If the sum of the 8th and 9th terms of an arithmetic progression is 72 and the 4th term is -6, find the common difference
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Correct Answer: Option D
Explanation:
Let the first term and common difference = a & d respectively.
\(T_{n} = \text{nth term} = a + (n - 1) d\) (A.P)
Given: \(T_{4} = -6 \implies a + 3d = -6 ... (i)\)
\(T_{8} + T_{9} = 72\)
\(\implies a + 7d + a + 8d = 72 \implies 2a + 15d = 72 ... (ii)\)
From (i), \(a = -6 - 3d\)
\(\therefore\) (ii) becomes \(2(-6 - 3d) + 15d = 72\)
\(-12 - 6d + 15d = 72 \implies 9d = 72 + 12 = 84\)
\(d = \frac{84}{9} = 9\frac{1}{3}\)
Let the first term and common difference = a & d respectively.
\(T_{n} = \text{nth term} = a + (n - 1) d\) (A.P)
Given: \(T_{4} = -6 \implies a + 3d = -6 ... (i)\)
\(T_{8} + T_{9} = 72\)
\(\implies a + 7d + a + 8d = 72 \implies 2a + 15d = 72 ... (ii)\)
From (i), \(a = -6 - 3d\)
\(\therefore\) (ii) becomes \(2(-6 - 3d) + 15d = 72\)
\(-12 - 6d + 15d = 72 \implies 9d = 72 + 12 = 84\)
\(d = \frac{84}{9} = 9\frac{1}{3}\)