There are 15 terms in an arithmetic progressi | vedclass

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(B) Let the first term be $a = 5$,the number of terms be $n = 15$,and the common difference be $d$.The sum of $n$ terms of an arithmetic progression i

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$26$

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(B) Let the first term be $a = 5$,the number of terms be $n = 15$,and the common difference be $d$.The sum of $n$ terms of an arithmetic progression is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.Given $S_{15} = 390$,we have $\frac{15}{2}[2(5) + (15 - 1)d] = 390$.$\frac{15}{2}[10 + 14d] = 390$.$15(5 + 7d) = 390$.$5 + 7d = \frac{390}{15} = 26$.$7d = 21$,so $d = 3$.The middle term of $15$ terms is the $\frac{15+1}{2} = 8^{th}$ term.The $n^{th}$ term is $a_n = a + (n - 1)d$.$a_8 = 5 + (8 - 1)3 = 5 + 7(3) = 5 + 21 = 26$.

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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