Question

the sum of the series in A. P. is 40 1/2, the common difference is 2 1/2 and the last term is 13. Find the first term and the number of terms in the series.

Answer 1

Answer :

First term is 0.5 and number of terms in series = 6

Step by Step working :

2 unknowns: A1 and N, we need 2 equations;

 

N *(A1 + 13)/2 = 40.5 <--- formula for the sum of arithmetic sequence; this is 1st equation

 

 

13 = A1 + (2.5)*(N-1) <---- last term = first term + (N-1)*(common difference);  this is 2nd equation

 

 

Multiplies 1st equation by 2 to clear the fraction:        N(A1+13) = 81

Multiplies 2nd equation by 10 to clear the decimal:  130 = 10*A1 + 25(N-1)

 

 

2nd equation says:  130 = 10*A1 + 25*N - 25

                            155 = 10*A1 + 25*N

 

                           155 - 10*A1 = 25*N

 

                         (155 - 10*A1)/25 = N

 

                         (31 - 2*A1)/5 = N <--- divides by 5

 

Substitutes this into the first equation:

N(A1+13) = 81

(31 - 2*A1)*(a1+13)/5 = 81

 

(31 - 2*A1)*(a1+13) = 405

 

To simplify the notation let X=A1

 

(31 - 2X)(X+13) = 405

 

31X + 403 - 2x^2 - 26x = 405

 

5x + 403 - 2x^2 = 405

 

0 = 2x^2 - 5x  + 2

 

0 = ( 2x     -   1)( x  -   2  )

 

2x-1 = 0  ---> x = 1/2

x-2 = 0  --->x = 2    

 

x=2 does not produce the correct sequence

 

x=1/2 is the first term

the sequence is { 1/2, 3, 5.5, 8, 10.5, 13}

 

N=6 terms in the sequence and their sum is in fact 40.5

6*(1/2 + 13)/2 = 3*13.5 = 3* (13 + 1/2) = 39 + 1.5 = 40.5