Question

Which value of Sn-Sn_1=?
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Answer 1

Answer

For example, let's consider an AP with a=2 and d=4.

S = (n/2)[2a + (n-1)d]

S = (n/2)[4 + (n-1)4]

S = n(2 + 2n - 2)

S = 2n^2

S(n) = 2n^2

By recursion,

S(n-1) = 2(n-1)^2

S(n-1) = 2n^2 - 4n + 2

Subtracting s(n-1) from S(n),

T(n) = 4n - 2

OR

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]