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Quantitative Aptitude Β· Logarithms and Progressions

Arithmetic and Geometric Progressions

Explanation

πŸ”΅ ARITHMETIC AND GEOMETRIC PROGRESSIONS 🟒 PART A: ARITHMETIC PROGRESSION (AP) 🟒 1. BASIC CONCEPT OF AP An Arithmetic Progression is a sequence in which the difference between consecutive terms is constant. Example:
2,5,8,11,14,… 2,5,8,11,14,\ldots
Here:
5βˆ’2=3 5-2=3
8βˆ’5=3 8-5=3
11βˆ’8=3 11-8=3
Therefore, the common difference is:
d=3 d=3
🟒 2. FIRST TERM OF AP The first term of an AP is denoted by:
a a
For example, in:
7,11,15,19,… 7,11,15,19,\ldots
the first term is:
a=7 a=7
🟒 3. COMMON DIFFERENCE The common difference is the difference between any two consecutive terms.
d=a2βˆ’a1 d=a_2-a_1
or:
d=anβˆ’anβˆ’1 d=a_n-a_{n-1}
🟒 4. GENERAL FORM OF AP An AP can be written as:
a,Β a+d,Β a+2d,Β a+3d,… a,\ a+d,\ a+2d,\ a+3d,\ldots
🟒 5. nth TERM OF AP The nth term of an AP is:
an=a+(nβˆ’1)d a_n=a+(n-1)d
where: a = first term d = common difference n = position of the term 🟒 6. LAST TERM OF AP If an AP has n terms, its last term is:
l=a+(nβˆ’1)d l=a+(n-1)d
🟒 7. FINDING THE NUMBER OF TERMS From:
l=a+(nβˆ’1)d l=a+(n-1)d
we get:
n=1+lβˆ’ad n=1+\frac{l-a}{d}
provided:
d≠0 d\neq0
🟒 8. SUM OF FIRST n TERMS OF AP The sum of the first n terms is:
Sn=n2[2a+(nβˆ’1)d] S_n=\frac{n}{2}[2a+(n-1)d]
🟒 9. SUM OF AP USING FIRST AND LAST TERMS If the first term is a and the last term is l:
Sn=n2(a+l) S_n=\frac{n}{2}(a+l)
🟒 10. AVERAGE OF TERMS OF AN AP The average of all terms of a finite AP is:
Average=a+l2 \text{Average} = \frac{a+l}{2}
🟒 11. SUM OF TERMS OF AN AP If the terms are:
a,Β a+d,Β a+2d,…,l a,\ a+d,\ a+2d,\ldots,l
then:
Sn=n2(a+l) S_n=\frac{n}{2}(a+l)
🟒 12. ARITHMETIC MEAN If A is the arithmetic mean between x and y:
x,A,y x,A,y
then:
A=x+y2 A=\frac{x+y}{2}
Therefore:
2A=x+y 2A=x+y
🟒 13. INSERTING ARITHMETIC MEANS To insert n arithmetic means between x and y, the common difference is:
d=yβˆ’xn+1 d=\frac{y-x}{n+1}
The resulting AP is:
x,Β x+d,Β x+2d,…,y x,\ x+d,\ x+2d,\ldots,y
🟒 14. PROPERTY OF TERMS EQUALLY DISTANT FROM THE ENDS In a finite AP, terms equally distant from the beginning and the end have the same sum. For example:
a1+an=a2+anβˆ’1=a3+anβˆ’2 a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}
🟒 15. MIDDLE TERM OF AN AP If an AP has an odd number of terms, the middle term is:
an+12 a_{\frac{n+1}{2}}
For an AP with an odd number of terms:
