59-09 Basics of Arithmetic & Geometric Progression

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Note: 

1) To find three consecutive terms in GP, consider the terms (a/r),  (a), and (ar).

2) To find four consecutive terms in GP, consider the terms (a/r³),  (a/r), (ar), and (ar³).

3) To find five consecutive terms in GP, consider the terms (a/r²),  (a/r), (a), (ar), and (ar²).

A) Find three consecutive three terms in GP if the sum of the first and the third term is 35 and the product of all three terms is 2744.

Solution:

1) Let " a " be the first term and " r " be the common ratio of GP.

2) Let three consecutive terms in GP be  (a/r),  (a), and (ar).

3) According to the problem, 

                 (a/r) (a)(ar) = 2744

                             (a³) = 4 x 686

                             (a³) = 4 x 2 x 343

                             (a³) = 4 x 2 x 7 x 49

                             (a³) = (2 x 7)³

                               (a) = (2 x 7)

                               (a) = (14)

4) Simillarly,

                        a/r + ar = 35

                     a (1/r + r) = 35

                   14 (1/r + r) = 35

                        (1/r + r) = 35/14

                        (1/r + r) = 5/2

                        (1/r + r) = 2 + 1/2

5) This shows that r = 2 or 1/2

6) Answer: Taking r = 2 and a = 14, the three consecutive terms are 7, 14, and 28.

                    Taking r = 1/2 and a = 14, the three consecutive terms are 28, 14, and 7.

Note:

1) Arithmetic Mean: The three terms p, q, and r in AP give " q " as the arithmetic means between p and q. So,

                 q - p = r - q

                 q + q = r + p

                    2 q = r + p

                       q = (r + p) / 2

2) Geometric Mean: The three terms p, q, and r in GP give " q " as the geometric mean between p and r. So,

                 q / p = r / q

                 q x q = r x p

                     q² = r x p

                       q = (r p)^1/2

^{Some Important Problems with AP & GP}

^{A) In AP  6, 12, 18, ....., how many terms show the sum as 7650.}

^Solution:

1)   Here, a = 6, d = 6 and S_n = 7650.

2)   We know that 

3) Answer: Total number of terms in AP having a sum of 7650 is 50.

Anil Satpute

58-08 Basics of Arithmetic & Geometric Progression

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Example:

A) If s₄  =  156 and s₈  =  97656, find a and r.

Solution:

  1) Let " a " be the 1st tern and " r " be the common ratio. So we have,

6) Answer: We have a = 1 when r = 5 and a = -3/2 when r = - 5

Note: Here the division of 97656 by 156 is complicated as we should not use the calculator. Please see the following simple method of calculation so that you will save a lot of time and you can't make any calculation mistakes. In the following method, you may observe that there are a few more steps to be done but all these steps are simple to understand the calculation. You may also notice that this calculation process is really very simple. So we will solve the same problem with another Simple Method.

Solution:

 1) Let " a " be the 1st tern and " r " be the common ratio. So we have,

 
 

6) Answer: We have a = 1 when r = 5 and a = -3/2 when r = - 5

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

Click here for the next basics.

Anil Satpute

57-07 Basics of Arithmetic & Geometric Progression

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

If t₁, t₂, t₃, ... t_n-1, t_n are the terms in GP then t_2/t_{1 =} t_3/t_{2 =} t_4/t_{3 = ..... =} r.
Find the n^th 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 the n^th term of the GP 2, 6, 18...

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

3) Answer: The nth term of the GP is t_n = 2 x (3) ^{(n – 1)}

^{Note: Above formula can be used to find any term in GP.}

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

Solution:
1) Here a = 1 and r = 2.
2) So      t_n = a x (r) ^{(n – 1)}
                t_n = 1 x (2) ^{(n – 1)}
                t_n = (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: 7^th and 13^th 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 1^st n terms of the GP.

Solution:
1) Let 1^st term be "a" and the Common Ratio be "r".
2) We know that 
                          s_n  =  a + a r + a r² + a r³ + ------ + a r ^{(n – 1)}
                        r s_n  =         a r + a r² + a r³ + ------ + a r ^{(n – 1)} + a r ^{(n)
                              ( - )                 ( - )      ( - )      ( - )                        ( - )              ( - )
            ------------------------------------------------------------------------------------------------------------}
             ( 1 - r )  s_n  =  a  + -------------------------------------------+ - a r ⁽ⁿ⁾ 
             ( 1 - r )  s_n  =  a  - a r ⁽ⁿ⁾ 
                          s_n  =  [a (1  -  rⁿ ) ] / ( 1 - r )
Note: 
1) If r < 1 then s_n  =  [a (1  -  rⁿ ) ] / ( 1 - r )
2) If r > 1 then s_n =  [a ( rⁿ  - 1 ) ] / ( r  - 1 )
3) Very Important:
     If r = 1 then we can't use the above formula as we get " 0 " in the denominator. So in this case, we have 
       s_n  =  a + a r + a r² + a r³ + ------ + a r ^{(n – 1)      put r = 1 we get,}
       s_n  =  a + a + a + ----- + n-times
       s_n  =  n x a

Examples of 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, s_n =  [a ( rⁿ  - 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 the Simple Method of calculations will be published in the Next Blog.

Click here for the next basics.

Anil Satpute

56-06 Basics of Arithmetic & Geometric Progression

Click here for the previous basics.

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 the middle 2 terms is -3.

Solution:

1) Let " a " be the first term & " d " be a 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   =  d^{2
                                        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 S_n =  S_m then find S_m+n.

Solution:

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

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

C) How many 3-digit natural numbers leave the remainder of 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_n = 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: The number of 3-digit natural numbers leaving the remainder of 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_m ]= [ n * t_n ] then find t_m+n

In the next part, we will see a few examples and some essential formulae.

Click here for the next basics.

Anil Satpute

55-05 Basics of Arithmetic & Geometric Progression

Click here for the previous basics.

Examples of Summation.

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

Solution:
1) For the natural number which is divisible by 6, a = 6, d = 6.
2) Here we can use the formula S_n = 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 numbers 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 t_n =  a + (n-1) d 
     So,   t_n =  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 S_n = n * [  a + t_n ] / 2
          S_n = 75 * [ 101 + 249 ] / 2
          S_n = 75 * [ 350 ] / 2
          S_n = 75 * [ 175 ]
          S_n = 13125
4) Answer: The sum of odd natural numbers between 100 & 250, is S_n = 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_n = 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 S_n using the formula S_n = n * [  a + t_n ] / 2.
                     S_n = 225 * [ 100 + 996 ] / 2
                     S_n = 225 * 4 * [ 25 + 249 ] / 2
                     S_n = 225 * 4 * [ 274 ] / 2
                     S_n = 225 * 4 * 137
                     S_n = 900 * 137
                     S_n = 123300
6) Answer: The sum of all 3-digit natural numbers which are divisible by 4 is S_n = 123300.

G) If t₁₇ = 59 and t₃₉ = 141 then find s_{55 .}
Solution:

1) Let " a " be the first term & " d " be a common difference.
2) We know that, t_n = 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    S_n = 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.

In the next part, we will see a few examples and some essential formulae.

Click here for the next basics.

Anil Satpute