8. Sequences and Series
hard

माना कि किसी समांतर श्रेणी के $n, 2 n,$ तथा $3 n$ पदों का योगफल क्रमशः $S _{1}, S _{2}$ तथा $S _{3}$ है तो दिखाइए कि $S _{3}=3\left( S _{2}- S _{1}\right)$

Option A
Option B
Option C
Option D

Solution

Let $a$ and $b$ be the first term and the common difference of the $A.P.$ respectively. Therefore,

$S_{1}=\frac{n}{2}[2 a+(n-1) d]$         ………$(1)$

$S_{2}=\frac{2 n}{2}[2 a+(2 n-1) d]=n[2 a+(2 n-1) d]$         …….$(2)$

$S_{3}=\frac{3 n}{2}[2 a+(3 n-1) d]$          ……….$(3)$

From $(1)$ and $(2),$ we obtain

$S_{2}-S_{1}=n[2 a+(2 n-1) d]-\frac{n}{2}[2 a+(n-1) d]$

$=n\left\{\frac{4 a+4 n d-2 d-2 a-n d+d}{2}\right\}$

$=n\left[\frac{2 a+3 n d-d}{2}\right]$

$=\frac{n}{2}[2 a+(3 n-1) d]$

$\therefore 3\left(S_{2}-S_{1}\right)=\frac{3 n}{2}[2 a+(3 n-1) d]=S_{3}$         [ From $(3)$ ]

Hence, the given result is proved.

Standard 11
Mathematics

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