Question

if 20th term of an ap is 1/40 and 40th term of the same AP is 1/20 . then the sum of 800 term is​

Answer 1

$Let \: 'a' \:and \: 'd' \: are \: first \:term \:and$

$Common \: difference \:an \:A.P$

$\boxed{\pink{ n^{th} \:term \: of \:A.P (a_{n}) = a+(n-1)d}}$

$i) 20^{th} \:term = \frac{1}{40} \:(given)$

$\implies a+19d = \frac{1}{40} \:---(1)$

$ii) 40^{th} \:term = \frac{1}{20} \:(given)$

$\implies a+39d = \frac{1}{20} \:---(2)$

/* Subtract equation (1) from equation (2), we get*/

$\implies 20d = \frac{1}{40}$

$\implies d = \frac{1}{800}\: --(3)$

/* Put value of d in equation (1) , we get */

$\implies a + 19 \times \frac{1}{800} = \frac{1}{40}$

$\implies a = \frac{1}{40} - \frac{19}{800}$

$\implies a = \frac{20 - 19}{800}$

$\implies a = \frac{1}{800}\: --(4)$

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

$Here, a = \frac{1}{800}, \:d = \frac{1}{800}$

$n = 800$

$S_{800} = \frac{800}{2}[2\times \frac{1}{800} + (800-1)\frac{1}{800} ]$

$= 400 [\frac{2}{800} + \frac{799}{800} ]$

$= 400 \times \frac{801}{800}$

$= \frac{801}{2}$

$= 400.5$

Therefore.,

$\red{ Sum \:of \:800 \: terms } \green {= 400.5}$

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Answer 2

Plz see the attached file

Note: 'a' denotes the first term of the A.P and 'd' denotes the common difference.