Sums of n terms of two arithmatic series are in ratio 2n + 3:6n + 5 , then ratio of their {13^{th}}

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8. Sequences and Series

medium

The sums of $n$ terms of two arithmatic series are in the ratio $2n + 3:6n + 5$, then the ratio of their ${13^{th}}$ terms is

$53 : 155$

$27 : 77$

$29 : 83$

$31 : 89$

Solution

(a) We have $\frac{{{S_{{n_1}}}}}{{{S_{{n_2}}}}} = \frac{{2n + 3}}{{6n + 5}}$

==> $\frac{{\frac{n}{2}[2{a_1} + (n – 1){d_1}]}}{{\frac{n}{2}[2{a_2} + (n – 1){d_2}]}} = \frac{{2n + 3}}{{6n + 5}}$

==> $\frac{{2\left[ {{a_1} + \left( {\frac{{n – 1}}{2}} \right)\,{d_1}} \right]}}{{2\left[ {{a_2} + \left( {\frac{{n – 1}}{2}} \right)\,{d_2}} \right]}} = \frac{{2n + 3}}{{6n + 5}}$

==> $\frac{{{a_1} + \left( {\frac{{n – 1}}{2}} \right)\,{d_1}}}{{{a_2} + \left( {\frac{{n – 1}}{2}} \right)\,{d_2}}} = \frac{{2n + 3}}{{6n + 5}}$

Put $n = 25$ then $\frac{{{a_1} + 12{d_1}}}{{{a_2} + 12{d_2}}} = \frac{{2(25) + 3}}{{6(25) + 3}}$

==> $\frac{{{T_{{{13}_1}}}}}{{{T_{{{13}_2}}}}} = \frac{{53}}{{155}}$.

Standard 11

Mathematics

If the variance of the terms in an increasing $A.P.$, $b _{1}, b _{2}, b _{3}, \ldots b _{11}$ is $90,$ then the common difference of this $A.P.$ is

medium

(JEE MAIN-2020)

If $1, \log _{10}\left(4^{x}-2\right)$ and $\log _{10}\left(4^{x}+\frac{18}{5}\right)$ are in
arithmetic progression for a real number $x$ then the value of the determinant $\left|\begin{array}{ccc}2\left(x-\frac{1}{2}\right) & x-1 & x^{2} \\ 1 & 0 & x \\ x & 1 & 0\end{array}\right|$ is equal to …… .

hard

(JEE MAIN-2021)

Let $a_1, a_2, a_3, \ldots$ be an arithmetic progression with $a_1=7$ and common difference $8$ . Let $T_1, T_2, T_3, \ldots$ be such that $T_1=3$ and $T_{n+1}-T_n=a_n$ for $n \geq 1$. Then, which of the following is/are $TRUE$ ?

$(A)$ $T_{20}=1604$

$(B)$ $\sum_{ k =1}^{20} T_{ k }=10510$

$(C)$ $T_{30}=3454$

$(D)$ $\sum_{ k =1}^{30} T_{ k }=35610$

hard

(IIT-2022)

The arithmetic mean of first $n$ natural number

easy

The number of terms in the series $101 + 99 + 97 + ….. + 47$ is