Let a_n be the n^{text{th}} term of an A.P. I | vedclass

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(C) Given $a_6 = a + 5d = 7$ $(i)$Given $S_7 = \frac{7}{2}(2a + 6d) = 7 \Rightarrow a + 3d = 1$ $(ii)$Subtracting $(ii)$ from $(i)$: $(a + 5d) - (a +

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$64$

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(C) Given $a_6 = a + 5d = 7$ $(i)$Given $S_7 = \frac{7}{2}(2a + 6d) = 7 \Rightarrow a + 3d = 1$ $(ii)$Subtracting $(ii)$ from $(i)$: $(a + 5d) - (a + 3d) = 7 - 1$ $\Rightarrow 2d = 6$ $\Rightarrow d = 3$.Substituting $d = 3$ in $(ii)$: $a + 3(3) = 1 \Rightarrow a = -8$.Given $S_n = \frac{n}{2}[2a + (n-1)d] = 700$.Substituting $a = -8$ and $d = 3$: $\frac{n}{2}[2(-8) + (n-1)3] = 700$.$\frac{n}{2}[-16 + 3n - 3] = 700$ $\Rightarrow n(3n - 19) = 1400$ $\Rightarrow 3n^2 - 19n - 1400 = 0$.Solving the quadratic equation: $(3n + 56)(n - 25) = 0$.Since $n$ must be a positive integer,$n = 25$.Thus,$a_n = a_{25} = a + 24d = -8 + 24(3) = -8 + 72 = 64$.

Let $a_n$ be the $n^{\text{th}}$ term of an $A.P.$ If $S_n = a_1 + a_2 + a_3 + \dots + a_n = 700$,$a_6 = 7$,and $S_7 = 7$,then $a_n$ is equal to:

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