 Knud Dec 24, 2020

# The sum of the series in AP is 82.5 thecommon different is 2.5 and the last termis 13, find thefind the 1st ferm & the number of termsIn the series​

The time for answering the question is over

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

328 Joanne Gonzalez
Dec 24, 2020 