Let a_1, a_2, a_3, ..., a_n be in an A.P. If | 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 the end is constant. Specifically,$a_k + a_{n-k+1} = a_1 + a_n$.Given the equatio

Vedclass · Accepted Answer

$306$

Vedclass · Answer

(A) In an $A.P.$,the sum of terms equidistant from the beginning and the end is constant. Specifically,$a_k + a_{n-k+1} = a_1 + a_n$.Given the equation: $a_3 + a_7 + a_{11} + a_{15} = 72$.We observe that the indices sum to $3+15 = 18$ and $7+11 = 18$.Using the property $a_m + a_n = a_p + a_q$ if $m+n = p+q$,we have:$(a_3 + a_{15}) = (a_1 + a_{17})$ 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 of an $A.P.$ is given by $S_{17} = \frac{17}{2}(a_1 + a_{17})$.Substituting the value $a_1 + a_{17} = 36$:$S_{17} = \frac{17}{2} 	imes 36 = 17 	imes 18 = 306$.

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

Similar Questions

Let $a_{1}, a_{2}, a_{3}, \ldots$ be a $G$.$P$. such that $a_{1} < 0$; $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$. If $\sum_{i=1}^{9} a_{i} = 4 \lambda$,then $\lambda$ is equal to:

If the roots of the cubic equation $ax^3 + bx^2 + cx + d = 0$ are in $G.P.$,then

If $a_1, a_2, a_3, . . . , a_n, . . .$ are in $A.P.$ such that $a_4 - a_7 + a_{10} = m$,then the sum of the first $13$ terms of this $A.P.$ is .............. $m$.

Given that $n$ $A$.$M$.'s are inserted between two sets of numbers $a, 2b$ and $2a, b$,where $a, b \in R$. Suppose further that the $m^{th}$ mean between these sets of numbers is the same,then the ratio $a:b$ equals

The value of $1^2 + 3^2 + 5^2 + ....... + 25^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module