Question 1 Report
If the sixth term of an Arithmetic Progression (A.P) is 37 and the sum of the first six terms is 147, find the
(a) first term;
(b) sum of the first fifteen terms.
For an A.P. with first term \(a\) and common difference \(d\):
Sixth term: \(a + 5d = 37\) ... (1)
Sum of first six terms: \(S_6 = \dfrac{6}{2}(2a + 5d) = 3(2a + 5d) = 147\), so \(2a + 5d = 49\) ... (2)
(a) Subtract (1) from (2):
\[(2a + 5d) - (a + 5d) = 49 - 37 \Rightarrow a = 12.\]
The first term is \(a = \mathbf{12}\).
From (1): \(12 + 5d = 37 \Rightarrow 5d = 25 \Rightarrow d = 5\).
(b) Sum of the first fifteen terms:
\[S_{15} = \frac{15}{2}\big(2a + 14d\big) = \frac{15}{2}\big(2(12) + 14(5)\big) = \frac{15}{2}(24 + 70) = \frac{15}{2}(94) = 705.\]
Therefore \(S_{15} = \mathbf{705}\).
Answer Details
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