Question

an ap consists of 50 terms of which 3rd term of an ap is 12 and last term is 106 find the 29th term

Answer 1

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Answer 2

Hello !!




Let a be the first term and d be the common difference.



3rd term = 12


A + 2D = 12 --------(1)


And,



50th term = 106



A + 49D = 106 --------(2)


From equation (1) we get,



A + 2D = 12


A = ( 12 - 2D ) -------(3)


Putting the value of A in equation (2)


A + 49D = 106


12 - 2D + 49D = 106



47D = 106-12



47D = 94


D = 94/47 = 2



Putting the value of D in equation (3)


A = 12-2D = 12 - 2 × 2



A = 12-4 = 8


★ First term ( A ) = 8


★ Common difference ( D ) = 2


Therefore,

29th term = A + 28D




=> 8 + 28 × 2


=> 8 + 56



=> 64


Hence,

★ 29 th term = 64.