Which term of the A.P. 40, 36, 32, ldots is 0 | vedclass

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(B) For the $A.P.$ $40, 36, 32, \ldots$,the first term $a = 40$ and common difference $d = 36 - 40 = -4$.To find which term is $0$,we use the formula

Vedclass · Accepted Answer

$11^{th}$ term,$21$ terms

Vedclass · Answer

(B) For the $A.P.$ $40, 36, 32, \ldots$,the first term $a = 40$ and common difference $d = 36 - 40 = -4$.To find which term is $0$,we use the formula $a_n = a + (n - 1)d$.Setting $a_n = 0$,we get $0 = 40 + (n - 1)(-4)$.$4(n - 1) = 40 \implies n - 1 = 10 \implies n = 11$.Thus,the $11^{th}$ term is $0$.To find the sum of $n$ terms equal to $0$,we use $S_n = \frac{n}{2}[2a + (n - 1)d]$.Setting $S_n = 0$,we get $0 = \frac{n}{2}[2(40) + (n - 1)(-4)]$.$0 = 80 - 4n + 4 \implies 4n = 84 \implies n = 21$.Thus,the sum of $21$ terms is $0$.

Which term of the $A.P.$ $40, 36, 32, \ldots$ is $0$? Also,find the sum of how many terms of this $A.P.$ is $0$?

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