Find the value of the middle most term(s) of | vedclass

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Question

(A) Given $AP$ is $-11, -7, -3, \ldots, 49$.Here,the first term $a = -11$ and the common difference $d = -7 - (-11) = 4$.The last term $a_n = 49$.Usin

Vedclass · Accepted Answer

$17, 21$

Vedclass · Answer

(A) Given $AP$ is $-11, -7, -3, \ldots, 49$.Here,the first term $a = -11$ and the common difference $d = -7 - (-11) = 4$.The last term $a_n = 49$.Using the formula $a_n = a + (n - 1)d$:$49 = -11 + (n - 1) 	imes 4$$60 = (n - 1) 	imes 4$$n - 1 = 15 \implies n = 16$.Since $n = 16$ is an even number,there are two middle terms: the $(\frac{n}{2})^{th}$ term and the $(\frac{n}{2} + 1)^{th}$ term.These are the $8^{th}$ term and the $9^{th}$ term.$a_8 = a + 7d = -11 + 7(4) = -11 + 28 = 17$.$a_9 = a + 8d = -11 + 8(4) = -11 + 32 = 21$.Thus,the middle terms are $17$ and $21$.

Find the value of the middle most term$(s)$ of the $AP:$ $-11, -7, -3, \ldots, 49$.

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