## Tuesday, April 30, 2013

### 57-07 Basics of Arithmetic & Geometric Progression (Grades 9 to 12) Part-07

Geometric Progression (GP) is a sequence: The terms in GP are obtained simply multiplying by Non-Zero constant. This constant is known as "Common Ratio". In short, 2nd term divided by 1st term is same as 3rd term divided by 2nd term and so on for all the terms of the sequence called as Geometric Progression (GP).

If t1, t2, t3, ... tn-1, tn are the terms in GP then t2/t1 = t3/t2 = t4/t3 = ..... = r.
Find nth term of GP.

Solution:
1) Let " a " be the 1st term and " r " be the common ratio.
2) By the definition of GP,

Now we will see some examples of GP.

A) Find nth term of the GP 2, 6, 18...

Solution:
1) Here a = 2 and r = 6/2 = 3.
2) We know that
tn  =  a x (r)( n – 1 )
tn  =  2 x (3)( n – 1 )

3) Answer: The nth term of the GP is tn = 2 x (3) (n – 1)

Note: Above formula can used to find any term in GP.

B) Find 7th and 13th term of the GP if a = 1 and r = 2

Solution:
1) Here a = 1 and r = 2.
2) So      tn = a x (r) (n – 1)
tn = 1 x (2) (n – 1)
tn = (2) (n – 1)
3)            t7 = (2) (7 – 1)
t7 = (2) (6)
t7 = 64

4)           t13 = (2) (13 – 1)
t13 = (2) (12)
t13 = 4096
5) Answer: 7th and 13th term of the GP are t7 = 64 and t13 = 4096.

Remember: For any GP, t1 = a, t2 = a r, t3 = a (r) (2), t4 = a (r) (2)  so on.

A) Find the sum of 1st n terms of the GP.

Solution:
1) Let 1st term  be "a" and Common Ratio be "r".
2) We know that
sn  =  a + a r + a ra r+ ------ + a r (n – 1)
r sn  =         a r + a ra r+ ------ + a r (n – 1) + a r (n)
( - )                 ( - )      ( - )      ( - )                        ( - )              ( - )
------------------------------------------------------------------------------------------------------------
( 1 - r )  sn  =  a  + -------------------------------------------+ - a r (n)
( 1 - r )  sn  =  a  - a r (n)
sn  =  [a (1  -  r) ] / ( 1 - r )
Note:
1) If r < 1 then sn  =  [a (1  -  r) ] / ( 1 - r )
2) If r > 1 then sn =  [a ( r - 1 ) ] / ( r  - 1 )
3) Very Important:
If r = 1 then we can't use above formula as we get " 0 " in the denominator. So in this case          we have
sn  =  a + a r + a ra r+ ------ + a r (n – 1)      put r = 1 we get,
sn  =  a + a + a + ----- + n-times
sn  =  n x a

Examples on Summation:

A) Find s10  of  the GP 2, 4, 8, 16 -----

Solution:
1) Here a = 2 and r = 4 / 2 = 2
2) Since r = 2 > 1, we can use the formula, sn =  [a ( r - 1 ) ] / ( r  - 1 )
3) So,           s10 =  [2 ( 210  - 1 ) ] / ( 2  - 1 )
s10 =  [2 ((25)2  - 1 ) ]
s10 =  [2 ((32)2  - 1 ) ]
s10 =  [2 (1024  - 1 ) ]
s10 =  [2 (1023) ]
s10 =  [2046
4) Answer: Here s10 =  [2046

Note: Please understand the simple method of finding the square of any two digit number:

The Simple method to find the square will be published in some other Blog.

Some more Problems with Simple Method of calculations will be published in the Next Blog.

Anil Satpute

## Saturday, April 27, 2013

### 56-06 Basics of Arithmetic & Geometric Progression (Grades 9 to 12) Part-06

Note:
1) To find 3 consecutive terms in an AP, let the terms be (a-d), (a), (a+d).
2) To find 5 consecutive terms in an AP, let the terms be (a-2 d), (a-d), (a), (a+d), (a+2 d).
3) To find 4 consecutive terms in an AP, let the terms be (a-3 d), (a-d), (a+d), (a+3 d).

Examples to find the consecutive terms in an AP...

A) Find 4 consecutive terms in an AP whose sum is 4 and the product of middle 2 terms is -3.

Solution:

1) Let " a " be the first term & " d " be the common difference.
2) As we need to find 4 consecutive terms of an AP, Let the terms be (a-3 d), (a-d), (a+d), (a+3 d).
3) According to the problem,
(a-3 d) + (a-d) + (a+d) + (a+3 d) = 4
4 a   = 4
a   = 1
4) According to the problem,
( a - d ) ( a + d ) = - 3
a2 – d2  = - 3
1 + 3   =  d2
4   =  d2
d = 2 or d = -2
5) Answer: Taking a = 1 & d = 2, the terms are - 5, - 1, 3, 7.
Taking a = 1 & d = - 2, the terms are 7, 3, - 1, - 5.

B) In an AP, if Sn =  Sm then find Sm+n.

