Let $a_1,a_2,\dots a_n$ be a given A.P. whose common difference is an integer and $S_n=a_1+a_2$ $+\dots +a_n$ . If $a_1=1,a_1=300$ and $15\le n\le 50$ , then the ordered pair $\left(S_{n-4},a_{n-4}\right)$ is equal to:

Given ${\left(2x^2+3x+4\right)}^{10}={\sum }_{r=0}^{20}?a_r{x}^r$

replace $x$ by $\frac{2}{x}$ in above identity :-

$\begin{array}{rr}& \frac{2^{10}{\left(2x^2+3x+4\right)}^{10}}{x^{20}}=\sum _{r=0}^{20}??\frac{a_r{2}^r}{x^r}\\ & \Rightarrow 2^{10}\sum _{r=0}^{20}??a_r{x}^r=\sum _{r=0}^{20}??a_r{2}^r{x}^{(20-r)}\left(\text{from}\right(i\left)\right)\end{array}$

now, comparing coefficient of $x^7$ from both sides

(take $r=7$ in L.H.S. and $r=13$ in R.H.S.)

$2^{10}{a}_7=a_{13}{2}^{13}\Rightarrow \frac{a_7}{a_{13}}=2^3=8$

$1+\left(1-2^2.1\right)+\left(1-4^2.3\right)+\dots \dots .+\left(1-{20}^2.19\right)$

$=\alpha -220\beta$

$=11-\left(2^2\cdot 1+4^2\cdot 3+\dots ..+{20}^2\cdot 19\right)$

$=11-2^2\cdot {\sum }_{r=1}^{10}?r^2(2r-1)=11-4\left(\frac{{110}^2}{2}-35\times 11\right)$

$=11-220\left(103\right)$

$\Rightarrow \alpha =11,\beta =103$

$a_1,a_2,a_3......$ are in A.P. (Let common difference is d₁)

$b_1,b_2,b_3....$  are in A.P. (Let common difference is d₂)

and a₁2, a₁₀ = 3, a₁b₁ = 1 = a₁₀b₁₀

$?a_1{b}_1=1$               

$\therefore b_1=\frac{1}{2}$

$a_{10}{b}_{10}=1$               

$\therefore b_{10}=\frac{1}{3}$              2026 Info edge India Ltd. All rights reserved.