- Home
- Standard 11
- Mathematics
8. Sequences and Series
hard
ધારો કે $S_n$ એ, સમાંતર શ્રેણી $3,7,11, \ldots . . .$. નાં $n$ પદોનો સરવાળો છે. જો $40<\left(\frac{6}{n(n+1)} \sum_{k=1}^n S_k\right)<42$ હોય,તો $n=$___________.
A
$9$
B
$8$
C
$10$
D
$7$
(JEE MAIN-2024)
Solution
$S_n= 3+7+11+………… n $ terams
${n}{2}(6+(n-1) 4)=3 n+2 n^2-2 n $
$ =2 n^2+n $
$ \sum_{k=1}^n S_k=2 \sum_{k=1}^n K^2+\sum_{k=1}^n K $
$ =2 \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2} $
$=n(n+1)\left[\frac{2 n+1}{3}+\frac{1}{2}\right]$
$=\frac{n(n+1)(4 n+5)}{6} $
Rightarrow $40<\frac{6}{n(n+1)} \sum_{k=1}^n S_k<42 $
$ 40<4 n+5<42 $
$ 35<4 n<37 $
$ n=9$
Standard 11
Mathematics
