The sum of the first 16 terms of the A.P. 1, | vedclass

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Question

(A) Given the $A.P.$ is $1, 1.5, 2, \ldots$Here,the first term $a = 1$.The common difference $d = 1.5 - 1 = 0.5$.The number of terms $n = 16$.The form

Vedclass · Accepted Answer

$76$

Vedclass · Answer

(A) Given the $A.P.$ is $1, 1.5, 2, \ldots$Here,the first term $a = 1$.The common difference $d = 1.5 - 1 = 0.5$.The number of terms $n = 16$.The formula for the sum of the first $n$ terms of an $A.P.$ is $S_n = \frac{n}{2}[2a + (n - 1)d]$.Substituting the values: $S_{16} = \frac{16}{2}[2(1) + (16 - 1)0.5]$.$S_{16} = 8[2 + 15 	imes 0.5]$.$S_{16} = 8[2 + 7.5]$.$S_{16} = 8 	imes 9.5 = 76$.

The sum of the first $16$ terms of the $A.P.$ $1, 1.5, 2, \ldots$ is:

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