The twenty-first term of an Arithmetic Progression is \(5\frac{1}{2}\) and the sum of the first twenty-one terms is \(94\frac{1}{2}\). Find the :
(a) first term ; (b) common difference ; (c) sum of the first thirty terms.
(a) first term ; (b) common difference ; (c) sum of the first thirty terms.
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Correct Answer: Option n
Explanation:
(a) \(T_{n} = a + (n - 1)d\) (terms of an AP)
\(T_{21} = a + 20d = 5\frac{1}{2}.... (1)\)
\(S_{n} = \frac{n}{2} (2a + (n - 1)d) = \frac{n}{2} (a + l)\)
Where a and l are the first and last terms respectively.
\(S_{21} = \frac{21}{2} (a + 5\frac{1}{2})\)\)
\(94\frac{1}{2} = \frac{21}{2} (a + 5\frac{21}{2})\)
\(189 = 21 (a + 5\frac{1}{2})\)
\(9 = a + 5\frac{1}{2} \implies a = 9 - 5\frac{1}{2} = 3\frac{1}{2}\)
(b) Put a in the equation (1),
\(3\frac{1}{2} + 20d = 5\frac{1}{2}\)
\(20d = 5\frac{1}{2} - 3\frac{1}{2} = 2\)
\(d = \frac{2}{20} = \frac{1}{10}\).
(c) \(S_{30} = \frac{30}{2} (2(3\frac{1}{2}) + (30 - 1)(0.1)\)
= \(15(9.9)\)
= \(148.5\)
(a) \(T_{n} = a + (n - 1)d\) (terms of an AP)
\(T_{21} = a + 20d = 5\frac{1}{2}.... (1)\)
\(S_{n} = \frac{n}{2} (2a + (n - 1)d) = \frac{n}{2} (a + l)\)
Where a and l are the first and last terms respectively.
\(S_{21} = \frac{21}{2} (a + 5\frac{1}{2})\)\)
\(94\frac{1}{2} = \frac{21}{2} (a + 5\frac{21}{2})\)
\(189 = 21 (a + 5\frac{1}{2})\)
\(9 = a + 5\frac{1}{2} \implies a = 9 - 5\frac{1}{2} = 3\frac{1}{2}\)
(b) Put a in the equation (1),
\(3\frac{1}{2} + 20d = 5\frac{1}{2}\)
\(20d = 5\frac{1}{2} - 3\frac{1}{2} = 2\)
\(d = \frac{2}{20} = \frac{1}{10}\).
(c) \(S_{30} = \frac{30}{2} (2(3\frac{1}{2}) + (30 - 1)(0.1)\)
= \(15(9.9)\)
= \(148.5\)