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

If $\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$ is the $A.M.$ between $a$ and $b,$ then find the value of $n$.

When the $9^{th}$ term of an $A.P.$ is divided by its $2^{nd}$ term,the quotient is $5$. When the $13^{th}$ term is divided by the $6^{th}$ term,the quotient is $2$ and the remainder is $5$. Find the first term of the $A.P.$

If $a_1, a_2, a_3, \dots, a_{21}$ are in $A.P.$ and $a_3 + a_5 + a_{11} + a_{17} + a_{19} = 10$,then the value of $\sum_{r=1}^{21} a_r$ is:

If $S_n = nP + \frac{1}{2}n(n - 1)Q$,where $S_n$ denotes the sum of the first $n$ terms of an $A.P.$,then the common difference is

The arithmetic mean of the nine numbers in the given set $\{9, 99, 999, \dots, 999999999\}$ is a $9$-digit number $N$,all whose digits are distinct. The number $N$ does not contain the digit:

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 $T_n = S_n - S_{n-1}$.