The 3 rd and 6 th terms of an arithmetic progression are 94 and 85 respectively. Find the value of the 1 5 th term.
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a 1 5 = a 2 + 1 2 ( 3 a 6 − a 3 ) = 9 4 + 1 2 ( − 3 ) = 5 8
Given: a 3 = 9 4 ; a 6 = 8 5
Formula: a n = a m + ( n − m ) d
We have
8 5 = 9 4 + ( 6 − 3 ) d
d = − 3
a 1 5 = a 3 + 1 2 d = 9 4 + 1 2 ( − 3 ) = 5 8
Good solution. There is no need of involving the first term then having two equations and two unknowns. You immediately calculated the value of d .
a 3 = 9 4 ⟹ a + 2 d = 9 4 ⟹ Eq.(1)
a 6 = 8 5 ⟹ a + 5 d = 8 5 ⟹ Eq.(2)
Eq.(2) - Eq.(1):
a + 5 d − ( a + 2 d ) = 8 5 − 9 4 3 d = − 9
a 1 5 = a + 1 4 d = a + 2 d + 1 2 d = a 3 + 4 ( 3 d ) = 9 4 + 4 ( − 9 ) = 9 4 − 3 6 = 5 8
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Let a be the first term and d be the common difference.
a 3 = a + 2 d a 6 = a + 5 d Subtracting 2 from 1 : − − 3 d d = − 3 Substituting d = -3 in 1 : − a + 2 ( − 3 ) a − 6 a ⟹ a 1 5 = 9 4 ⟶ 1 = 8 5 ⟶ 2 = 9 = 9 4 = 9 4 = 1 0 0 = a + 1 4 d = 1 0 0 + 1 4 ( − 3 ) = 1 0 0 − 4 2 = 5 8