if sn denotes sum of n terms of an ap whose first term is a and common difference is d. find s_n-1+s_n-2

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Solution

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

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if sn denotes sum of n terms of an ap whose first term is a and common difference is d. find sn-1+sn-2

Sn=n2[2a+(n-1)d]Sn-1=n-12[2a+(n-2)d]Sn-2=n-22[2a+(n-3)d]Sn-1+Sn-2=n-12[2a+(n-2)d]+n-22[2a+(n-3)d]

if sn denotes sum of n terms of an ap whose first term is a and common difference is d. find s_n-1+s_n-2

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