Find the sum of the first 17 terms of an AP w | vedclass

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(B) Let the first term be $a$ and the common difference be $d$ for the $AP$.The $n^{	ext{th}}$ term of an $AP$ is given by $T_n = a + (n - 1)d$.Given

Vedclass · Accepted Answer

$-510$

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(B) Let the first term be $a$ and the common difference be $d$ for the $AP$.The $n^{	ext{th}}$ term of an $AP$ is given by $T_n = a + (n - 1)d$.Given,$T_4 = -15$:$a + 3d = -15$  --- $(i)$Given,$T_9 = -30$:$a + 8d = -30$  --- $(ii)$Subtracting equation $(i)$ from equation $(ii)$:$(a + 8d) - (a + 3d) = -30 - (-15)$$5d = -15$$d = -3$Substituting $d = -3$ in equation $(i)$:$a + 3(-3) = -15$$a - 9 = -15$$a = -6$The sum of the first $n$ terms is $S_n = \frac{n}{2}[2a + (n - 1)d]$.For $n = 17$:$S_{17} = \frac{17}{2}[2(-6) + (17 - 1)(-3)]$$S_{17} = \frac{17}{2}[-12 + 16(-3)]$$S_{17} = \frac{17}{2}[-12 - 48]$$S_{17} = \frac{17}{2}[-60]$$S_{17} = 17 	imes (-30) = -510$.Thus,the sum of the first $17$ terms is $-510$.

Find the sum of the first $17$ terms of an $AP$ whose $4^{\text{th}}$ and $9^{\text{th}}$ terms are $-15$ and $-30$ respectively.

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