Question

23 no question plz its urgent anyone know how to do

Answer 1

Answer:

Hope it helps you.....

Answer 2

Question:

If $S_n$ denotes the sum of $n$ terms of an AP whose common difference is $d$ and first term is $a,$ find $S_n-2S_{n-1}+S_{n-2}.$

Answer:

$\mathbf {d}$

Solution:

Before, we may note that,

$S_1=a_1\\\\S_2=a_1+a_2=S_1+a_2\\\\S_3=a_1+a_2+a_3=S_1+a_2+a_3=S_2+a_3$

Randomly, as an example,

$S_{10}=S_9+a_{10}=S_8+a_9+a_{10}=S_7+a_8+a_9+a_{10}=\dots$

Algebraically,

$S_n=S_{n-1}+a_{n-1}\\\\S_n=S_{n-2}+a_{n-1}+a_n\\\\S_n=S_{n-3}+a_{n-2}+a_{n-1}+a_n\\\\\dots\dots\dots$

Generally, we can say that,

$\large\boxed {\displaystyle S_n=S_{n-k}+\sum_{r=0}^{k-1}a_{n-r}}$

where $a_i$ is the i'th term of the AP.

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So we can have,

$S_n=S_{n-2}+a_{n-1}+a_n\\\\S_{n-1}=S_{n-2}+a_{n-1}$

Or it is also true that,

$S_n-S_{n-1}=a_n\quad\iff\quad S_n=S_{n-1}+a_n\\\\S_{n-1}-S_{n-2}=a_{n-1}\quad\iff\quad S_{n-1}=S_{n-2}+a_{n-1}$

Then, finally,

$S_n-2S_{n-1}+S_{n-2}\\\\=S_{n-2}+a_{n-1}+a_n-2[S_{n-2}+a_{n-1}]+S_{n-2}\\\\=S_{n-2}+a_{n-1}+a_n-2S_{n-2}-2a_{n-1}+S_{n-2}\\\\=a_{n-1}+a_n-2a_{n-1}\\\\=a_n-a_{n-1}\\\\=\mathbf{d}$

Or,

$S_n-2S_{n-1}+S_{n-2}\\\\=S_n-[S_{n-1}+S_{n-1}]+S_{n-2}\\\\=S_n-S_{n-1}-S_{n-1}+S_{n-2}\\\\=[S_n-S_{n-1}]-[S_{n-1}-S_{n-2}]\\\\=a_n-a_{n-1}\\\\=\mathbf{d}$

[Note that $a_p-a_q=(p-q)d$ ]

Thus the common difference d is the answer.

$\large\boxed {\mathbf{S_n-2S_{n-1}+S_{n-2}=d}}$

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