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First if u give me correct answer i will mark u as brainliest
Plsee
only if answer is given
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[tex]\rule{200}{2}[/tex]
QUESTION:
[tex]\textsf{If $S_n$ denotes the sum of n terms of an AP} \\ \textsf{whose common difference is d and first term is a} \\ \textsf{Find $S_n$ + 2 $S_{n - 1}$ + $S_{n - 1}$}.[/tex]
[tex]\rule{200}{2}[/tex]
ANSWER:
[tex]\rule{200}{2}[/tex]
GIVEN:
[tex]\rule{200}{2}[/tex]
TO FIND:
[tex]\rule{200}{2}[/tex]
TERMS USED:
[tex] \boxed{ \bold{\large{S_n = \frac{n}{2} (2a + (n - 1)d)}}}[/tex]
[tex]2S_{n-1} = 2\bigg(\dfrac{n - 1}{2} \bigg)( 2a + (n-1-1)d)[/tex]
[tex]2S_{n-1} = (n - 1)(2a + (n-2)d)[/tex]
[tex]S_{n-2} = \bigg(\dfrac{n - 2}{2} \bigg)2a + (n-2 - 1)d[/tex]
[tex]S_{n-2} = \dfrac{n - 2}{2}(2a + (n-3)d)[/tex]
NOTE : REFER ATTACHMENT FOR CALCULATION.
[tex]\rule{200}{2}[/tex]
Answer:
Sn = (n/2)[ 2a + ( n -1) d]
then Sn - 2Sn-1 + Sn + 2
⇒ (n/2)[ 2a + ( n -1) d] - 2(n - 1)/2)[ 2a + ( n - 1 -1) d] + (n +2) / 2)[ 2a + ( n + 2 -1) d]
⇒ (1/2)[ 2an + n( n -1) d] + [ 4a(n - 1) + 2(n - 1)( n - 2) d] +[ 2a(n + 2)+ ( n + 1) (n + 2) d]
⇒ (1/2)[ 2a[ n - 2n + 2 + n + 2] + d [ n2 - n - 2n2 + 6n - 4 + n2 + 3n + 2] ]
⇒ (1/2)[ 2a(4) + d(8n - 2) ]
= [ 4a + (4n - 1)d]