Q5 of 105 Page 146

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

Second term (a2) = 14; Third term (a3) = 18

a2 = a+(2-1)d


14 = a+d ⇒ 1


a3 = a+(3-1)d


18 = a+2d ⇒ 2


By subtracting both the equations


14 -18 = a+d –(a+2d)


-4 = -d


d = 4


By substituting d in the equation 1


a + d = 14


a + 4 = 14


a = 10


The sum of first 51 terms is given by (sn) = [2a + (n-1) d]


S51 = [2×10 + (51-1) 4]


= 5610


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