Question

Find the 12^th term from the end of the following arithmetic progressions:

(i) 3, 5, 7, 9, ..... 201

(ii) 3, 8, 13, ....., 253

(iii) 1, 4, 7, 10, ......., 88

Answer 1

(i) 3, 5, 7, 9, ..... 201

a = 3, d = 5 - 3 = 2, a_n = 201

a_n = a +(n -1)d

201 = 3 +2n – 2

N = 100

Now, we have to find 12^th term from the last that means,

100^th – 11 = 89^th term

Then,

a₈₉ = a + (89 – 1)d

= 3 + 88 x 2

= 179

Hence, the 12^th term from the end of the A.P. is 179.

(ii) 3, 8, 13, ....., 253

a = 3, d = 8 - 3 = 5

a_n = 253

a + (n - 1)d = 253

3 + (n - 1)5 = 253

n = 51

Now, we have to find 12th term from the last that means,

51^th -11 = 40^th term

Then,

a₄₀ = a + (n -1)d

= 3 + 39(5)

= 198

Hence, the 12^th term from the end of the A.P. is 198

(iii) 1, 4, 7, 10, ......., 88

a = 1, d = 4 -1 = 3

a_n = 88

a + (n -1)d = 88

1 + (n -1)3 = 88

n = 30

Now, we have to find 12^th term from the last that means,

30^th - 11 = 19^th term

a₁₉ = a + 18d

= 1 + 18(3)

= 55

Hence, the 12^th term from the end of the A.P. is 198.