MiddleΒ Term=a+l2 \text{Middle Term} = \frac{a+l}{2}
🟒 16. SUM OF FIRST n NATURAL NUMBERS The numbers:
1,2,3,…,n 1,2,3,\ldots,n
form an AP. Their sum is:
1+2+3+β‹―+n=n(n+1)2 1+2+3+\cdots+n = \frac{n(n+1)}{2}
🟒 17. SUM OF FIRST n ODD NUMBERS
1+3+5+β‹―+(2nβˆ’1)=n2 1+3+5+\cdots+(2n-1) = n^2
🟒 18. SUM OF FIRST n EVEN NUMBERS
2+4+6+β‹―+2n=n(n+1) 2+4+6+\cdots+2n = n(n+1)
🟒 19. AP WITH NEGATIVE COMMON DIFFERENCE An AP can decrease when:
d<0 d<0
Example:
20,17,14,11,… 20,17,14,11,\ldots
Here:
d=βˆ’3 d=-3
🟒 20. CONSTANT SEQUENCE If:
d=0 d=0
then all terms are equal. Example:
5,5,5,5,… 5,5,5,5,\ldots
🟑 AP KEY POINTS 1. AP has a constant difference. 2. nth term:
an=a+(nβˆ’1)d a_n=a+(n-1)d
3. Last term:
l=a+(nβˆ’1)d l=a+(n-1)d
4. Sum of n terms:
Sn=n2[2a+(nβˆ’1)d] S_n=\frac{n}{2}[2a+(n-1)d]
5. Sum using first and last terms:
Sn=n2(a+l) S_n=\frac{n}{2}(a+l)
6. Arithmetic mean:
A=x+y2 A=\frac{x+y}{2}
7. To find the common difference:
d=yβˆ’xn+1 d=\frac{y-x}{n+1}
when n means are inserted between x and y. 8. Equally distant terms from the ends have equal sums. 9. Always identify a, d, n and l before applying an AP formula. 🟒 PART B: GEOMETRIC PROGRESSION (GP) 🟒 21. BASIC CONCEPT OF GP A Geometric Progression is a sequence in which the ratio between consecutive terms is constant. Example:
2,6,18,54,… 2,6,18,54,\ldots
Here:
62=3 \frac{6}{2}=3
186=3 \frac{18}{6}=3
5418=3 \frac{54}{18}=3
Therefore, the common ratio is:
r=3 r=3
🟒 22. FIRST TERM OF GP The first term of a GP is denoted by:
a a
Example:
5,10,20,40,… 5,10,20,40,\ldots
Here:
a=5 a=5
🟒 23. COMMON RATIO The common ratio is:
r=SecondΒ TermFirstΒ Term r=\frac{\text{Second Term}}{\text{First Term}}
or:
r=ananβˆ’1 r=\frac{a_n}{a_{n-1}}
provided the denominator is non-zero. 🟒 24. GENERAL FORM OF GP A GP can be written as:
a,Β ar,Β ar2,Β ar3,… a,\ ar,\ ar^2,\ ar^3,\ldots
🟒 25. nth TERM OF GP The nth term of a GP is:
an=arnβˆ’1 a_n=ar^{n-1}
🟒 26. LAST TERM OF GP If a GP has n terms, its last term is:
l=arnβˆ’1 l=ar^{n-1}
🟒 27. SUM OF FIRST n TERMS OF GP For:
r≠1 r\neq1
the sum of the first n terms is:
Sn=a(rnβˆ’1)rβˆ’1 S_n = \frac{a(r^n-1)}{r-1}
An equivalent form is:
Sn=a(1βˆ’rn)1βˆ’r S_n = \frac{a(1-r^n)}{1-r}
🟒 28. GP WHEN r = 1 If:
r=1 r=1
then all terms are equal. Therefore:
Sn=na S_n=na
🟒 29. SUM OF INFINITE GP For an infinite GP, if:
∣r∣<1 |r|<1
then the sum to infinity is:
S∞=a1βˆ’r S_\infty=\frac{a}{1-r}
🟒 30. CONDITION FOR INFINITE GP A finite sum exists for an infinite GP only when:
∣r∣<1 |r|<1
