Class 11 Math Algebra Sequences and series

Find the value of N if

$1+4+7+10+.......$to n terms $=590$

$1+4+7+10+.......$to n terms $=590$

(a)

20

Correct answer: (a)

Solution: we have

$1+4+7+10+.......$to n terms $=590$

so, here $a=1$

$d=3$

$s_{n}=590$

$⇒[2×1+(n−1)3]=590$

$⇒[2+3n−3]=590$

$⇒n[3n−1]=1180$

$⇒3n_{2}−n−1180−0$

$⇒3n(n−20)+59(n−20)=0$

$⇒(n−20)(3n+59)=0$

$⇒n−20=0$

$⇒n=20$

$1+4+7+10+.......$to n terms $=590$

so, here $a=1$

$d=3$

$s_{n}=590$

$⇒[2×1+(n−1)3]=590$

$⇒[2+3n−3]=590$

$⇒n[3n−1]=1180$

$⇒3n_{2}−n−1180−0$

$⇒3n(n−20)+59(n−20)=0$

$⇒(n−20)(3n+59)=0$

$⇒n−20=0$

$⇒n=20$

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