There are 15 terms in an arithmetic progressi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(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

Similar Questions

The number of terms of an $A.P.$ is even; the sum of all the odd terms is $24$,the sum of all the even terms is $30$ and the last term exceeds the first by $\frac{21}{2}$. Then the number of terms which are integers in the $A.P.$ is:

If $\log _{5} 2, \log _{5}(2^{x}-3)$ and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$,then the value of $x$ is:

Find the mean of the first $n$ terms of the $A$.$P$. $a, (a + d), (a + 2d), \dots$

If $100$ times the $100^{th}$ term of an Arithmetic Progression with a non-zero common difference is equal to $50$ times its $50^{th}$ term,then what is its $150^{th}$ term?

The sum of $24$ terms of the following series $\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module