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The general form of an A.P is a,a+d,a+2d,a+3d ........ Therefore the second term is a+d ,which is equal to 14. Therefore the fourth term is a+3d,which is equal to 18. By solving the above equations , we get d=4 and a=10. therefore the sum of first 51 terms =(51/2)*(20+200)=5610. Therefore the required answer is 5610.

Difference b/w 2nd & 3rd term is 4, there fore 1st term is 14 -4 =10
51st term is 10 + 50 * 4 = 210
Sum = 51(10 + 210)/2 = 5610

In an arithmatic progression, consecutive terms have the same difference. In this case, consecutive terms have a difference of 4. As the difference is 4, and the second and third terms are even, all the numbers are even, and therefore, the sum of any amount of numbers cannot be odd. Additionally, all the terms are factors of 2 that are not factors of 4, as 14 and 18 are factors of 2 that are not factors of 4, and the difference being 4, there is no way for there to be a factor of 4 in the sequence. Therefore, the only way for the sum of the numbers to result in a factor of 4 is if there is an even number of numbers being added together. As the questions asks for the first 51 numbers, and odd number, the sum will not not be a factor of 4, leaving 5610 to be the only possible solution.