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Sia ? 6 years, 6 months ago
Let n = 4q + 1 (an odd integer)
{tex}\therefore \quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}
{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \quad \text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}
{tex}= 16{q^2} + 8q{/tex}
{tex}= 8 \left( 2 q ^ { 2 } + q \right){/tex}
= 8m, which is divisible by 8.
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Sia ? 6 years, 6 months ago
Let a be the first term and d be the common difference of the given A.P. Then,
Sm = Sn
{tex}\Rightarrow \quad \frac { m } { 2 } \{ 2 a + ( m - 1 ) d \} = \frac { n } { 2 } \{ 2 a + ( n - 1 ) d \}{/tex}
{tex} \Rightarrow{/tex} 2a (m - n) + {m (m - 1) - n (n - 1)} d = 0
{tex} \Rightarrow{/tex} 2a (m - n) + {m2 - m - n2 + n}d = 0
{tex} \Rightarrow{/tex} 2a (m - n) + {(m2 - n2) - (m - n)} d = 0
{tex} \Rightarrow{/tex}2a (m - n) + {(m - n) (m +n) - (m - n)} d = 0
{tex} \Rightarrow{/tex} (m - n) {2a + (m + n - 1)d} = 0
{tex} \Rightarrow{/tex} 2a + (m + n - 1)d = 0 {tex} [ \because m - n \neq 0 ]{/tex} ...(i)
Now, {tex} S _ { m + n } = \frac { m + n } { 2 } \{ 2 a + ( m + n - 1 ) d \} = \frac { m + n } { 2 } \times {/tex}0 = 0 [Using (i)]
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Sanjana Yadav 7 years, 9 months ago
1Thank You