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First term and common difference of an A.P. are 1 and 2 respectively; find S10. Given: a = 1, d = 2 find S10 = ? Sn = n2[2a+(n-1)□] S10 = 102[2×1+(10-1)□] = 102[2+9×□] = 102[2+□] = 102×□=□ - Algebra

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प्रश्न

First term and common difference of an A.P. are 1 and 2 respectively; find S10.

Given: a = 1, d = 2

find S10 = ?

Sn = `n/2 [2a + (n - 1) square]`

S10 = `10/2 [2 xx 1 + (10 - 1)square]`

= `10/2[2 + 9 xx square]`

= `10/2 [2 + square]`

= `10/2 xx square = square`

योग

उत्तर

Given: a = 1, d = 2

find S10 = ?

Sn = `n/2 [2a + (n - 1)bbd]`

S10 = `10/2 [2 xx 1 + (10 - 1)bb2]`

= `10/2[2 + 9 xx bb2]`

= `10/2 [2 + bb18]`

= `10/2 xx bb20` = 100

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