If a_1, a_2, ..., a_{24} are in an arithmetic | vedclass

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

Question

(A) Given that $a_1, a_2, ..., a_{24}$ are in an arithmetic progression.We know that in an arithmetic progression,the sum of terms equidistant from th

Vedclass · Accepted Answer

$900$

Vedclass · Answer

(A) Given that $a_1, a_2, ..., a_{24}$ are in an arithmetic progression.We know that in an arithmetic progression,the sum of terms equidistant from the beginning and the end is constant and equal to the sum of the first and last terms,i.e.,$a_k + a_{n-k+1} = a_1 + a_n$.Therefore,$a_1 + a_{24} = a_5 + a_{20} = a_{10} + a_{15}$.Given $a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{24} = 225$.Substituting the property,we get $3(a_1 + a_{24}) = 225$.So,$a_1 + a_{24} = 75$.The sum of the first $n$ terms is $S_n = \frac{n}{2}(a_1 + a_n)$.For $n = 24$,$S_{24} = \frac{24}{2}(a_1 + a_{24}) = 12 	imes 75 = 900$.

If $a_1, a_2, ..., a_{24}$ are in an arithmetic progression and $a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{24} = 225$,then what is the sum of the first $24$ terms of this arithmetic progression?

Similar Questions

If the $m^{th}$ terms of the sequences $63, 65, 67, 69, \dots$ and $3, 10, 17, 24, \dots$ are equal,then $m = \dots$

The $8^{th}$ term of the series $2\sqrt{2} + \sqrt{2} + 0 + \dots$ will be: (in $\sqrt{2}$)

The sum of the series $1 + 3 + 5 + 7 + \dots$ up to $n$ terms is equal to:

The sum of $1 + 3 + 5 + 7 + \dots$ up to $n$ terms is

Let $9 < x_1 < x_2 < \ldots < x_7$ be in an $A.P.$ with common difference $d$. If the standard deviation of $x_1, x_2, \ldots, x_7$ is $4$ and the mean is $\overline{x}$,then $\overline{x} + x_6$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module