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How to calculate two APs with the same first term and last term but different common difference. Two APs have the same first and last terms. The first AP has 21 terms with a common difference of 9. How many terms has the other AP if its common difference is 4?  Solution Since the first term and the last term of the two APs are the same So let their equal first and last terms be x and y Recall the nth term general formula of AP Tn = a +(n-1)d Solve for the first AP Tn = y (last term) a = x (First term) n = 21 (This is the number of terms) d = 9 (This is the common difference between terms)  Substitute y, a, 21, and 9 for Tn, a, n, and d in the equation formula below Tn = a +(n-1)d y = x + (21-1)9 y = x +20 x 9   y  = x + 180   ---------- eqn(1)  Solve for the Second AP Tn =  y  (last term) a =  x  (First term) n = ? d = 4 (This is the common difference between terms)  Substitute  y,  a,  and 4 for Tn, a, and d in the equation formula below Tn = a +(n-1)d y  =  x +  (n-1)4 y  =  x  +4n

Question   Find the number of terms in an Ap given that its first and last terms are a and 37a respectively and that its common difference is 4a. Solution nth term of AP Tn = a +(n-1)d From the question First term a = a Last term Tn = 37a Common difference d = 4a Number of terms n =? Substitute a, Tn and d into the nth term formula Tn = a +(n-1)d 37a = a +(n-1)4a Collect like terms 37a-a = (n-1)4a 36a = (n-1)4a Divide through by 4a 36a/4a = (n-1)4a/4a 9 = n-1 make n subject formula n = 9+1 n = 10 The AP as 10 terms

 How to find the common difference of AP when the first and last term are given Question: The first and last terms of an AP are 6.7  and 17.1 respectively. If there are 14 terms in the sequence, find its common difference.  Solution First term of the sequence a = 6.7 Last term of the sequence Tn = 17.1 Since the sequence has 14 terms, it means that the 14th term is the last term Therefore:  Tn = T14 = 17.1  Recall the nth term formula Tn = a + (n-1)d Substitute Tn for T14, a for 6.7 and n for 14 to get the formula below T14 = 6.7 + (14-1)d T14 = 6.7 + 13d Substitute T14 for 17.1 T14 = 6.7 + 13d 17.1 = 6.7 + 13d Subtract 6.7 from both sides to get 17.1 - 6.7 = 6.7-6.7 +13d 10.4 = 13d Divide through by 13 10.4/13 = 13d/13 d = 104/130 The common difference is 104/130 CHECK Substitute 104/130 for d in T14 = 6.7 + 13d  T14 = 6.7 + 13*104/130        = (6.7*130 + 13*104)/130        = (871 + 1352)/130        = 2223/130       

How to calculate the common difference when the first term and the nth term is know.  Question: The 28th term of an AP is -5. Find the common difference if its first term is 31. Solution T1 = a ( First term of the sequence) a = 31 T28 = -5 (28th term of the sequence) d = ? (common difference) Recall the nth term formula of Ap i.e  Tn = a + (n-1)d        Substitute 28 for n to get the following equation        T28 = a + (28-1)d          T28     = a + 27d        Substitute T28 for -5 and 31 for a  in the equation above               -5     = 31 + 27d          Subtract 31 from both sides of the equation         -5 -31 = 27d          -36 = 27d        Divide through by 27        -36/27  =  27d/27        d = -36/27        d = -4/3 The common difference is  -4/3 CHECK Substitute the first term and the common difference into the equation below,   T28     = a + 27d    T28    = 31 + 27 * -4/3 Divide 27 by 3 to get 9     T28   = 31 - 9 * 4     T28   = 31 - 36     T28  = - 5

Solving Simultaneous Equation By Elimination  Method 3 x + 2 y = 6   5 x  -  y = 8 In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other 3 x + 2 y = 6 ,  5 x − y = 8 5 × 3 x + 5 × 2 y = 5 × 6 , 3 × 5 x + 3 ( − 1 ) y = 3 × 8 Simplify 1 5 x + 1 0 y = 3 0 , 1 5 x − 3 y = 2 4 Subtract 1 5 x − 3 y = 2 4 from  1 5 x + 1 0 y = 3 0  by subtracting like terms on each side of the equal sign 1 5 x − 1 5 x + 1 0 y + 3 y = 3 0 − 2 4 Cancel out 15 x - 15 x to get  1 0 y + 3 y = 3 0 − 2 4 Add  1 0 y  to  3 y   1 3 y = 3 0 − 2 4 Subtract 24 from 30 to get 1 3 y = 6 Divide both side by 13 y = 1 3 6 ​ 5 x − 1 3 6 ​ = 8 5 x = 1 3 1 1 0 ​ Divide through by 5 x = 1 3 2 2 ​ x = 1 3 2 2 ​ , y = 1 3 6 ​

Solving Simultaneous equation  using matrix  Method Question: Find the values of  x and y in the equations below 3 x + 2 y = 6 5 x − y = 8 Using matrix method Put the equations in standard form and then use matrices to solve the system of equations. 3 x + 2 y = 6 , 5 x − y = 8 Write the equations in matrix form ( 3 5 ​ 2 − 1 ​ ) ( x y ​ ) = ( 6 8 ​ ) Left multiply the equation by the inverse matrix of  ( 3 5 ​ 2 − 1 ​ ) . i n v e r s e ( ( 3 5 ​ 2 − 1 ​ ) ) ( 3 5 ​ 2 − 1 ​ ) ( x y ​ ) = i n v e r s e ( ( 3 5 ​ 2 − 1 ​ ) ) ( 6 8 ​ ) The product of a matrix and its inverse is the identity matrix ( 1 0 ​ 0 1 ​ ) ( x y ​ ) = i n v e r s e ( ( 3 5 ​ 2 − 1 ​ ) ) ( 6 8 ​ ) Multiply the matrices on the left hand side of the equal sign. ( x y ​ ) = i n v e r s e ( ( 3 5 ​ 2 − 1 ​ ) ) ( 6 8 ​ ) For the  2 × 2  matrix  ( a c ​ b d ​ ) , the inverse matrix is  ( a d − b c d ​ a d − b c − c ​ ​ a d − b c − b ​ a d − b c a ​ ​ ) , so the matrix equation can be rewritten as a matrix multiplication pr