Question

2. Find the 40th term for the arithmetic sequence in which
t8 =60 and t12= 48​

Answer 1

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