🟒 31. INSERTING GEOMETRIC MEANS If n geometric means are inserted between x and y, then the common ratio is:
r=(yx)1n+1 r=\left(\frac{y}{x}\right)^{\frac{1}{n+1}}
assuming the relevant real root exists. 🟒 32. GEOMETRIC MEAN The geometric mean between two positive numbers x and y is:
G=xy G=\sqrt{xy}
🟒 33. THREE TERMS IN GP If three positive terms are in GP, they can be written as:
ar,Β a,Β ar \frac{a}{r},\ a,\ ar
Their middle term satisfies:
a2=(ar)(ar) a^2 = \left(\frac{a}{r}\right)(ar)
🟒 34. RELATION BETWEEN THREE TERMS IN GP If:
a,b,c a,b,c
are in GP, then:
b2=ac b^2=ac
🟒 35. PRODUCT OF TERMS IN A FINITE GP For a finite GP with n terms:
a,Β ar,Β ar2,…,arnβˆ’1 a,\ ar,\ ar^2,\ldots,ar^{n-1}
the product is:
P=anrn(nβˆ’1)2 P=a^n r^{\frac{n(n-1)}{2}}
🟒 36. GP WITH NEGATIVE COMMON RATIO A GP can alternate between positive and negative terms. Example:
2,βˆ’4,8,βˆ’16,… 2,-4,8,-16,\ldots
Here:
r=βˆ’2 r=-2
🟒 37. GP WITH FRACTIONAL COMMON RATIO If:
0<r<1 0<r<1
the terms decrease toward zero. Example:
16,8,4,2,1,… 16,8,4,2,1,\ldots
Here:
r=12 r=\frac12
🟒 38. RELATION BETWEEN AP AND GP AP:
a,Β a+d,Β a+2d,… a,\ a+d,\ a+2d,\ldots
The difference is constant. GP:
a,Β ar,Β ar2,… a,\ ar,\ ar^2,\ldots
The ratio is constant. 🟑 GP KEY POINTS 1. GP has a constant ratio. 2. nth term:
an=arnβˆ’1 a_n=ar^{n-1}
3. Sum of n terms:
Sn=a(rnβˆ’1)rβˆ’1 S_n=\frac{a(r^n-1)}{r-1}
4. Sum to infinity:
S∞=a1βˆ’r S_\infty=\frac{a}{1-r}
provided:
∣r∣<1 |r|<1
5. Geometric mean:
G=xy G=\sqrt{xy}
for positive xx and yy. 6. If a, b and c are in GP:
b2=ac b^2=ac
7. For r = 1:
Sn=na S_n=na
8. Always identify a, r and n before applying a GP formula. πŸ”΄ EXAM TIP For AP, look for a constant DIFFERENCE. For GP, look for a constant RATIO. AP:
an=a+(nβˆ’1)d a_n=a+(n-1)d
GP:
an=arnβˆ’1 a_n=ar^{n-1}

Example

πŸ”΅ ARITHMETIC PROGRESSION β€” EXAMPLES 🟠 EXAMPLE 1: IDENTIFY THE COMMON DIFFERENCE QUESTION: Find the common difference of the AP:
7,12,17,22,… 7,12,17,22,\ldots
🟒 SOLUTION:
d=12βˆ’7 d=12-7
d=5 d=5
Check:
17βˆ’12=5 17-12=5
Therefore:
d=5 d=5
βœ… ANSWER:
55
🟠 EXAMPLE 2: FIND THE nth TERM QUESTION: Find the 15th term of the AP:
3,7,11,15,… 3,7,11,15,\ldots
🟒 SOLUTION: First term:
a=3 a=3
Common difference:
d=4 d=4
Using:
an=a+(nβˆ’1)d a_n=a+(n-1)d
For n = 15:
a15=3+(15βˆ’1)4 a_{15} = 3+(15-1)4
=3+56 =3+56
=59 =59
βœ… ANSWER:
5959
🟠 EXAMPLE 3: FIND A PARTICULAR TERM QUESTION: Find the 20th term of the AP:
10,7,4,1,… 10,7,4,1,\ldots
🟒 SOLUTION:
a=10 a=10
d=βˆ’3 d=-3
