How many terms of the AP: -15, -13, -11, ldot | vedclass

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(A) Let $n$ be the number of terms required to make the sum $-55$.Here,the first term $(a) = -15$ and the common difference $(d) = -13 - (-15) = 2$.Th

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$5, 11$

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(A) Let $n$ be the number of terms required to make the sum $-55$.Here,the first term $(a) = -15$ and the common difference $(d) = -13 - (-15) = 2$.The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Substituting the values: $-55 = \frac{n}{2}[2(-15) + (n-1)2]$.$-55 = \frac{n}{2}[-30 + 2n - 2] = \frac{n}{2}[2n - 32] = n(n - 16)$.$n^2 - 16n + 55 = 0$.Factoring the quadratic equation: $n^2 - 11n - 5n + 55 = 0 \Rightarrow n(n - 11) - 5(n - 11) = 0$.$(n - 5)(n - 11) = 0$,so $n = 5$ or $n = 11$.The reason for the double answer is that the sum of the terms from the $6^{th}$ term to the $11^{th}$ term is zero. Specifically,the terms are $-15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5$. The sum of the $6^{th}$ to $11^{th}$ terms is $(-5) + (-3) + (-1) + 1 + 3 + 5 = 0$. Thus,adding these terms to the sum of the first $5$ terms does not change the total sum.

How many terms of the $AP: -15, -13, -11, \ldots$ are needed to make the sum $-55$? Explain the reason for the double answer.

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