Q- IF Sn DENOTES THE SUM OF n TERMS OF AN AP WHOSE COMMON DIFFERENCE IS d AND FIRST TERM IS a, FIND Sn - 2Sn-1 + Sn-2....ITS OF AP.....GIVE FULL ANSWER WITH STEPS OR ELSE WILL BE REPORTED!!
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Hello sis ,,,
I think this is the right answer.,!!!!
Given a is first term and d be the common difference.
Sn = (n/2)[ 2a + ( n -1) d]
Now 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]
Hope it helps you
Happy Learning.,!! :-D
I think this is the right answer.,!!!!
Given a is first term and d be the common difference.
Sn = (n/2)[ 2a + ( n -1) d]
Now 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]
Hope it helps you
Happy Learning.,!! :-D
thanks
Your most welcome sis.,!!! :-)
hey can u answer the other ques which I hav asked?
Answered by
3
hope it is helpful for you
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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]