There are 15 terms in an arithmetic progressi | vedclass

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(B) Given that the number of terms $n = 15$,the first term $a = 5$,and the sum of terms $S_{15} = 390$.Using the sum formula for an arithmetic progres

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$26$

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(B) Given that the number of terms $n = 15$,the first term $a = 5$,and the sum of terms $S_{15} = 390$.Using the sum formula for an arithmetic progression: $S_n = \frac{n}{2} [2a + (n - 1)d]$.Substituting the values: $390 = \frac{15}{2} [2(5) + (15 - 1)d]$.$390 = \frac{15}{2} [10 + 14d]$.$390 = 15(5 + 7d)$.$26 = 5 + 7d$.$7d = 21$,so $d = 3$.The middle term of an arithmetic progression with $15$ terms is the $\frac{15+1}{2} = 8^{th}$ term.The $n^{th}$ term formula is $a_n = a + (n - 1)d$.For $n = 8$: $a_8 = 5 + (8 - 1)(3) = 5 + 7(3) = 5 + 21 = 26$.

There are $15$ terms in an arithmetic progression. Its first term is $5$ and their sum is $390$. The middle term is

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