The sum of how many terms of the A.P. 31, 36, | vedclass

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Question

(B) Given $A.P.$ is $31, 36, 41, \ldots$Here,the first term $a = 31$ and the common difference $d = 36 - 31 = 5$.Let the sum of $n$ terms be $S_n = 53

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) Given $A.P.$ is $31, 36, 41, \ldots$Here,the first term $a = 31$ and the common difference $d = 36 - 31 = 5$.Let the sum of $n$ terms be $S_n = 535$.The formula for the sum of $n$ terms of an $A.P.$ is $S_n = \frac{n}{2} [2a + (n - 1)d]$.Substituting the values: $535 = \frac{n}{2} [2(31) + (n - 1)5]$.$1070 = n [62 + 5n - 5]$.$1070 = n [57 + 5n]$.$5n^2 + 57n - 1070 = 0$.Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{-57 \pm \sqrt{57^2 - 4(5)(-1070)}}{2(5)}$.$n = \frac{-57 \pm \sqrt{3249 + 21400}}{10}$.$n = \frac{-57 \pm \sqrt{24649}}{10}$.$n = \frac{-57 \pm 157}{10}$.Since $n$ must be positive,$n = \frac{100}{10} = 10$.Thus,the sum of $10$ terms is $535$.

Column $A$	Column $B$
$(A_{1}) \quad 2, -2, -6, -10, \ldots$	$(B_{1}) \quad \frac{2}{3}$
$(A_{2}) \quad a = -18, n = 10, a_{n} = 0$	$(B_{2}) \quad -5$
$(A_{3}) \quad a = 0, a_{10} = 6$	$(B_{3}) \quad 4$
$(A_{4}) \quad a_{2} = 13, a_{4} = 3$	$(B_{4}) \quad -4$
	$(B_{5}) \quad 2$
	$(B_{6}) \quad \frac{1}{2}$
	$(B_{7}) \quad 5$

The sum of how many terms of the $A.P.$ $31, 36, 41, \ldots$ is $535$?

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