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8. Sequences and Series
easy
If $\frac{{3 + 5 + 7 + ..........{\rm{to}}\;n\;{\rm{terms}}}}{{5 + 8 + 11 + .........{\rm{to}}\;10\;{\rm{terms}}}} = 7$, then the value of $n$ is
A
$35$
B
$36$
C
$37$
D
$40$
Solution
(a) We have $\frac{{3 + 5 + 7 + ……{\rm{upto}}\;n\;{\rm{terms}}}}{{5 + 8 + 11 + ……..{\rm{upto}}\;10\;{\rm{terms}}}} = 7$
$ \Rightarrow \frac{{\frac{n}{2}[6 + (n – 1)2]}}{{\frac{{10}}{2}[10 + (10 – 1)3]}} = 7 $
$\Rightarrow \frac{{n(2n + 4)}}{{10 \times 37}} = 7$
$ \Rightarrow $${n^2} + 2n – 1295 = 0 $
$\Rightarrow (n + 37)(n – 35) = 0$
Hence $n = 35$.
Standard 11
Mathematics