3. Find tn for an Arithmetic Progression where t3 = 22, t17 = – 20. (3 marks)
Given : For an A.P. t3 = 22 and t17 = – 20
Find : tn.
Sol. tn = a + (n – 1) d
∴ t3 = a + (3 – 1) d
22 = a + 2d
∴ a + 2d = 22 ......(i)
∴ a = 22 – 2d ......(i)
∴ t17 = a + (17 – 1) d
– 20 = a + 16d
∴ a + 16d = – 20 ......(ii)
substituting eq. (i) in eq. (ii)
a + 16d = – 20 ......(ii)
∴ 22 – 2d + 16d = - 20 [from eq. 1]
∴ 14d = - 20 – 22
∴ 14d = - 42
∴ d = -42/14
∴ d = -3
Substituting d = – 3 in (i),
a = 22 – 2(-3)
a = 22 + 6
∴ a = 28
Now,
tn = a + (n – 1) d
∴ tn = 28 + (n – 1) (– 3)
∴ tn = 28 – 3n + 3
∴ tn = 31 – 3n
