An arithmetic progression is written in the following way The sum of all the terms of the …

Question

An arithmetic progression is written in the following way

The sum of all the terms of the 10th row is \_\_\_\_\_\_\_\_\_.

Answer: 1505

Accepted Answer

First term is each row form pattern

$\begin{gathered} \Rightarrow {T}_{n}=a{n}^{2}+bn+c \ \Rightarrow {T}_{1}=a+b+c=2 \ \Rightarrow {T}_{2}=4a+2b+c=5 \ \Rightarrow {T}_{3}=9a+3b+c=11 \ \Rightarrow 3a+b=3 \ 5a+b=6 \ \Rightarrow 2a=3\Rightarrow a=\frac{3}{2},\, b=\frac{-3}{2}\Rightarrow c=2 \ \Rightarrow {T}_{n}=\frac{3}{2}{n}^{2}-\frac{3(n)}{2}+2\Rightarrow \frac{3{n}^{2}-3n+4}{2} \ {T}_{10}=\frac{3	imes 100-3	imes 10+4}{2}=\frac{274}{2}=137 \end{gathered}$

Terms in ${10}^{}$ row is 10 with 3 differences

$\begin{gathered} \Rightarrow 137,140,143\ldots \ \Rightarrow \, {S}_{10}=\frac{10}{2}(2	imes 137+(10-1)	imes 3) \ =5(274+27)=5	imes 301=1505 \end{gathered}$

Solutions · 1