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8. Sequences and Series
hard
श्रेणियों $3+7+11+15+\ldots$ तथा $1+6+11+16+\ldots \ldots$, के बीच उभयनिष्ठ प्रथम $20$ पदों का योग है
A
$4000$
B
$4020$
C
$4200$
D
$4220$
(JEE MAIN-2014)
Solution
Given $n = 20;\,\,{S_{20}} = ?$
Series $\left( 1 \right) \to 3,7,11,15,19,23,27,31,35,$
$39,43,47,$
$51,55,59…$
Series $\left( 2 \right) \to 1,6,11,16,21,26,31,36,41,$
$46,51,56,$
$61,66,71.$
the common terms between both the series are
$11,13,51,71…$
Above series froms an Arithmetic progression $(A.P)$
Therefore, first term$(a)=11$ and
common difference $(d)=20$
Now, ${S_n} = \frac{n}{2}\left[ {2a + \left( {n – 1} \right)d} \right]$
${S_{20}} = \frac{{20}}{2}\left[ {2 \times 11 + \left( {20 – 1} \right)20} \right]$
${S_{20}} = 10\left[ {22 + 19 \times 20} \right]$
${S_{20}} = 10 \times 402 = 4020$
$\therefore {S_{20}} = 4020$
Standard 11
Mathematics
