Find the sum of the first 25 terms of the A.P | vedclass

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(D) The given $A.P.$ is $3, \frac{9}{2}, 6, \frac{15}{2}, \ldots$Here,the first term $a = 3$.The common difference $d = \frac{9}{2} - 3 = \frac{9-6}{2

Vedclass · Accepted Answer

$525$

Vedclass · Answer

(D) The given $A.P.$ is $3, \frac{9}{2}, 6, \frac{15}{2}, \ldots$Here,the first term $a = 3$.The common difference $d = \frac{9}{2} - 3 = \frac{9-6}{2} = \frac{3}{2}$.The number of terms $n = 25$.The sum of the first $n$ terms of an $A.P.$ is given by the formula $S_n = \frac{n}{2} [2a + (n-1)d]$.Substituting the values: $S_{25} = \frac{25}{2} [2(3) + (25-1)(\frac{3}{2})]$.$S_{25} = \frac{25}{2} [6 + 24(\frac{3}{2})]$.$S_{25} = \frac{25}{2} [6 + 12(3)]$.$S_{25} = \frac{25}{2} [6 + 36] = \frac{25}{2} [42]$.$S_{25} = 25 	imes 21 = 525$.

Find the sum of the first $25$ terms of the $A.P.$ $3, \frac{9}{2}, 6, \frac{15}{2}, \ldots$

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