Let a_1, a_2, a_3, dots, a_n be in A.P. If a_ | vedclass

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

Question

(A) In an $A.P.$,the sum of terms equidistant from the beginning and end is constant. Specifically,$a_k + a_{n-k+1} = a_1 + a_n$.Given $a_3 + a_7 + a_

Vedclass · Accepted Answer

$306$

Vedclass · Answer

(A) In an $A.P.$,the sum of terms equidistant from the beginning and end is constant. Specifically,$a_k + a_{n-k+1} = a_1 + a_n$.Given $a_3 + a_7 + a_{11} + a_{15} = 72$.We know that $a_3 + a_{15} = a_1 + a_{17}$ and $a_7 + a_{11} = a_1 + a_{17}$.Substituting these into the given equation:$(a_1 + a_{17}) + (a_1 + a_{17}) = 72$$2(a_1 + a_{17}) = 72$$a_1 + a_{17} = 36$.The sum of the first $17$ terms is given by $S_{17} = \frac{17}{2}(a_1 + a_{17})$.$S_{17} = \frac{17}{2} 	imes 36 = 17 	imes 18 = 306$.

Let $a_1, a_2, a_3, \dots, a_n$ be in $A.P$. If $a_3 + a_7 + a_{11} + a_{15} = 72$,then the sum of its first $17$ terms is equal to:

Similar Questions

The first term of an $A.P.$ is $2$ and the common difference is $4$. The sum of its $40$ terms will be:

The maximum sum of the series $20 + 19\frac{1}{3} + 18\frac{2}{3} + \dots$ is

If $x, y, z \in R^+$ such that $x + y + z = 4$,then the maximum possible value of $xyz^2$ is -

Let $a_n$ be the $n^{\text{th}}$ term of an $A.P.$ If $S_n = a_1 + a_2 + a_3 + \dots + a_n = 700$,$a_6 = 7$,and $S_7 = 7$,then $a_n$ is equal to:

The sum of integers from $1$ to $50$ that are divisible by both $2$ and $3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module