Gujarati
8. Sequences and Series
medium

यदि ${S_n}$ समान्तर श्रेणी के $n$ पदों का योगफल दर्शाता हो, तो $({S_{2n}} - {S_n})$ का मान है

A

$2{S_n}$

B

${S_{3n}}$

C

$\frac{1}{3}{S_{3n}}$

D

$\frac{1}{2}{S_n}$

Solution

(c) ${S_{2n}} – {S_n} = \frac{{2n}}{2}\{ 2a + (2n – 1)d\}  – \frac{n}{2}\{ 2a + (n – 1)d\} $

$ = \frac{n}{2}\{ 4a + 4nd – 2d – 2a – nd + d\}  = \frac{n}{2}\{ 2a + (3n – 1)d\} $

 $ = \frac{1}{3}.\frac{{3n}}{2}\{ 2a + (3n – 1)d\}  = \frac{1}{3}{S_{3n}}$.

Standard 11
Mathematics

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