This is the basic and conceptual question from arithmetic progression chapter in which we have given the second and third terms of an AP are 14 and 18 respectively and we have to find the sum of its first 51 terms.

# Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

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As we know nth term, an=a+(n−1)d

& Sum of first n terms, Sn=n/2(2a+(n−1)d), where a & d are the first term amd common difference of an AP.

Since, Second term =a2=a+d=14 …(1)

## Third term =a3=a+2d=18 …(2)

On subtracting (1) from (2), we get

d=4

Putting d=4 in (1), we get

a=14−4=10

Now,

## Required sum=n/2[2a+(n−1)d]

## =51/2[2×10+(51−1)×4]

=51×110=5610