In previous post "Number System: Important Facts and Formulae" you learn about some important fact and formula that is simple but very important and essential for student or any person. A student learn how to do simple to complex calculation with help of simple knowledge of number system and formula whereas common people make life simple. How? if you ask this that means you did not understand, "Did you?"
Yes, you did not understand? Okay, don't worry, let me explain. Everybody do some shopping in everyday life, whether it is small or big it doesn't matter. Suppose if he hasn't knowledge of simple formula or simple math, how difficult, his life would be? He will be cheated (most of time) because he will have to pay whatever shopkeeper say, but if he has knowledge then the shopkeeper cannot cheat him. He can immediately, count and calculated with help of simple formula and check out what is right. This is very simple example of use of simple facts and formula in daily life and to understand life without math is impossible.
Today, we will start from there where we left in last session or post. Are you ready, "aren't you?" Okay, In this section, you will get information about simple introduction of Progression and it's use by lots of solved example. This is different from Rational number, or Irrational number,
X. PROGRESSION: Simple Introduction
A
succession of numbers formed and arranged in a definite order according to
certain
definite
rule, is called a progression.
1. Arithmetic Progression (A.P.) : If each term of a progression differs from its
preceding
term by a constant, then such a progression is called an arithmetical
progression.
This constant difference is called the common difference of the A.P.
An
A.P. with first term a and common difference d is given by a, (a + d), (a +
2d),(a +
3d),.....
The
nth term of this A.P. is given by Tn =a (n - 1) d.
The sum of n terms of this A.P.
Sn
= n/2 [2a + (n - 1) d] =
n/2 (first term + last term).
SOME IMPORTANT RESULTS :
(i)
(1 + 2 + 3 +…. + n) =n(n+1)/2
(ii)
(l2 + 22 + 32 + ... + n2) = n (n+1)(2n+1)/6
(iii)
(13 + 23 + 33 + ... + n3) =n2(n+1)2
2. Geometrical Progression (G.P.) : A progression of numbers in which every term
bears
a constant ratio with its preceding term, is called a geometrical progression.
The
constant ratio is called the common ratio of the G.P. A G.P. with first term a
and
common
ratio r is :
a,
ar, ar2,
In
this G.P. Tn = arn-1
sum
of the n terms, Sn= a(1-rn)
(1-r)
On the basis of these formula there are some solved exaple.
SOLVED EXAMPLES:-
Ex. 1. Simplify : (i) 8888 + 888 + 88 + 8
(ii) 11992 - 7823 - 456
Sol. i ) 8888
ii)
11992 - 7823 - 456 = 11992 - (7823 + 456)
888
= 11992 - 8279
= 3713 - 88
7823 11992 +
8 + 456 - 8279
= 9872
8279 3713
Ex. 2, What value will replace the question mark in each of the
following equations
?
(i) ? - 1936248 = 1635773 (ii) 8597 - ? = 7429 - 4358
Sol. (i) Let x -
1936248=1635773.
Then, x = 1635773 + 1936248
=3572021.
(ii)
Let 8597 - x = 7429 - 4358.
Then,
x = (8597 + 4358) - 7429
= 12955 - 7429
= 5526.
Ex. 3. What could be the maximum value of Q in the following
equation? 5P9
+ 3R7 + 2Q8 = 1114
Sol. We may analyse the given
equation as shown : 1 2
Clearly,
2 + P + R + Q = ll. 5 P 9
So,
the maximum value of Q can be 3 R 7
(11
- 2) i.e., 9 (when P = 0, R = 0);
2 Q 8 11
1 4
Ex. 4. Simplify : (i) 5793405 x 9999 (ii) 839478 x 625
Sol. (i)5793405x9999=5793405(10000-1)
=57934050000-5793405
=57928256595.b
(ii)
839478 x 625 = 839478 x 54
= 8394780000
= 524673750.16
Ex. 5. Evaluate : (i) 986 x 237 + 986 x 863 (ii) 983 x 207 - 983 x
107
Sol. (i) 986 x 137 + 986 x 863
= 986 x (137 + 863)
= 986 x 1000 = 986000.
(ii)
983 x 207 - 983 x 107 = 983 x (207 - 107)
= 983 x 100 = 98300.
Ex. 6. Simplify : (i) 1605 x 1605 ii) 1398 x 1398
Sol. (i) 1605 x 1605 = (1605)2
=
(1600 + 5)2
= (1600)2 + (5)2 + 2 x 1600 x 5
=
2560000 + 25 + 16000
= 2576025.
(ii)
1398 x 1398 - (1398)2 = (1400 - 2)2
= (1400)2 + (2)2 - 2 x 1400 x 2
=1960000
+ 4 - 5600
= 1954404.
Ex. 7. Evaluate : (313 x 313 + 287 x 287).
