How many terms of the A.P. 1, 4, 7, ldots are | vedclass

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(B) Given the $A.P.$ is $1, 4, 7, \ldots$ where the first term $a = 1$ and the common difference $d = 4 - 1 = 3$.Let the sum of $n$ terms be $S_n = 71

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(B) Given the $A.P.$ is $1, 4, 7, \ldots$ where the first term $a = 1$ and the common difference $d = 4 - 1 = 3$.Let the sum of $n$ terms be $S_n = 715$.The formula for the sum of $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$.Substituting the values: $715 = \frac{n}{2}[2(1) + (n - 1)3]$.$1430 = n[2 + 3n - 3]$.$1430 = n[3n - 1]$.$3n^2 - n - 1430 = 0$.Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{1 \pm \sqrt{(-1)^2 - 4(3)(-1430)}}{2(3)}$.$n = \frac{1 \pm \sqrt{1 + 17160}}{6} = \frac{1 \pm \sqrt{17161}}{6} = \frac{1 \pm 131}{6}$.Since $n$ must be a positive integer,$n = \frac{1 + 131}{6} = \frac{132}{6} = 22$.

How many terms of the $A.P.$ $1, 4, 7, \ldots$ are needed to give the sum $715$?

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