Question

the first and the last term of an AP are 5 and 45 .if the sum of all its terms is 400 .find the comman difference​

Answer 1

see attached file for the answer

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Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Common\:difference=\frac{8}{3}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies First \: term(a) = 5 \\ \\ \tt: \implies Last \: term(l) = 45 \\ \\ \tt: \implies Sum \: of \:all \: terms(s_{n}) = 400 \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \tt: \implies Common \: difference(d ) =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies s_{n} = \frac{n}{2}(a + l) \\ \\ \tt: \implies 400 = \frac{n}{2} (5 + 45) \\ \\ \tt: \implies 400 \times 2 = n \times 50 \\ \\ \tt: \implies n = \frac{800}{50} \\ \\ \tt: \implies n = 16 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies l = a + (n - 1)d \\ \\ \tt: \implies 45 = 5 + (16 - 1)d \\ \\ \tt: \implies 45 - 5 = 15 \times d \\ \\ \tt: \implies 40 = 15 \times d \\ \\ \tt: \implies d = \frac{40}{15} \\ \\ \green{\tt: \implies d = \frac{8}{3} }\\\\\green{\tt{\therefore{Common\:difference\:is\:\frac{8}{3}}}}$