SOLUTION: Please solve this too The 14th term of an arithmetic progression is 96 while the 25th term is 173. find the a.19th term b.sum of 13th and 56th terms

Algebra ->  Sequences-and-series -> SOLUTION: Please solve this too The 14th term of an arithmetic progression is 96 while the 25th term is 173. find the a.19th term b.sum of 13th and 56th terms      Log On


   

Question 1083979: Please solve this too
The 14th term of an arithmetic progression is 96 while the 25th term is 173. find the
a.19th term
b.sum of 13th and 56th terms
Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
1.  a%5B14%5D = a%5B1%5D+%2B+13%2Ad =  96,    (1)
    a%5B25%5D = a%5B1%5D+%2B+24%2Ad = 173.    (2)

    ====> Subtract (1) from (2) (both sides). You will get

    24d - 13d = 173 - 96  ---->  11d = 77  ====>  d = 7.


2.  Then a%5B19%5D = a%5B14%5D%2B+5%2Ad = 96 + 5*7 = calculate.


3.  a%5B13%5D+%2B+a%5B56%5D = %28a%5B14%5D-d%29 + %28a%5B14%5D+%2B+%2856-14%29%2Ad%29 = (96 - 7) + (96 + 42*7) = calculate.

Solved.


There is a bunch of lessons on arithmetic progressions in this site:
    - Arithmetic progressions
    - The proofs of the formulas for arithmetic progressions
    - Problems on arithmetic progressions
    - Word problems on arithmetic progressions
    - One characteristic property of arithmetic progressions
    - Solved problems on arithmetic progressions


Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Arithmetic progressions".