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Sequence And Series-II
Arithmetic And Geometric Progressions
CPT Section D Quantitative Aptitude Chapter 6
CA Loveneesh Kapoor
Geometric Progressions
(G.P. Series)
Geometric Progression-Meaning
In a sequence of terms, if each term is constant
multiple of the preceding term, then the sequence is
called as Geometric Progression(G.P.)
The constant multiplier is called the common ratio.
Example: 1,2,4,8,16,32….here the common ratio is 2.
nth term of the series
5, 15, 45, …..next term will be 135 following pattern of multiplication of 3.
t1 = 5 = 5.r0 = a.r0
t2 = 15 = 5.(3)1 = a.r1
t3 = 45 = 5.(3)2 = a.r2
tn = 5.(3)n-1 = a.rn-1
Thus, in a G.P. Series
nth term i.e, tn = a.rn-1
And the series is
a, ar, ar2, ar3, ….arn-1
Example 1
Q: Find the 6th term of the series 3+1+1/3+….
Sol: Here the given series is G.P. series with
a = 3, r = t2 /t1 = 1/3 , n = 6
Tn = arn-1
T6 = ar5 = 3.(1/3)5
= 3.(1/243)
= 1/81
Thus 6th term is 1/81.
Example 2
Q: Which term of the progression 1,2,4,8….is 256.
Sol: Here a = 1, r = 2, tn = 256.
tn = arn-1 = 1.(2)n-1 = 256
= (2)n-1 = 28
= n - 1= 8
= n = 8+1 = 9
Thus 256 is the 9th term of the series i.e, t9
Example 3
Q: Find the last term of the series 1,-3,27,-81..to 7 terms
Sol: Here a = 1, n = 7
r = t2 /t1 = -3 / 1 = -3,
tn = arn-1
t7 = ar6
t7 = (1)(-3)6
t7 = 729
Example 4
Q: Find the G.P. series where 4th term is 8 and 8th term is 128/625.
Sol : t4 = ar3 = 8
t8 = ar7 = 128/625
Dividing the two terms
t8 /t4 = ar7 / ar3 = 128 / 625*1 / 8
= r4 = 16 / 625
= r4 = (2 / 5)4
=r=2/5
Now ar4 = a (2 / 5)4 = 8
Solving , a = 125
Thus a = 125, r = 2/5, the G.P Series is 125, 50, 20, 8, ….
Geometric Mean (G.M.)
If a, b, c are in G.P, the middle term i.e, b is said to be
the Geometric Mean (G.M.) between a & c.
Also b2 = ac
Thus in a GP series 3,9,27
9 is the middle term and is Geometric Mean between
3 & 27.
Example 5
Q: Insert three Geometric Means between 1/9 and 9.
Sol: G.P. Series 1/9, --, --, --, --, 9
Here t1 = a = 1/9
t5 = a.r4 = 9
Now, t5 = 1/9.r4 = 9
= r4 = 81
= r4 = 34
=r=3
t2 = ar = 1/9*3 = 1/3
t3 = ar2 = 1/9*32 = 1
t4 = ar3 = 1/9*33 = 3
Thus the series 1/9, 1/3, 1, 3, 9.
Sum of G.P. Series
If S be the sum of n terms of a G.P. series, with ‘a’ be
the 1st term and ‘r’ be the common ratio.
Sn =
Sn =
(
)
a r n −1
r −1
a 1− rn
(
1− r
)
when r > 1
when r < 1
Example 6
Q: Find the sum of 1st 8 terms of G.P series
1+2+4+8+……
Sol: Here a = 1, r = 2, n = 8
Sn = a.(rn -1) /(r - 1) when r >1
S8 = 1.(28 - 1) / (2 - 1)
= 1( 256 - 1) = 255
Thus S8 = 255
Example 7
Q: Find the sum of the series -2, 6, -18…..7 terms?
Sol: Here a = -2, r = -3, n = 7
Sn = a.(1 - rn ) / ( 1- r ) when r<1
S7 = (-2) [1 - (-3)7 ] / [ 1 – (-3)]
= (-2)(1 + 2187) / 4
= (-2)(2188) / 4
S7 = -1094
Example 8
Q: In a G.P. the product of the 1st three terms 27/8. The
middle term is
Sol: Let the three terms of GP are a/r, a, ar
Now Product of terms
a/r *a*ar = 27/8
a3 = 27/8
a3 = (3/2)3
a = 3/2
Thus the middle term, a = 3/2
Example 9
Q: If you save 1 paisa today, 2 paise the next day and 4 paise the succeeding
day and so on, then your total savings in two weeks will be.
Sol: Here the pattern of savings shows the G.P series 0.01, 0.02, 0.04, …..
