For an A.P.,the 12^{th} term is -13 and the s | vedclass

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

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(A) Let the first term be $a$ and the common difference be $d$.The $n^{th}$ term of an $A.P.$ is given by $a_n = a + (n-1)d$.Given $a_{12} = -13$,so $a + 11d = -13$ --- (Equation $1$).The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n-1)d]$.Given $S_4 = 24$,so $\frac{4}{2}[2a + 3d] = 24$,which simplifies to $2(2a + 3d) = 24$,or $2a + 3d = 12$ --- (Equation $2$).Multiply Equation $1$ by $2$: $2a + 22d = -26$ --- (Equation $3$).Subtract Equation $2$ from Equation $3$: $(2a + 22d) - (2a + 3d) = -26 - 12$,which gives $19d = -38$,so $d = -2$.Substitute $d = -2$ into Equation $2$: $2a + 3(-2) = 12$,so $2a - 6 = 12$,$2a = 18$,$a = 9$.Now,find the sum of the first $10$ terms $S_{10} = \frac{10}{2}[2(9) + (10-1)(-2)]$.$S_{10} = 5[18 + 9(-2)] = 5[18 - 18] = 5(0) = 0$.

For an $A.P.$,the $12^{th}$ term is $-13$ and the sum of its first $4$ terms is $24$. Find the sum of the first $10$ terms of the $A.P.$

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