If the 12^{th} term of an A.P. is -13 and the | vedclass

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(A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$The $n^{th}$ term is given by $a_n = a + (n - 1)d$.Given $a_{12} = -13$,we

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(A) Let $a$ be the first term and $d$ be the common difference of the $A.P.$The $n^{th}$ term is given by $a_n = a + (n - 1)d$.Given $a_{12} = -13$,we have $a + 11d = -13$  --- $(1)$.The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$.Given $S_4 = 24$,we have $\frac{4}{2}[2a + 3d] = 24$,which simplifies to $2a + 3d = 12$ --- $(2)$.Multiply equation $(1)$ by $2$ to get $2a + 22d = -26$ --- $(3)$.Subtract equation $(2)$ from equation $(3)$:$(2a + 22d) - (2a + 3d) = -26 - 12$$19d = -38$,so $d = -2$.Substitute $d = -2$ into equation $(1)$:$a + 11(-2) = -13$$a - 22 = -13$,so $a = 9$.Now,calculate the sum of the first $10$ terms $(S_{10})$:$S_{10} = \frac{10}{2}[2(9) + (10 - 1)(-2)]$$S_{10} = 5[18 + 9(-2)]$$S_{10} = 5[18 - 18] = 5(0) = 0$.

If the $12^{th}$ term of an $A.P.$ is $-13$ and the sum of the first four terms is $24,$ then what is the sum of the first $10$ terms?

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