JEE MAIN - Mathematics (2024 - 8th April Evening Shift - No. 29)
An arithmetic progression is written in the following way
The sum of all the terms of the 10th row is _________.
Explanation
First term is each row form pattern
$$\begin{aligned} & \Rightarrow T_n=a n^2+b n+c \\ & \Rightarrow T_1=a+b+c=2 \\ & \Rightarrow T_2=4 a+2 b+c=5 \\ & \Rightarrow T_3=9 a+3 b+c=11 \\ & \Rightarrow 3 a+b=3 \\ & 5 a+b=6 \\ & \Rightarrow 2 a=3 \Rightarrow a=\frac{3}{2}, \quad b=\frac{-3}{2} \Rightarrow c=2 \\ & \Rightarrow T_n=\frac{3}{2} n^2-\frac{3(n)}{2}+2 \Rightarrow \frac{3 n^2-3 n+4}{2} \\ & T_{10}=\frac{3 \times 100-3 \times 10+4}{2}=\frac{274}{2}=137 \end{aligned}$$
Terms in $$10^{\text {th }}$$ row is 10 with 3 differences
$$\begin{aligned} & \Rightarrow 137,140,143 \ldots \\ & \Rightarrow \quad S_{10}=\frac{10}{2}(2 \times 137+(10-1) \times 3) \\ & =5(274+27)=5 \times 301=1505 \end{aligned}$$
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