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

Given the sum of the first $n$ terms of an $A.P.$ is $S_n = 2n + 3n^2$. Another $A.P.$ is formed with the same first term and double the common difference. The sum of $n$ terms of the new $A.P.$ is:

If the sum and product of the first three terms in an $A.P.$ are $33$ and $1155$,respectively,then a value of its $11^{th}$ term is

If $a^2, b^2, c^2$ are in $A.P.$,then $(b + c)^{-1}, (c + a)^{-1}$ and $(a + b)^{-1}$ will be in

If the first term of an $A.P.$ is $10$,the last term is $50$,and the sum of all the terms is $300$,then the number of terms is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module