Question

find the sum first 10 terms of an AP whose first term is 100 and the 10 term is 1000​

Answer 1

$Given \: first \:term \: of \: A.P : a = 100\: --(1)$

$and \: 10^{th} \:term = 1000$

$\implies a + 9d = 1000$

$\boxed{\pink{ \because a_{n} = a+(n-1)d }}$

$\implies 100 + 9d = 1000$

$\implies 9d = 1000 - 100$

$\implies 9d = 900$

$\implies d = \frac{900}{9}$

$\implies d = 100\: --(2)$

/* We know that */

$\boxed{\pink{Sum \: of \: first \: 'n' \:terms (S_{n}) = \frac{n}{2} [ 2a + (n-1)d ] }}$

$Here, a = 100, d = 100, n = 10$

$S_{10} = \frac{10}{2}[2\times 100+(10-1)100]$

$= 5( 200 + 900 )$

$= 5 \times 1100$

$= 5500$

Therefore.,

$\red{ Sum \:of \: first \:10\: terms } \green {= 5500}$

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