If An^th term is 2n+1 then prove that the ratio of the sum of odd no. and even no. is n+1:n.

If An^th term is 2n+1 then prove that the ratio of the sum of odd no.

1.	If An^th term is 2n+1 then prove that the ratio of the sum of odd no. and even no. is n+1:n.
Answer» Let a be the first term and d be the\xa0common difference respectively of the given A.P. Let ak denote the kth terms of the given A.P. Then,ak\xa0= a + (k - 1) dNow, S1 = Sum of odd terms{tex}\\Rightarrow{/tex}\xa0S1 = a1 + a3 + a5 +... + a2n + 1{tex}\\Rightarrow \\quad S _ { 1 } = \\frac { n + 1 } { 2 } \\left\\{ a _ { 1 } + a _ { 2 n + 1 } \\right\\}{/tex}{tex}\\Rightarrow \\quad S _ { 1 } = \\frac { n + 1 } { 2 } \\{ a + a + ( 2 n + 1 - 1 ) d ){/tex}[because,a2n+1 = a + (2n + 1 - 1)d]{tex}\\Rightarrow{/tex}\xa0S1 = (n + 1) (a + nd)and, S2 = Sum of even terms{tex}\\Rightarrow{/tex}\xa0S2 = a2 + a4 + a6 + ..... + a2n{tex}\\Rightarrow \\quad S _ { 2 } = \\frac { n } { 2 } \\left[ a _ { 2 } + a _ { 2 n } \\right]{/tex}{tex}\\Rightarrow \\quad S _ { 2 } = \\frac { n } { 2 } [ ( a + d ) + \\{ a + ( 2 n - 1 ) d \\} ]{/tex}{tex}\\left[ \\because a _ { 2 n } = a + ( 2 n - 1 ) d \\right]{/tex}{tex}\\Rightarrow \\quad S _ { 2 } = n ( a + n d ){/tex}Therefore, S1:S2 = (n + 1)(a + nd):n(a + nd) = (n + 1):n