If the sum of the 8th and 9th terms of an arithmetic prog...
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
4
8
6\(\frac{2}{3}\)
9\(\frac{1}{3}\)
Correct answer is D
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 \(2(-6 - 3d) + 15d = 72\)
\(-12 - 6d + 15d = 72 \implies 9d = 72 + 12 = 84\)
\(d = \frac{84}{9} = 9\frac{1}{3}\)
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