Answer :

Let a be the first term and d be the common difference.

Given: a5 = 16


a13 = 4 a3


Now, Consider a5 = 16


a + (5 - 1)d = 16


a + 4d = 16 ………………….(1)


Consider a13 = 4 a3


a + 12d = 4(a + 2d)


a + 12d = 4a + 8d


3a - 4d = 0 ………………….(2)


Now, adding equation (1) and (2), we get,


4a = 16


a = 4


from equation (2), we get,


4d = 3a


4d = 12


d = 3


Now, Sum of first n terms of an AP is


Sn = [2a + (n - 1)d]


Sum of first 10 terms is given by:


S10 = [2(4) + (10 - 1)(3)]


= 5 × [8 + 27]


= 5 × 35


= 175


S10=175


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