Find tn for an Arithmetic Progression where t3 = 22, t17 = – 20.

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