Here a = 0.01, r = 2, n = 14
Sn = a.(rn -1) / (r -1) when r >1
S14 = 0.01( 214 -1) / ( 2 -1 )
= 0.01(16384 - 1) / 1
= 0.01*16383
S14 = 163.83
Thus the total savings in 14 days would be Rs 163.83.
Sum of Infinite G. P. Series
The sum of an Infinite Geometric Series
a, ar, ar2, ar3, ……..∞ when r < 1.
S∞ = a / (1 - r)
Where, a is the 1st term,
r is the common ratio
Example 10
Q: The sum of the Infinite G.P Series
1 - 1/3 + 1/9 - 1/27 …..
Sol: Here a = 1, r = (-1/3)
S∞ = a/(1 - r) = 1 / [1 - ( -1/3)]
= 1/[1+1/3]
= 1/[4/3]
= 3/4
= 0.75
MCQ-1
Q : The nth term of the series 16, 8, 4 ….is 1/217 . The
value of n is
a) 20
b) 21
c) 22
d) 24
Answer : (c) 22
Hint: a = 16= 24 , r = 1/2= 2-1 tn = 1/217 = 2-17
MCQ-2
Q : The sum of an infinite series 1 +2/3 + 4/9 + …..
a)1/3
b) 3
c) 2/3
d)3/2
Answer : (b) 3
Hint : a = 1, r = 2/3
MCQ-3
Q : The A.M of two positive numbers is 40 and their G.M. is 24. Find
the numbers
a) (72, 8)
b) (70, 10)
c) (60,20)
d) (40, 40)
Answer : (a) (72,8)
Hint : (a +b)/2= 40, a * b = 242
MCQ-4
Q :Four Geometric Means between 4 and 972 are
a) 12, 30, 100, 324
b) 12, 24, 108, 320
c) 10, 36, 108, 320
d) 12, 36, 108, 324
Answer : (d) 12, 36,108, 324
Hint : t1 = a = 4, t6 = ar5 = 972
MCQ-5
Q : The sum of the series 1+3+9+27….is 364.
Find the number of terms.
a) 5
b) 6
c) 11
d) 9
Answer : (b) 6
Hint : a = 1, r = 3
MCQ-6
Q : The number of terms to be taken so that
1+2+4+8+….will be 8191
a) 10
b) 13
c) 12
d) 14
Answer : (b) 13
Hint : a = 1, r = 2, Sn = 8191
MCQ-7
Q : The 2nd term of a G.P. is 24 and the fifth term is 81. The
series is
a) 16, 36, 24, 54
b) 24, 36, 53, 60
c) 16, 24, 36, 54
d) None of these
Answer : (c) 16, 24, 36, 54
Hint : t2 = ar = 24, t5 = ar4 = 81
MCQ-8
Q : Find three numbers in G.P. whose sum is 19 and
product is 216.
a) 4, 6, 9
b) 6, 9, 12
c) 5, 6, 8
d) None of these
Answer : (a) 4, 6, 9
Hint : Three Numbers are a/r, a, ar
MCQ-9
Q :The numbers x, 8, y are in G.P and the numbers x, y, -8
are in A.P. Find the value of x and y
a) (8, 8)
b) (16, 4)
c) (14, 6)
d) None of these
Answer : (b) 16, 4
Hint : (x – 8) / 2 = y & x*y = 82
MCQ-10
Q : The last term of the series x2 , x, 1, ….to 31 terms is
a) X28
b) 1/x
c)1/x28
d)None of these
Answer : (c) 1/x28
Hint : t31 = ar30
MCQ-11
Q : t4 of a G.P. is x, t10 = y, t16 = z, then
a) x2 = yz
b) y2 = xz
c) z2 = xy
d) None of these
Answer : (b) y2 = xz
Hint : Solve t10 / t6 & t16 / t10
MCQ-12
Q : Find the first term of a G.P. whose sum to infinite terms
is 8 and second term is 2.
a) 4
b) 6
c) 2
d) 3/2
Answer : (a) 4
Hint : S∞ = a/(1-r) = 8, a.r = 2
Lesson SummaryArithmetic Progression(A.P.)
nth term of the series i.e, tn = a + (n-1) d
Arithmetic Mean (A.M.) = ( a + b )/2
Sum of terms of A.P. series= n/2 [2a + (n-1)d]
Geometric Progression (G.P.)
nth term of the series i.e, tn = a.(r)n-1
Geometric Mean Mean (G.M.) = ( a*b )1/2
Sum of terms of G.P. series =
a.(rn -1) / (r - 1) when r >1 & a.(1-rn) / ( 1 - r) when r<1
Sum of Infinite G.P Series= a / (1 - r )
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All the Best !!!