Q8 of 49 Page 112

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

We have; a2 = 14, a3 = 18 and n = 51


We know,


a3 – a2 = 18 – 14 = 4


Therefore, d = 4


a2 – a = 4


14 – a = 4


a = 10


Now, sum can be calculated as follows:




= 51*(10 + 100)


= 51*110


= 5610


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