Sol. (a2 + b2) = 1/2 [(a + b)2
+ (a- b)2]
(313)2
+ (287)2 = 1/2 [(313 + 287)2 + (313 - 287)2]
= ½[(600)2 + (26)2]
=
1/2 (360000 + 676)
= 180338.
Ex. 8. Which of the following are prime numbers ?
(i) 241 (ii) 337 (Hi) 391 (iv) 571
Sol.(i) Clearly, 16 > Ö241.
Prime numbers less than 16 are 2, 3, 5, 7, 11, 13.
241
is not divisible by any one of them.
241
is a prime number.
(ii)
Clearly, 19>Ö337. Prime numbers less than 19 are 2, 3, 5, 7, 11,13,17.
337
is not divisible by any one of them.
337
is a prime number.
(iii)
Clearly, 20 > Ö39l". Prime numbers less than 20 are 2, 3, 5, 7, 11, 13,
17, 19.
We
find that 391 is divisible by 17.
391
is not prime.
(iv)
Clearly, 24 > Ö57T. Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17,
19, 23.
571
is not divisible by any one of them.
571
is a prime number.
Ex. 9. Find the unit's digit in the product (2467)163 x (341)72.
Sol. Clearly, unit's digit in
the given product = unit's digit in 7153 x 172.
Now,
74 gives unit digit 1.
7152
gives unit digit 1,
7153
gives unit digit (l x 7) = 7. Also, 172 gives unit digit 1.
Hence,
unit's digit in the product = (7 x 1) = 7.
Ex. 10. Find the unit's digit in (264)102 + (264)103
Sol. Required unit's digit =
unit's digit in (4)102 + (4)103.
Now,
42 gives unit digit 6.
(4)102
gives unjt digit 6.
(4)103
gives unit digit of the product (6 x 4) i.e., 4.
Hence,
unit's digit in (264)m + (264)103 = unit's digit in (6 + 4) = 0.
Ex. 11. Find the total number of prime factors in the expression
(4)11 x (7)5 x (11)2.
Sol. (4)11x (7)5 x (11)2 = (2 x
2)11 x (7)5 x (11)2
= 211 x 211 x75x 112
= 222 x 75 x112
Total
number of prime factors = (22 + 5 + 2) = 29.
Ex.12. Simplify : (i) 896 x 896 - 204 x 204
(ii) 387 x 387 + 114 x 114 + 2 x 387 x 114
(iii) 81 X 81 + 68 X 68-2 x 81 X 68.
Sol.(i) Given exp = (896)2 -
(204)2
= (896 + 204) (896 - 204)
= 1100 x 692
= 761200.
(ii)
Given exp = (387)2+ (114)2+ (2 x 387x 114)
=
a2 + b2 + 2ab, where a
= 387,b=114
=
(a+b)2 = (387 + 114 )2
= (501)2 = 251001.
(iii)
Given exp = (81)2 + (68)2 – 2x 81 x 68
= a2 + b2 – 2ab,
Where a =81,b=68
=
(a-b)2 = (81 –68)2 = (13)2 = 169.
Ex.13. Which of the following numbers is divisible by 3 ?
(i) 541326 (ii) 5967013
Sol.(i) Sum of digits in
541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3.
Hence,
541326 is divisible by 3.
(ii)
Sum of digits in 5967013 =(5+9 + 6 + 7 + 0+1 +3) = 31, which is not divisible
by 3.
Hence,
5967013 is not divisible by 3.
Ex.14.What least value must be assigned to * so that the number
197*5462 is r 9 ?
Sol. Let the missing digit be
x.
Sum
of digits = (1 + 9 + 7 + x + 5 + 4 + 6 +»2) = (34 + x).
For
(34 + x) to be divisible by 9, x must be replaced by 2 .
Hence,
the digit in place of * must be 2.
Ex. 15. Which of the following numbers is divisible by 4 ?
(i) 67920594 (ii) 618703572
Sol.(i) The number formed by
the last two digits in the given number is 94, which is not
divisible
by 4.
Hence,
67920594 is not divisible by 4.
(ii)
The number formed by the last two digits in the given number is 72, which is
divisible
by 4.
Hence,
618703572 is divisible by 4.
Ex. 16. Which digits should come in place of * and $ if the number
62684*$ is
divisible by both 8 and 5 ?
Sol. Since the given number is
divisible by 5, so 0 or 5 must come in place of $. But, a
number
ending with 5 is never divisible by 8. So, 0 will replace $.
Now,
the number formed by the last three digits is 4*0, which becomes divisible by
8, if
*
is replaced by 4.
Hence,
digits in place of * and $ are 4 and 0 respectively.
Ex. 17. Show that 4832718 is divisible by 11.
Sol. (Sum of digits at odd
places) - (Sum of digits at even places)
=
(8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11.