Therefore:
a20=10+(20βˆ’1)(βˆ’3) a_{20} = 10+(20-1)(-3)
=10βˆ’57 =10-57
=βˆ’47 =-47
βœ… ANSWER:
βˆ’47-47
🟠 EXAMPLE 4: FIND THE NUMBER OF TERMS QUESTION: How many terms are there in the AP:
5,9,13,…,81 5,9,13,\ldots,81
🟒 SOLUTION:
a=5 a=5
d=4 d=4
l=81 l=81
Using:
l=a+(nβˆ’1)d l=a+(n-1)d
81=5+(nβˆ’1)4 81=5+(n-1)4
76=4(nβˆ’1) 76=4(n-1)
19=nβˆ’1 19=n-1
n=20 n=20
βœ… ANSWER:
20terms20 terms
🟠 EXAMPLE 5: SUM OF FIRST n TERMS QUESTION: Find the sum of the first 20 terms of the AP:
2,5,8,11,… 2,5,8,11,\ldots
🟒 SOLUTION:
a=2 a=2
d=3 d=3
n=20 n=20
Using:
Sn=n2[2a+(nβˆ’1)d] S_n = \frac{n}{2}[2a+(n-1)d]
S20=202[2(2)+(20βˆ’1)3] S_{20} = \frac{20}{2}[2(2)+(20-1)3]
=10[4+57] =10[4+57]
=10Γ—61 =10\times61
=610 =610
βœ… ANSWER:
610610
🟠 EXAMPLE 6: SUM USING FIRST AND LAST TERMS QUESTION: Find the sum of the AP:
5,10,15,…,50 5,10,15,\ldots,50
🟒 SOLUTION: First term:
a=5 a=5
Last term:
l=50 l=50
Common difference:
d=5 d=5
Number of terms:
n=1+50βˆ’55 n = 1+\frac{50-5}{5}
=10 =10
Using:
Sn=n2(a+l) S_n=\frac{n}{2}(a+l)
S10=102(5+50) S_{10} = \frac{10}{2}(5+50)
=5Γ—55 =5\times55
=275 =275
βœ… ANSWER:
275275
🟠 EXAMPLE 7: FIND THE ARITHMETIC MEAN QUESTION: Find the arithmetic mean between 12 and 28. 🟒 SOLUTION:
A=12+282 A=\frac{12+28}{2}
=402 =\frac{40}{2}
=20 =20
βœ… ANSWER:
2020
🟠 EXAMPLE 8: INSERT ARITHMETIC MEANS QUESTION: Insert 3 arithmetic means between 5 and 21. 🟒 SOLUTION: There are 3 means, so the total number of intervals is:
3+1=4 3+1=4
Common difference:
d=21βˆ’54 d=\frac{21-5}{4}
=4 =4
Therefore, the AP is:
5,9,13,17,21 5,9,13,17,21
The three arithmetic means are:
9,13,17 9,13,17
βœ… ANSWER:
9,13and179, 13 and 17
🟠 EXAMPLE 9: SUM OF FIRST n NATURAL NUMBERS QUESTION: Find:
1+2+3+β‹―+50 1+2+3+\cdots+50
🟒 SOLUTION: Using:
Sn=n(n+1)2 S_n=\frac{n(n+1)}{2}
S50=50(51)2 S_{50} = \frac{50(51)}{2}
=25Γ—51 =25\times51
=1275 =1275
βœ… ANSWER:
1,2751,275
🟠 EXAMPLE 10: SUM OF ODD NUMBERS QUESTION: Find the sum of the first 20 odd numbers. 🟒 SOLUTION: Using:
1+3+5+β‹―+(2nβˆ’1)=n2 1+3+5+\cdots+(2n-1)=n^2
For n = 20:
S=202 S=20^2
=400 =400
βœ… ANSWER:
400400
🟠 EXAMPLE 11: SUM OF EVEN NUMBERS QUESTION: Find the sum of the first 15 even numbers. 🟒 SOLUTION: Using:
2+4+6+β‹―+2n=n(n+1) 2+4+6+\cdots+2n=n(n+1)
For n = 15:
S=15(16) S=15(16)
=240 =240
βœ… ANSWER:
240240
🟠 EXAMPLE 12: FIND THE MISSING TERM QUESTION: If 5, x, 17 are consecutive terms of an AP, find x. 🟒 SOLUTION: In an AP, the middle term is the arithmetic mean of the first and third terms.
