The sum of the first ten terms of an Arithmetic Progression (A.P.) is 130. If the fifth term is 3 times the first term, find the:
- common difference;
- first term;
- number of terms of the A.P. if the last term is 28.
Explanation
(a) it was expected that they substitute into the formula for sum of an Arithmetic Progression to have 130 = 1\(\frac{1}{2}\) [2a+(10 - 1)d] and simplifying will yield 2a + 9d = 26 -------(1).
Also, U5 = a + (5 – 1)d = 3a and simplifying will yield a = 2d --------------------------------(2)
Substituting for a in equation (1) will yield 2(2d) + 9d = 26 and solving for d, d = 2.
(b), Substituting for d in equation (2) to get the first term:a = 2(2) = 4
(c) The number of terms required can be obtained by solving 28 = 4 + (n – 1)2 which resulted to n = 13.