Find the sum of first 51 terms of an appointment whose second and third terms are 14 and 18 respectively
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Step-by-step explanation:
common difference (d) = 18-14 = 4
the second term = a+d = 14
so, a = 14-4 = 10
SUM
number of terms (n) = 51
a = 10
d = 4
Sum=n/2 [2a + (n-1)d]
= 51/2 [20 + (50 x 4)]
= 51/2 [20+200]
= 51 x 110
= 5610
therefore, the sum of 51 terms of AP is 5610
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