2d)+a+(a+d)+(a+2d)+(a+3d) = 6.
= > 6a + 3d = 6.
= > 0 + 3d = 6.
= > d = 2.
So a+3d = 6.
6^{th} term is 6.
Q. Sum of 3 terms of an AP is 30. If we add 6 in product of these 3 terms then it becomes a perfect square. Find all 3 terms of AP.
Solution :
Let three terms are ad, a, a+d.
= > (ad)+a+(a+d) = 30.
= > 3a = 30.
= > a = 10.
(10 – d) x 10 x (10+d) + 6 = k^2.
= > 10(100 – d^2)+6 = k^2.
d = 1,2,3,4,5,6,7,8,9.
d^2 = 1,4,9,16,25,36,49,64,81.
100 – d^2 = 99,96,91,84,75,64,51,36,19.
10(100 – d^2)+6 = 996,966,916,846,756,646,516,366,196.
196 is the only number here which is perfect square.
So d = 9.
3 terms of AP are = (109),10,(10+9)
= 1,9,19.
Q. Sum of first 4 terms of an AP is 42. Sum of square of 4 terms are 566.Find all the 4 terms.
Solution :
Let terms are a2d, ad , a+d , a+2d.
(a3d)+(ad)+(a+d)+(a+3d) = 42.
= > 4 a = 42.
a = 10.5.
(a3d)^2 + (ad)^2 + (a+d)^2 + (a+3d)^2 = 566.
= > a^2 + 9d^2 – 6ad + a^2 + d^2 – 2ad + a^2 + d^2 + 2ad + a^2 + 9d^2 + 6ad = 566.
= > 4a^2 + 20 d^2 = 566.
= > 4 x (10.5)^2 + 20 d^2 = 566.
= > 441 + 20d^2 = 566.
= > 20 d^2 = 125.
= > d^2 = 6.25
= > d = 2.5 .
Numbers are 3,8,13,18.
Q. From the first 20 natural numbers how many Arithmetic Progressions of five terms can be formed such that the common difference is a factor of the fifth term ?
Solution :
For d=1, total = 16.
(1,2,3,4,5)(2,3,4,5,6)………………………..(16,17,18,19,20)
For d=2, total = 6.
(2,4,6,8,10)(4,6,8,10,12)…….(12,14,16,18,20).
For d=3, total = 2.
(3,6,9,12,15)(6,9,12,15,18 )
For d=4, total = 1.
(4,8,12,16,20)
Total = 16+6+2+1 = 25.
Q. Ratio of sum of two APs are (n+1 : 2n).Find ratio of 6^{th} terms of both series.
Solution :
To find ratio of any term m.
We need to put n = 2m – 1.
m = middle term.
Here m = 6.
n = 11.
So ratio of 6^{th} terms = 11 + 1 / 2 x 11 = 12 / 2 x 11
= 6 / 11.
Q. Find total no of APs with 5 distinct terms that can be formed from the first 50 natural numbers.
Solution :
For d = 1, Total 46.
(1,2,3,4,5),(2,3,4,5,6)…………………….(46,47,48,49,50)
For d=2, total 42.
(1,3,5,7,9),(2,4,6,8,10)…………………(42,44,46,48,50)
For d=3, total 38.
(1,4,7,10,13),(2,5,8,11,14)………….(38,41,44,47,50)
For d= 12, total 2.
(1,13,25,37,49)(2,14,26,38,50).
Total = 46+42+38…………….2.
Total = (12/2) (2+46) = 288.
Q. If log_{3}2 + log_{3}(2^{x} – 5) and log_{3}(2^{x} – 7/2) are in AP, then find the value of x.
Solution n:
All nos are in AP.
2 log_{3}(2^{x} – 5) = log_{3}2 + log_{3}(2^{x} – 7/2)
Suppose 2^{x } = t.
2 log_{3}(t – 5) = log_{3}2 + log_{3}(t – 7/2)
= > log_{3}(t5)^{2 }= log_{3}(2t – 7)
= > (t – 5 )^{2 }= 2t – 7.
= > t^{2 }– 10t + 25 = 2t – 7
= > t^{2 }– 12t + 32 = 0.
= > t = 4,8
2^{x }= 4 , 8
x = 2,3.
Q. A sequence of terms is defined such that
2a_{n } = a_{n+1 }+ a_{n1};
a_{0 }= 1; a_{1 }= 3.
What is the value of a_{0 }+ a_{1 }+ a_{2 ... }a_{50}.
Solution :
It shows property of an AP.
a_{0 }=1, a_{1 }= 3.
d = 2.
a_{50} = a_{0 }+ 50 d = 1 + 50 x 2 = 101.
Sum = (51/2) (1+101) = 51 x 51 = 2601.
Geometric Progression
T_{n }= ar^{n1 }
S_{n }= a (r^{n }– 1) / (r – 1) if r > 0
= a ( 1 – r^{n })/ (1 – r ) if r < 0.