x=5+172 x=\frac{5+17}{2}
=222 =\frac{22}{2}
=11 =11
βœ… ANSWER:
1111
πŸ”΅ GEOMETRIC PROGRESSION β€” EXAMPLES 🟠 EXAMPLE 13: IDENTIFY THE COMMON RATIO QUESTION: Find the common ratio of the GP:
3,9,27,81,… 3,9,27,81,\ldots
🟒 SOLUTION:
r=93 r=\frac{9}{3}
=3 =3
Check:
279=3 \frac{27}{9}=3
Therefore:
r=3 r=3
βœ… ANSWER:
33
🟠 EXAMPLE 14: FIND THE nth TERM OF GP QUESTION: Find the 8th term of the GP:
2,6,18,54,… 2,6,18,54,\ldots
🟒 SOLUTION:
a=2 a=2
r=3 r=3
Using:
an=arnβˆ’1 a_n=ar^{n-1}
a8=2(3)8βˆ’1 a_8 = 2(3)^{8-1}
=2(37) =2(3^7)
=2(2187) =2(2187)
=4374 =4374
βœ… ANSWER:
4,3744,374
🟠 EXAMPLE 15: FIND A PARTICULAR TERM OF GP QUESTION: Find the 6th term of the GP:
5,10,20,40,… 5,10,20,40,\ldots
🟒 SOLUTION:
a=5 a=5
r=2 r=2
Using:
an=arnβˆ’1 a_n=ar^{n-1}
a6=5(2)5 a_6 = 5(2)^5
=5Γ—32 =5\times32
=160 =160
βœ… ANSWER:
160160
🟠 EXAMPLE 16: FIND THE COMMON RATIO QUESTION: The first term of a GP is 4 and the second term is 12. Find the common ratio. 🟒 SOLUTION:
r=124 r=\frac{12}{4}
=3 =3
βœ… ANSWER:
33
🟠 EXAMPLE 17: FIND THE SUM OF A GP QUESTION: Find the sum:
2+6+18+54+162 2+6+18+54+162
🟒 SOLUTION:
a=2 a=2
r=3 r=3
n=5 n=5
Using:
Sn=a(rnβˆ’1)rβˆ’1 S_n = \frac{a(r^n-1)}{r-1}
S5=2(35βˆ’1)3βˆ’1 S_5 = \frac{2(3^5-1)}{3-1}
=2(243βˆ’1)2 = \frac{2(243-1)}{2}
=242 =242
βœ… ANSWER:
242242
🟠 EXAMPLE 18: SUM OF FIRST n TERMS OF GP QUESTION: Find the sum of the first 6 terms of the GP:
3,6,12,24,… 3,6,12,24,\ldots
🟒 SOLUTION:
a=3 a=3
r=2 r=2
n=6 n=6
Using:
Sn=a(rnβˆ’1)rβˆ’1 S_n = \frac{a(r^n-1)}{r-1}
S6=3(26βˆ’1)2βˆ’1 S_6 = \frac{3(2^6-1)}{2-1}
=3(64βˆ’1) =3(64-1)
=3Γ—63 =3\times63
=189 =189
βœ… ANSWER:
189189
🟠 EXAMPLE 19: SUM TO INFINITY QUESTION: Find the sum to infinity:
8+4+2+1+β‹― 8+4+2+1+\cdots
🟒 SOLUTION: First term:
a=8 a=8
Common ratio:
r=48 r=\frac{4}{8}
=12 =\frac12
Since:
∣r∣<1 |r|<1
the sum to infinity exists. Using:
S∞=a1βˆ’r S_\infty=\frac{a}{1-r}
S∞=81βˆ’12 S_\infty = \frac{8}{1-\frac12}
=812 =\frac{8}{\frac12}
=16 =16
βœ… ANSWER:
1616
🟠 EXAMPLE 20: GEOMETRIC MEAN QUESTION: Find the geometric mean between 4 and 25. 🟒 SOLUTION:
G=4Γ—25 G=\sqrt{4\times25}
=100 =\sqrt{100}
=10 =10
βœ… ANSWER:
1010
🟠 EXAMPLE 21: INSERT GEOMETRIC MEANS QUESTION: Insert 2 geometric means between 2 and 54. 🟒 SOLUTION: The GP is:
2,Β 2r,Β 2r2,Β 2r3 2,\ 2r,\ 2r^2,\ 2r^3
