Question

1+3+5+7+9+11+....101=?
4+6+8+....104=?​

Answer 1

$Answer \: \: \\ \\ Question \: \: Number \: \: 01 \\ \\ 1 + 3 + 5 + 7 + 9 + 11 + ... + 101 \\ \\ this \: series \: is \: in \: ap \: becoz \: common \\ difference \: is \: same \\ \\ an = a + (n - 1)d \: \\ where \: an \: is \: nth \: term \: a \: is \: ist \: \\ term \: n \: is \: number \: of \: terms \: and \: \\ d \: is \: common \: difference \\ \\ here \: an \: = 101 \: \: a \: = 1 \: \: d \: = 3 - 1 = 2 \\ \\ 101 = 1 + (n - 1)2 \\ \\ 101 - 1 = (n - 1)2 \\ \\ 100 = (n - 1)2 \\ \\ (n - 1) = \frac{100}{2} \\ \\ (n - 1) = 50 \\ \\ n = 51 \\ \\ so \: the \: number \: of \: terms \: is \: 51 \\ \\ sum \: of \: this \: series \: can \: be \: calculated \: as \: \\ \\ s = \frac{n}{2} (an + a) \\ \\ s = \frac{51}{2} (101 + 1) \\ \\ s = \frac{51}{2} (102) \\ \\ s = 51 \times 51 \\ \\ s = 2601 \\ \\ \\ Question \: Number \: \: \: 02 \\ \\ 4 + 6 + 8 + 10 + ... + 104 \\ this \: series \: is \: also \: in \: ap \: as \: the \: common \\ difference \: is \: same \: \\ \\ an = a + (n - 1)d \\ where \: the \: symbols \: have \: their \: usually \: meaning \: \\ \\ 104 = 4 + (n - 1)2 \\ \\ 104 - 4 = 2(n - 1) \\ \\ 100 = 2(n - 1) \\ \\ \frac{100}{2} = (n - 1) \\ \\ 50 = (n - 1) \\ \\ n = 51 \\ \\ its \: sum \: is \: given \: by \: \\ \\ s = \frac{51}{2} (104 + 4) \\ \\ s = \frac{51}{2} (108) \\ \\ s = 51 \times 54 \\ \\ s = 2754$

Answer 2

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