With respect to the usual notations of an A.P | vedclass

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(A) Given: $T_{n} = a + (n - 1)d = 4$ and $S_{n} = \frac{n}{2}[a + T_{n}] = -14$. Substituting $T_{n} = 4$ in the sum formula: $\frac{n}{2}[a + 4] = -

Vedclass · Accepted Answer

$n = 7, a = -8$

Vedclass · Answer

(A) Given: $T_{n} = a + (n - 1)d = 4$ and $S_{n} = \frac{n}{2}[a + T_{n}] = -14$. Substituting $T_{n} = 4$ in the sum formula: $\frac{n}{2}[a + 4] = -14 \implies n(a + 4) = -28$. From the first equation: $a = 4 - (n - 1)d = 4 - 2n + 2 = 6 - 2n$. Substitute $a$ in the sum equation: $n(6 - 2n + 4) = -28 \implies n(10 - 2n) = -28$. $10n - 2n^{2} = -28 \implies 2n^{2} - 10n - 28 = 0 \implies n^{2} - 5n - 14 = 0$. Factoring the quadratic: $(n - 7)(n + 2) = 0$. Since $n$ must be a positive integer,$n = 7$. Now,find $a$: $a = 6 - 2(7) = 6 - 14 = -8$. Thus,$n = 7$ and $a = -8$.

With respect to the usual notations of an $A.P.$,if $T_{n} = 4$,$d = 2$,and $S_{n} = -14$,find $n$ and $a$.

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