2. Find the 40th term for the arithmetic sequence in which
t8 =60 and t12= 48
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Solution!!
t₈ = 60
t₁₂ = 48
t₈ = a + 7d
t₁₂ = a + 11d
60 = a + 7d...(1)
60 = a + 7d...(1)48 = a + 11d...(2)
Subtracting (1) and (2)
60 - 48 = a - a + 7d - 11d
12 = -4d
d = -3
Putting the vale of d in (1)
60 = a + 7d
60 = a + 7(-3)
60 = a - 21
a = 81
aₙ = a + (n - 1)d
a₄₀ = 81 + (40 - 1)(-3)
a₄₀ = 81 + (39)(-3)
a₄₀ = 81 - 117
a₄₀ = -36
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