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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