Suppose that the number of terms in an A.P. i | vedclass

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

Question

(A) Let the $A.P.$ be $a_1, a_2, a_3, \ldots, a_{2k}$.Given the sum of odd-positioned terms: $\sum_{r=1}^{k} a_{2r-1} = 40$.Given the sum of even-posi

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Let the $A.P.$ be $a_1, a_2, a_3, \ldots, a_{2k}$.Given the sum of odd-positioned terms: $\sum_{r=1}^{k} a_{2r-1} = 40$.Given the sum of even-positioned terms: $\sum_{r=1}^{k} a_{2r} = 55$.The difference between the sum of even and odd terms is $\sum_{r=1}^{k} (a_{2r} - a_{2r-1}) = 55 - 40 = 15$.Since $a_{2r} - a_{2r-1} = d$,we have $k 	imes d = 15$,so $d = \frac{15}{k}$.The last term is $a_{2k} = a_1 + (2k-1)d$. Given $a_{2k} - a_1 = 27$,we have $(2k-1)d = 27$.Substituting $d = \frac{15}{k}$ into the equation: $(2k-1) \frac{15}{k} = 27$.$15(2k-1) = 27k \Rightarrow 30k - 15 = 27k$.$3k = 15 \Rightarrow k = 5$.

Suppose that the number of terms in an $A.P.$ is $2k$,where $k \in N$. If the sum of all odd-positioned terms of the $A.P.$ is $40$,the sum of all even-positioned terms is $55$,and the last term of the $A.P.$ exceeds the first term by $27$,then $k$ is equal to:

Similar Questions

If the $m^{th}$ term of an arithmetic progression is $1/n$ and the $n^{th}$ term is $1/m$,then the sum of the first $mn$ terms is:

Given an $A.P.$ whose terms are all positive integers. The sum of its first nine terms is greater than $200$ and less than $220$. If the second term is $12$,then its $4^{th}$ term is:

If the sum of three numbers in an arithmetic sequence is $15$ and the sum of their squares is $83$,then the numbers are

What is the sum of all integers between $81$ and $719$ that are divisible by $5$?

The sum of three terms of an Arithmetic Progression $(AP)$ is $18$ and the sum of their squares is $158$. The largest term is.......

Vedclass Test Series

Exam Paper Generator

Online Exam Module