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Question

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


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Solution

Solving A.P with the given terms:

Step 1:Calculate the equation of 3rd term and last term.

Given that an A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106.

Assume that the first term is aand the common difference is d.

Know that the nthterm is given as a+(n1)d.

So, the 3rd term is given as

12=a+(31)d12=a+2da=122d.......(1)

Similarly, the last term is given as

106=a+(501)d106=a+49da=10649d.......(2)

Step 2: Calculate the common difference and the first term.

Equate equations (1) and (2).

122d=10649d49d2d=1061247d=94d=2

Apply d=2 in equation (2).

So,

a=10649(2)a=10698a=8

Step 3: Calculate the 29thterm.

Use a=8 and d=2 to find the 29th term.

Therefore,

a29=8+(291)2a29=8+28×2a29=8+56a29=64

Hence, the 29thterm is 64.


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