Sum of infinite terms = a / (1 – r).
Q. T_{n }is the nth term of geometric progression such that T_{n }= T_{n1 }+ T_{n2 }for n > 2. If the first term and common ratio of the progression are positive real numbers, then find the common ratio.
Solution :
T_{3 }= T_{2 + }T_{1.}
ar^{2 }= ar + a.
r^{2 }= r + 1.
= > r^{2 }– r – 1 = 0.
By using Shridhar acharya’s method
r = 1+ root (1+4) / 2
r = (1+ root 5 )/ 2.
Q. If the roots of the equation ax^{3}+bx^{2}+cx+d = 0 are in geometric Progression, then which of the following relation is true?
(a) ac^{2 }= bd (b) ac^{3}=b^{3}d
(c) a^{2}c = bd^{2} (d) a^{3}c=bd^{3}
Solution :
Let roots are 1,2 & 4.
a = 1.
b = (1+2+4) = 7.
c = 1 x 2 + 2 x 4 +1 x 4 = 2+8+4=14.
d = (1 x 2 x 4) =  8.
(c / b)^{3} = 8.
d / a = 8.
= > (c/b)^{3} = d/a.
= > ac^{3 } = b^{3}d.
Harmonic Progression
Q. ‘M’ and ‘N’ are natural numbers such that by M = (5N – 4)(5N + 1).
If 1 < = N < = 200, what is the harmonic mean of all possible values of M ?
Solution :
N =1, M = 1 x 6.
N = 2, M = 6 x 11.
N = 200, M =996 x 1001.
Total numbers / Harmonic mean = 1 / (1 x 6) + 1 / (6 x 11) ………………………….. 1/996 x 1001.
= > 200 / Harmonic mean = (1 / 5) ( 1 – 1/6 + 1/6 – 1/11…………… 1/1001)
= > 200 / Harmonic mean = (1/5) ( 1 – 1/1001)
= > 200 / Harmonic mean = (1 / 5) (1000 / 1001)
= > Harmonic mean = 1001.
Q. A person leaves his home at 6 AM at speed of 20 km/hr and reaches his office at 9 AM. When his speed is 30 km/hr, he reaches at 8 AM.
If he has reach at 8:30 AM, what should be his speed?
Solution :
Here we see time is in AP.
So Speed must be in HP.
We need to find harmonic mean of speed 20 & 30.
Speed = 2 x 20 x 30 / (20 + 30) = 24 km/hr.
Method – 2
D = 20 x 3 = 30 x 2 = 60.
Time = 2.5 hr.
Speed = 60 / 2.5 = 24 km/hr.
Miscellaneous Problems
Q. If S = 1^{2 }  2^{2} + 3^{2}  4^{2} ... + 2001^{2}.
Then what is the value of S.
Solution :
Method 1
S = (1^{2}2^{2})+(3^{2}4^{2})……………………………….(1999^{2 }– 2000^{2}) + 2001^{2}
S = (3)+(7)………………………………..(3999)+2001^{2}.
Total terms from (3 to 3999) = (3999 – 3)/4 + 1 = 1000.
S = (1000 / 2) (33999) + 2001^{2}
S = 2001000 + 4004001 = 2003001.
Method 2
S = (1^{2}+2^{2}+3^{2}………………..2001^{2}) – 2(2^{2}+4^{2}+6^{2}…………………….2000^{2})
S = (1^{2}+2^{2}…………………2001^{2}) – 8(1^{2}+2^{2}…………..1000^{2})
S = (2001 x 4003 x 2002 / 6) – 8(1000 x 2001 x 1001 / 6)
S = 2003001.
Q. a_{1 }= 1, a_{2 }= 2 and a_{n+2} = a_{n}(a_{n+1} – 1),what is the value a_{1000} ?
Solution :
a_{3} = a_{1}(a_{2}1).
a_{3}=1(2 – 1) = 1.
a_{4}=2(1 – 1) = 0.
a_{5}= 1(0 – 1) = 1.
a_{6 }=0(101)= 0.
a_{7 }= 1(01)=1.
a_{8}=0.
a_{9 }= 1.
a_{10} = 0.
We see, after a_{1 }and a_{2} cyclicity of 4 in rest terms.
a_{998 }= 0.
a_{999 }= 1.
a_{1000 }= 0.
Q. 4x + 2y + z =35, where ‘x’,’y’ and ‘z’ are positive real numbers.
Find maximum value of x^{1/2 } x y^{1/4} x z^{1/8}.
Solution :
4x = k/2.
2y = k/4.
z = k/8.
(k/2)+(k/4)+(k/8 ) = 35.
= > 4k+2k+k/8 = 35.
= > 7 k/8 = 35
= > k = 40.
x = 5, y = 5, z=5.
Max value = 5^{7/8 }