Saturday, April 27, 2013

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

Blog-56

Dear Students,

Download following files for learning Formulas more effectively. Take the print out of these files and write your answers daily to improve your scores in 10th standard/grade.


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1) Algebra Formulas
2) Geometry Formulas


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 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

Wish you Happy "Happy working"

Anil Satpute