Suppose that all the terms of an arithmetic p | vedclass

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

Question

(D) Let the first term be $a$ and the common difference be $d$. The sum of the first $n$ terms is given by $S_n = \frac{n}{2} [2a + (n-1)d]$.Given $\f

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) Let the first term be $a$ and the common difference be $d$. The sum of the first $n$ terms is given by $S_n = \frac{n}{2} [2a + (n-1)d]$.Given $\frac{S_7}{S_{11}} = \frac{6}{11}$,we have $\frac{\frac{7}{2}(2a + 6d)}{\frac{11}{2}(2a + 10d)} = \frac{6}{11}$.Simplifying this,$\frac{7(a + 3d)}{11(a + 5d)} = \frac{6}{11}$ $\Rightarrow 7a + 21d = 6a + 30d$ $\Rightarrow a = 9d$.The seventh term $a_7 = a + 6d = 9d + 6d = 15d$.Given $130 Since all terms are natural numbers,$d$ must be an integer. Thus,$d = 9$.

Suppose that all the terms of an arithmetic progression $(A.P.)$ are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6:11$ and the seventh term lies between $130$ and $140$,then the common difference of this $A.P.$ is

Similar Questions

If $a, b, c, d, e, f$ are in $A.P.$,then the value of $e - c$ is

Let $a_1, a_2, a_3, \ldots, a_n$ be positive real numbers. Then the minimum value of $\frac{a_1}{a_2}+\frac{a_2}{a_3}+\ldots+\frac{a_n}{a_1}$ is

If the sum of the first $n$ terms of an arithmetic progression is $cn^2$,what is the sum of the squares of these $n$ terms?

Find the sum of all numbers between $200$ and $400$ which are divisible by $7.$

If $S_1, S_2, S_3, \dots, S_m$ are the sums of $n$ terms of $m$ $A.P.s$ whose first terms are $1, 2, 3, \dots, m$ and common differences are $1, 3, 5, \dots, 2m - 1$ respectively,then $S_1 + S_2 + S_3 + \dots + S_m = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module