Rajasthan 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 nth 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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