Solution:

1)  Let " a " be the first term & " d " be the common difference.
2)      Sn  =  (n/2) * [ 2 a + ( n - 1 ) d ]       ----------------  (1)
Sm = (m/2) * [ 2 a + ( m - 1 ) d ]      ----------------- (2)
3)  According to the problem, as Sn =  Sm,

4) Sm+n  =  (m+n/2) * [ 2 a + ( m + n - 1 ) d ]
Sm+n  =  (m+n/2) * [ 0 ]
Sm+n  =  0
5) Answer: Here Sm+n  =  0

C) How many 3-digit natural numbers leave the remainder 3 when divided by 4.

Solution:

1) We know that the lowest 3 digit number is 100 & the largest one is 999.
2) So the numbers leaving 3 as the remainder when divided by 4 between 100 & 999 are
107, 111, 115, ----, 999.
3) Here a = 107, d = 4 and last term = 999.
4) First we will find the total number of terms, so here,
t= a + ( n - 1 ) d, we have,
999 = 107 + ( n - 1 ) * 4
4 *  ( n - 1 )  =  999 - 107
4 *  ( n - 1 )  =  892
( n - 1 )  =  892 / 4
( n - 1 )  =  223
n    =  223 + 1
n    =  224
6) Answer: Number of 3-digit natural numbers leaving the remainder 3 when divided by 4 is n =  224.

Note: Please solve the following problem as "Happy Work". You may take the help of problem (B)  to solve the following problem. You may ask your doubt about this problem simply by emailing me at anil@7pute.com

Problem :
If [ m * t]= [ n * t] then find tm+n

Anil Satpute

## Thursday, April 25, 2013

### 55-05 Basics of Arithmetic & Geometric Progression (Grades 9 to 12) Part-05

Examples on Summation.

D) Find the sum of first 100 natural numbers which are divisible by 6.

Solution:
1) For the natural number which are divisible by 6, a = 6, d = 6.
2) Here we can use the formula Sn = n * [ 2 a + (n -1) d ) ] / 2
S100 = 100 * [ 2 (6) + (100 -1) (6) ) ] / 2
S100 = 6 * 100 * [ 2 + 99 ] / 2
S100 = 3 * 100 * [ 101 ]
S100 = 100 * [ 303 ]
S100 =  30300
3) Answer: The sum of the first 100 natural numbers which are divisible by 6, is S100 = 30300.

E) Find the sum of all odd natural numbers between 100 and 250.

Solution:
1) For odd natural number between 100 and 250, a = 101, d = 2.
2) First, we will find the total number of terms between 100 & 250 which are odd.
Here we use the formula tn =  a + (n-1) d
So,   tn =  101 + (n -1) * 2
249   =  101 + (n -1) * 2
(n - 1)   =  (249 - 101) / 2
(n - 1)   =  (148) / 2
(n - 1)   =  74
n        =  75
3) Now we will use the formula Sn = n * [  a + tn ] / 2
Sn = 75 * [ 101 + 249 ] / 2
Sn = 75 * [ 350 ] / 2
Sn = 75 * [ 175 ]
Sn = 13125
4) Answer: The sum of odd natural numbers between 100 & 250, is Sn = 13125.

F) Find the sum of all 3-digit natural numbers which are divisible by 4.

Solution:

1) We know that the lowest 3 digit number is 100 & the largest one is 999.
2) So the numbers divisible by 4 between 100 & 999 are
100, 104, 108, ----, 992, 996.
3) Here a = 100, d = 4 and last term = 996.
4) First we will find the total number of terms, so here,
t= a + ( n - 1 ) d, we have,
996 = 100 + ( n - 1 ) * 4
4 *  ( n - 1 )  =  996 - 100
4 *  ( n - 1 )  =  896
( n - 1 )  =  896 / 4
( n - 1 )  =  224
n    =  224 + 1
n    =  225
5) Now we will find Sn using the formula Sn = n * [  a + tn ] / 2.
Sn = 225 * [ 100 + 996 ] / 2
Sn = 225 * 4 * [ 25 + 249 ] / 2
Sn = 225 * 4 * [ 274 ] / 2
Sn = 225 * 4 * 137
Sn = 900 * 137
Sn = 123300
6) Answer: The sum of all 3-digit natural numbers which are divisible by 4 is Sn = 123300.

G) If t17 = 59 and t39 = 141 then find s55 .
Solution:

1) Let " a " be the first term & " d " be the common difference.
2) We know that, t= a + ( n - 1 ) d
t17 = a + ( 17 - 1 ) d = 59 and  t39 = a + ( 39 - 1 ) d = 141,
So we have,
a + ( 16 ) d =    59
a + ( 38 ) d =  141
-----------------------------------------
2 a + ( 54 ) d = 200            --------------------- (1)

3) Here    Sn = n * [ 2 a + (n -1) d ) ] / 2
S55 = 55 * [ 2 a + (54) d ) ] / 2
S55 = 55 * [ 200 ] / 2
S55 = 55 * [ 100 ]
S55 = 5500
4) Answer: Here S55 = 5500.

Few more Problems on AP will be published in the next Blog.

Anil Satpute