Question

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

Answer 1

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Given that,

3rd term, a3 = 12

50th term, a50 = 106

We know that,

an = a+(n−1)d

a3 = a+(3−1)d

12 = a+2d ……………………………. (i)

In the same way, 

a50 = a+(50−1)d

106 = a+49d …………………………. (ii)

On subtracting equation (i) from (ii), we get

94 = 47d

d = 2 = common difference

From equation (i), we can write now,

12 = a+2(2)

a = 12−4 = 8

a29 = a+(29−1) d

a29 = 8+(28)2

a29 = 8+56 = 64

Therefore, 29th term is 64.

Answer 2

Step-by-step explanation:

Given that,

3rd term, a3 = 12

50th term, a50 = 106

We know that,

an = a+(n−1)d

a3 = a+(3−1)d

12 = a+2d ……………………………. (i)

In the same way,

a50 = a+(50−1)d

106 = a+49d …………………………. (ii)

On subtracting equation (i) from (ii), we get

94 = 47d

d = 2 = common difference

From equation (i), we can write now,

12 = a+2(2)

a = 12−4 = 8

a29 = a+(29−1) d

a29 = 8+(28)2

a29 = 8+56 = 64

Hence ,

29th term is 64.