Question

An AP consists of 50 terms of which 3rd term is 12 and the last term is
106. Find the 29th term​

Answer 1

Answer:

29th term is 64. hope it helps you

Answer 2

Given:-

• An AP consists of 50
• 3rd term is 12 and the last term is 106.

To find:-

• Find the 29th term...?

Solutions:-

• The 3rd term of Ap is 12
• The last term of Ap is 106

we know that;

The 3rd term of Ap is 12.

an = a + (n - 1)d

a3 = a + (3 - 1)d

12 = a + 2d .............(i).

The last term of Ap is 106.

=> an = a + (n - 1)d

=> a50 = a + (50 - 1)d

=> 106 = a + 49d .............(ii).

Now, Subtracting Eq. (ii) and (i) we get,

${a} \: + \: {49d} \: \: = \: \: {106} \\ {a} \: + \: {2d} \: \: = \: \: {12} \\ \underline{ - \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: = \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: } \\ \: \: \: \: \: \: \: \: {47d} \: \: \: \: \: \: \: \: = \: \: \: {94}$

=> y = 94/47

=> y = 2

Now, putting the value of y in Eq. (i).

=> a + 2d = 12

=> a + 2(2) = 12

=> a + 4 = 12

=> a = 12 - 4

=> a = 8

So,

=> a29 = a + (29 - 1)d

=> a29 = a + 28d

=> a29 = 8 + 28 × 2

=> a29 = 8 + 56

=> a29 = 64

Hence the 29th term of Ap is 64.