Six numbers are in an AP such that their sum | vedclass

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

Question

(D) Let the six terms of the $AP$ be $a-5d, a-3d, a-d, a+d, a+3d, a+5d$ with common difference $2d$. Sum of the terms $= (a-5d) + (a-3d) + (a-d) + (a+

Vedclass · Accepted Answer

-$4$

Vedclass · Answer

(D) Let the six terms of the $AP$ be $a-5d, a-3d, a-d, a+d, a+3d, a+5d$ with common difference $2d$. Sum of the terms $= (a-5d) + (a-3d) + (a-d) + (a+d) + (a+3d) + (a+5d) = 6a$. Given $6a = 3$, so $a = \frac{1}{2}$. Given $T_1 = 4T_3$, where $T_1 = a-5d$ and $T_3 = a-d$. $a-5d = 4(a-d)$ $\Rightarrow a-5d = 4a-4d$ $\Rightarrow -3a = d$. Substituting $a = \frac{1}{2}$, we get $d = -3 	imes \frac{1}{2} = -\frac{3}{2}$. The fifth term is $T_5 = a+3d$. $T_5 = \frac{1}{2} + 3(-\frac{3}{2}) = \frac{1}{2} - \frac{9}{2} = -\frac{8}{2} = -4$.

Six numbers are in an $AP$ such that their sum is $3$. The first term is $4$ times the third term. Then, the fifth term is:

Similar Questions

The $n^{th}$ term of an $A.P.$ is $3n - 1$. Choose from the following the sum of its first five terms.

$150$ workers were engaged to finish a job in a certain number of days. $4$ workers dropped out on the second day,$4$ more workers dropped out on the third day,and so on. It took $8$ more days to finish the work. Find the number of days in which the work was completed.

If the roots of the equation $x^3+ax^2+bx+c=0$ are in arithmetic progression,then

Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 390$ and the ratio of the tenth and the fifth terms is $15:7$,then $S_{15} - S_5$ is equal to:

If the $7^{th}$ term of an $A.P.$ is $40$,then the sum of the first $13$ terms is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module