CBSE BoardEnglish MediumSTD 11 ScienceMathsArithmetic Progressions3 Marks
Question
Solve:$1 + 4 + 7 + 10 + ... + x = 590.$
✓
Answer
$1 + 4 + 7 + 10 + ... + x = 590.$
Here,
$\text{a}=1$
$\text{d}=4-1=3$
Let there be n terms so the $n^{th} $ term $= x$
$\Rightarrow\text{x}=1+(\text{n}-1)3$ $[\therefore\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow\text{x}=3\text{n}-2\ .....(1)$
and
$\text{s}_{\text{n}}=590 [$given$]$
$\Rightarrow\frac{\text{n}}{2}[\text{a}+\text{l}]=590$
$\Rightarrow\frac{\text{n}}{2}[1+3\text{n}-2]$ $[\because\text{l}=\text{x}=\text{3n}-2]$
$\Rightarrow3\text{n}^2-\text{n}-1080=0$
$\Rightarrow3\text{n}^2-60\text{n}+59(\text{n}-20)=0$
$\Rightarrow3\text{n}(\text{n}-20)+59(\text{n}-20)=0$
$\Rightarrow\text{n}=2 .....(2)$
from$(1)$ and $(2)$
$\text{x}=3\text{n}-2$
$=3(20)-2$
$=58$
$\text{x}=58$
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