The sum of all those terms of the arithmetic | vedclass

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(D) The given arithmetic progression is $3, 8, 13, \ldots, 373$.Here,$a = 3$,$d = 5$. The $n$-th term is $a_n = a + (n-1)d = 3 + (n-1)5 = 373$.$5(n-1)

Vedclass · Accepted Answer

$9525$

Vedclass · Answer

(D) The given arithmetic progression is $3, 8, 13, \ldots, 373$.Here,$a = 3$,$d = 5$. The $n$-th term is $a_n = a + (n-1)d = 3 + (n-1)5 = 373$.$5(n-1) = 370 \implies n-1 = 74 \implies n = 75$.The total sum $S_{75} = \frac{75}{2}(3 + 373) = \frac{75}{2}(376) = 75 	imes 188 = 14100$.Terms divisible by $3$ are of the form $3 + (k-1)5$ such that $3 + 5k - 5$ is a multiple of $3$,i.e.,$5k - 2 = 3m \implies 5k \equiv 2 \pmod 3 \implies 2k \equiv 2 \pmod 3 \implies k \equiv 1 \pmod 3$.So $k = 1, 4, 7, \ldots$. The terms are $3, 18, 33, \ldots$.The last term is $363$ (since $363 = 3 + (m-1)15 \implies 360 = (m-1)15 \implies m-1 = 24 \implies m = 25$).Sum of terms divisible by $3$ is $S' = \frac{25}{2}(3 + 363) = \frac{25}{2}(366) = 25 	imes 183 = 4575$.Required sum $= 14100 - 4575 = 9525$.

The sum of all those terms of the arithmetic progression $3, 8, 13, \ldots, 373$ which are not divisible by $3$ is equal to $.......$.

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