If 4m+8,2m²+3m+6 & 3m² + 4m +4 are 3 concecutive terms of an AP. Fin

1.	If 4m+8,2m²+3m+6 & 3m² + 4m +4 are 3 concecutive terms of an AP. Find m.
Answer» First term = 4m+8Second term =\xa0{tex}2m^2+3m+6{/tex}Third term =\xa0{tex}3m^2+4m+4{/tex}Common Difference d = second term - first termd =\xa0{tex}(2m^2+3m+6) - (4m+8){/tex}d =\xa0{tex}2m^2-m-2{/tex}Common Difference = third term - second term\xa0d =\xa0{tex}(3m^2+4m+4)-(2m^2+3m+6){/tex}d =\xa0{tex}m^2+m-2{/tex}So,\xa0{tex}2m^2-m-2 = m^2+m-2{/tex}{tex}m^2-2m=0{/tex}m(m-2)= 0This gives m= 0 and m-2=0m=0 or m=2\xa0

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