The number of terms of the A.P. 3, 7, 11, 15, | vedclass

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(D) The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.Here,$a = 3$,$d = 4$,and $S_n = 406$.Substituting the v

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$14$

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(D) The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.Here,$a = 3$,$d = 4$,and $S_n = 406$.Substituting the values: $406 = \frac{n}{2}[2(3) + (n - 1)4]$.$406 = \frac{n}{2}[6 + 4n - 4]$.$406 = \frac{n}{2}[4n + 2]$.$406 = n(2n + 1)$.$2n^2 + n - 406 = 0$.Using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$n = \frac{-1 \pm \sqrt{1^2 - 4(2)(-406)}}{2(2)}$.$n = \frac{-1 \pm \sqrt{1 + 3248}}{4}$.$n = \frac{-1 \pm \sqrt{3249}}{4}$.$n = \frac{-1 \pm 57}{4}$.Since $n$ must be positive,$n = \frac{56}{4} = 14$.

The number of terms of the $A.P. 3, 7, 11, 15, ...$ to be taken so that the sum is $406$ is

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