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

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Answer 2

Solution:

Given:

$\sf{\implies a_{3}=12}$

$\sf{\implies a_{50}=106}$

To Find:

=> 29th term

Formula used:

$\sf{\implies a_{n}=a+(n-1)d}$

So, we know that

$\sf{\implies a_{3}=a+2d}$

$\sf{\implies a+2d=12\;\;\;..........(1)}$

$\sf{\implies a_{50}=a+49d}$

$\sf{\implies a+49d=106\;\;\;..........(2)}$

By using substitution method we will find value of a and d.

=> a + 2d = 12      .......(1)

=> a + 49d = 106     .......(2)

=> a + 2d = 12

=> a = 12 - 2d

Put the value of a in Equation (2), we get

=> a + 49d = 106

=> (12 - 2d) + 49d = 106

=> 12 - 47d = 106

=> 47d = 106 - 12

=> 47d = 94

=> d = 94/47

=> d = 2

Put the value of d in Equation (1), we get

=> a + 2d = 12

=> a + 4 = 12

=> a = 8

Now, we will find 29st term,

$\sf{\implies a_{29}=a+(n-1)d}$

=> 8 + (29 - 1)2

=> 8 + 56

=> 64

So, 29th term of an AP is 64.