Hence,
4832718 is divisible by 11.
Ex. 18. Is 52563744 divisible by 24 ?
Sol. 24 = 3 x 8, where 3 and 8
are co-primes.
The
sum of the digits in the given number is 36, which is divisible by 3. So, the
given
number is divisible by 3.
The
number formed by the last 3 digits of the given number is 744, which is
divisible
by 8. So, the given number is divisible by 8.
Thus,
the given number is divisible by both 3 and 8, where 3 and 8 are co-primes.
So,
it is divisible by 3 x 8, i.e., 24.
Ex. 19. What least number must be added to 3000 to obtain a number
exactly
divisible by 19 ?
Sol. On dividing 3000 by 19, we
get 17 as remainder.
Number
to be added = (19 - 17) = 2.
Ex. 20. What least number must be subtracted from 2000 to get a
number exactly
divisible by 17 ?
Sol. On dividing 2000 by 17, we
get 11 as remainder.
Required
number to be subtracted = 11.
Ex. 21. Find the number which is nearest to 3105 and is exactly
divisible by 21.
Sol. On dividing 3105 by 21, we
get 18 as remainder.
Number
to be added to 3105 = (21 - 18) - 3.
Hence,
required number = 3105 + 3 = 3108.
Ex. 22. Find the smallest number of 6 digits which is exactly
divisible by 111.
Sol. Smallest number of 6
digits is 100000.
On
dividing 100000 by 111, we get 100 as remainder.
Number
to be added = (111 - 100) - 11.
Hence,
required number = 100011.-
Ex. 23. On dividing 15968 by a certain number, the quotient is 89
and the remainder
is 37. Find the divisor.
Dividend
- Remainder 15968-37
Sol. Divisor =
-------------------------- = ------------- = 179.
.Quotient
89
Ex. 24. A number when divided by 342 gives a remainder 47. When
the same
number ift divided by 19, what would be the remainder ?
Sol. On dividing the given
number by 342, let k be the quotient and 47 as remainder.
Then,
number – 342k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9.
The
given number when divided by 19, gives (18k + 2) as quotient and 9 as
remainder.
Ex. 25. A number being successively divided by 3, 5 and 8 leaves
remainders 1, 4
and 7 respectively. Find the respective remainders if the order of
divisors be
reversed,
Sol.3 X - 5
y - 18
z - 4
1
- 7
z
= (8 x 1 + 7) = 15; y = {5z + 4) = (5 x 15 + 4) = 79; x = (3y + 1) = (3 x 79 +
1) = 238.
Now,
8
238
5
29 - 6
3
5 - 4
1
- 9,
Respective
remainders are 6, 4, 2.
Ex. 26. Find the remainder when 231 is divided by 5.
Sol. 210 = 1024. Unit digit of
210 x 210 x 210 is 4 [as 4 x 4 x 4 gives unit digit 4].
Unit
digit of 231 is 8.
Now,
8 when divided by 5, gives 3 as remainder.
Hence,
231 when divided by 5, gives 3 as remainder.
Ex. 27. How many numbers between 11 and 90 are divisible by 7 ?
Sol. The required numbers are
14, 21, 28, 35, .... 77, 84.
This
is an A.P. with a = 14 and d = (21 - 14) = 7.
Let
it contain n terms.
Then,
Tn = 84 => a + (n - 1) d = 84
=>
14 + (n - 1) x 7 = 84 or n = 11.
Required
number of terms = 11.
Ex. 28. Find the sum of all odd numbers upto 100.
Sol. The given numbers are 1,
3, 5, 7, ..., 99.
This
is an A.P. with a = 1 and d = 2.
Let
it contain n terms. Then,
1
+ (n - 1) x 2 = 99 or n = 50.
Required
sum = n (first term + last term)
2 =
50 (1 + 99) = 2500.
2
Ex. 29. Find the sum of all 2 digit numbers divisible by 3.
Sol. All 2 digit numbers
divisible by 3 are :
12,
51, 18, 21, ..., 99.
This
is an A.P. with a = 12 and d = 3.
Let
it contain n terms. Then,
12
+ (n - 1) x 3 = 99 or n = 30.
Required
sum = 30 x (12+99) = 1665.
2
Ex.30.How many terms are there in 2,4,8,16……1024?
Sol.Clearly 2,4,8,16……..1024
form a GP. With a=2 and r = 4/2 =2.
Let
the number of terms be n . Then
2
x 2n-1 =1024
or 2n-1 =512 = 29.
n-1=9
or n=10.
Ex. 31. 2 + 22 + 23 + ... + 28 = ?
Sol. Given series is a G.P.
with a = 2, r = 2 and n = 8.
sum
= a(rn-1) = 2 x (28 –1) = (2 x 255)
=510 (r-1)
(2-1)
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