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:
(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.
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