Statement-I: If the ratio of the sum of n ter | vedclass

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

Question

(D) Let the two arithmetic progressions have first terms $a_1, a_2$ and common differences $d_1, d_2$ respectively.The sum of $n$ terms is given by $S

Vedclass · Accepted Answer

Statement-$I$ is false. Statement-$II$ is true.

Vedclass · Answer

(D) Let the two arithmetic progressions have first terms $a_1, a_2$ and common differences $d_1, d_2$ respectively.The sum of $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.The ratio of sums is $\frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a_1 + (n-1)d_1]}{\frac{n}{2}[2a_2 + (n-1)d_2]} = \frac{2a_1 + (n-1)d_1}{2a_2 + (n-1)d_2} = \frac{7n+1}{4n+17}$.The $n^{th}$ term is $T_n = a + (n-1)d$.To find the ratio of $n^{th}$ terms,we replace $(n-1)$ with $(2n-2)$ in the sum ratio formula,effectively setting $n = 2n-1$.Ratio of $n^{th}$ terms = $\frac{a_1 + (n-1)d_1}{a_2 + (n-1)d_2} = \frac{2a_1 + (2n-2)d_1}{2a_2 + (2n-2)d_2} = \frac{7(2n-1)+1}{4(2n-1)+17} = \frac{14n-6}{8n+13}$.This is not $7:4$. Thus,Statement-$I$ is false.Statement-$II$ is a standard property of sequences where the $n^{th}$ term is the difference between the sum of $n$ terms and the sum of $(n-1)$ terms. This is true.Therefore,Statement-$I$ is false and Statement-$II$ is true.

Statement-$I$: If the ratio of the sum of $n$ terms of two arithmetic progressions is $(7n + 1) : (4n + 17)$,then the ratio of their $n^{th}$ terms is $7 : 4$.
Statement-$II$: If $S_n = an^2 + bn + c$,then $T_n = S_n - S_{n-1}$.

Similar Questions

The sum of the first $p, q,$ and $r$ terms of an $A.P.$ are $a, b,$ and $c,$ respectively. Prove that $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$.

If $p(\frac{1}{q}+\frac{1}{r}), q(\frac{1}{r}+\frac{1}{p}), r(\frac{1}{p}+\frac{1}{q})$ are in $AP$,then $p, q, r$:

Let $a_1, a_2, a_3, \ldots$ be terms of an $A.P.$ If $\frac{a_1 + a_2 + \ldots + a_p}{a_1 + a_2 + \ldots + a_q} = \frac{p^2}{q^2}$ for $p \ne q$,then $\frac{a_6}{a_{21}}$ equals:

If the sum and product of the first three terms in an $A.P.$ are $33$ and $1155$,respectively,then a value of its $11^{th}$ term is

The sequence $\frac{5}{\sqrt{7}}, \frac{6}{\sqrt{7}}, \sqrt{7}, \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Statement-$I$: If the ratio of the sum of $n$ terms of two arithmetic progressions is $(7n + 1) : (4n + 17)$,then the ratio of their $n^{th}$ terms is $7 : 4$.Statement-$II$: If $S_n = an^2 + bn + c$,then $T_n = S_n - S_{n-1}$.