Since the last term is 54:
2r3=54 2r^3=54
r3=27 r^3=27
r=3 r=3
Therefore:
2,6,18,54 2,6,18,54
The two geometric means are:
6,18 6,18
βœ… ANSWER:
6and186 and 18
🟠 EXAMPLE 22: THREE TERMS IN GP QUESTION: If x, 12 and 48 are consecutive terms of a GP, find x. 🟒 SOLUTION: For three consecutive terms in GP:
122=xΓ—48 12^2=x\times48
144=48x 144=48x
x=3 x=3
βœ… ANSWER:
33
🟠 EXAMPLE 23: GP WITH A FRACTIONAL RATIO QUESTION: Find the 5th term of the GP:
32,16,8,4,… 32,16,8,4,\ldots
🟒 SOLUTION:
a=32 a=32
r=12 r=\frac12
Using:
an=arnβˆ’1 a_n=ar^{n-1}
a5=32(12)4 a_5 = 32\left(\frac12\right)^4
=32Γ—116 =32\times\frac{1}{16}
=2 =2
βœ… ANSWER:
22
🟠 EXAMPLE 24: GP WITH NEGATIVE RATIO QUESTION: Find the 5th term of the GP:
2,βˆ’4,8,βˆ’16,… 2,-4,8,-16,\ldots
🟒 SOLUTION:
a=2 a=2
r=βˆ’2 r=-2
Using:
an=arnβˆ’1 a_n=ar^{n-1}
a5=2(βˆ’2)4 a_5 = 2(-2)^4
=2Γ—16 =2\times16
=32 =32
βœ… ANSWER:
3232
🟠 EXAMPLE 25: FIND THE NUMBER OF TERMS IN GP QUESTION: How many terms are there in the GP:
3,6,12,…,384 3,6,12,\ldots,384
🟒 SOLUTION: Using:
an=arnβˆ’1 a_n=ar^{n-1}
Here:
a=3 a=3
r=2 r=2
an=384 a_n=384
Therefore:
384=3(2)nβˆ’1 384=3(2)^{n-1}
128=2nβˆ’1 128=2^{n-1}
Since:
128=27 128=2^7
we get:
nβˆ’1=7 n-1=7
n=8 n=8
βœ… ANSWER:
8terms8 terms
🟠 EXAMPLE 26: AP VS GP QUESTION: Determine whether the sequence below is an AP or GP:
5,10,15,20,… 5,10,15,20,\ldots
🟒 SOLUTION: Check the differences:
10βˆ’5=5 10-5=5
15βˆ’10=5 15-10=5
20βˆ’15=5 20-15=5
The difference is constant. Therefore, it is an AP. It is not a GP because the ratios are not constant. βœ… ANSWER:
ArithmeticProgressionArithmetic Progression
🟠 EXAMPLE 27: FIND THE SUM OF AN AP QUESTION: Find the sum of the first 25 terms of:
4,7,10,13,… 4,7,10,13,\ldots
🟒 SOLUTION:
a=4 a=4
d=3 d=3
n=25 n=25
Using:
Sn=n2[2a+(nβˆ’1)d] S_n = \frac{n}{2}[2a+(n-1)d]
S25=252[2(4)+24(3)] S_{25} = \frac{25}{2}[2(4)+24(3)]
=252[8+72] = \frac{25}{2}[8+72]
=252Γ—80 = \frac{25}{2}\times80
=1000 =1000
βœ… ANSWER:
1,0001,000
🟠 EXAMPLE 28: SUM TO INFINITY QUESTION: Find the sum to infinity of:
12+6+3+32+β‹― 12+6+3+\frac32+\cdots
🟒 SOLUTION:
a=12 a=12
r=612 r=\frac{6}{12}
=12 =\frac12
Since:
∣r∣<1 |r|<1
we can use:
S∞=a1βˆ’r S_\infty = \frac{a}{1-r}
=121βˆ’12 = \frac{12}{1-\frac12}
=1212 =\frac{12}{\frac12}
=24 =24
βœ… ANSWER:
2424