The first 3 terms of an ap are respectively (3y-1), (3y+5) and (5y+1)

1.	The first 3 terms of an ap are respectively (3y-1), (3y+5) and (5y+1) find your.
Answer» In given AP,\xa0{tex}{a_2} - {a_1} = {a_3} - {a_2}{/tex}3y + 5 - 3y + 1 = 5y + 1 - 3y - 5=> 6 = 2y - 4=> y = 5Then the given AP will be 14, 20, 26, .....Here common difference is 20 - 14 = 6Thereforem the fourth term is 26 + 6 = 32 The first three terms (3y—1), (3y+5) and (5y+1) are in A.P, so the common difference will be sameThat is d1=d2(3y+5)—(3y—1) = (5y+1)—(3y+5)3y+5—3y+1=5y+1—3y—56=2y—46+4 =2y10=2y5=y Given an AP in whicha1 = 3y - 1a2 = 3y + 5a3 = 5y + 1we know that term 2 - term 1 = term 3 - term 2 = common difference d a2 - a1 = a3 - a2 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5 6 = 2y - 4 10 = 2y y = 5

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