If lim n -r-1na1-3-5-2 r-1-a-2 r n-n -r-1n2 r-1-4- then the value of a isA- 3B- 4 - 6D

The correct option is C 6n →∞lim∑_r = 1^nar^2 + (a 2)rnn(n^2)=4 n →∞lim∑_r = 1^n1/n[a(r/n)^2 + (a 2) r/n] = 4 ⇒∫^1_0[ax^2 + (a 2)x]dx = 4 ⇒a/3 + a 2/2 = 4 ⇒ ...

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Standard XII

Mathematics

Common Difference

If lim n →∞∑r...

Question

If $lim n \to \infty n \sum r = 1 \frac{a (1 + 3 + 5 + \dots + (2 r - 1)) + (a - 2) r n}{n n \sum r = 1 (2 r - 1)} = 4$ , then the value of $a$ is

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Solution

The correct option is C $6$
$lim n \to \infty n \sum r = 1 \frac{a r^{2} + (a - 2) r n}{n (n^{2})} = 4 lim n \to \infty n \sum r = 1 \frac{1}{n} [a (\frac{r}{n})^{2} + (a - 2) \frac{r}{n}] = 4 \Rightarrow 1 \int 0 [a x^{2} + (a - 2) x] d x = 4 \Rightarrow \frac{a}{3} + \frac{a - 2}{2} = 4 \Rightarrow a = 6$

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If limn→∞n∑r=1a(1+3+5+…+(2r−1))+(a−2)rnnn∑r=1(2r−1)=4, then the value of a is

The correct option is C 6limn→∞n∑r=1ar2+(a−2)rnn(n2)=4limn→∞n∑r=11n[a(rn)2+(a−2)rn]=4⇒1∫0[ax2+(a−2)x]dx=4⇒a3+a−22=4⇒a=6

If $lim n \to \infty n \sum r = 1 \frac{a (1 + 3 + 5 + \dots + (2 r - 1)) + (a - 2) r n}{n n \sum r = 1 (2 r - 1)} = 4$ , then the value of $a$ is