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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
Let $a_n, n \geq 1$, be an arithmetic progression with first term $2$ and common difference $4$ . Let $M_n$ be the average of the first $n$ terms. Then the sum $\sum \limits_{n=1}^{10} M_n$ is
  • ✓
    $110$
  • B
    $335$
  • C
    $770$
  • D
    $1100$
✓

Answer

Correct option: A.
$110$
a
(a)

The sum of first $n, n \geq 1$ terms of arithmetic progression with first term $2$ and common difference $4$, is

$S_n=\frac{n}{2}[4+(n-1) 4]=2 n^2$

So, the average of the first $n$ terms

$M_n=\frac{S_n}{n}=2 n$

Now, $\sum\limits_{n=1}^{10} M_n=2 \sum\limits_{n=1}^{10} n$

$=2 \times\left(\frac{10 \times 11}{2}\right)=110$

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