Question

the sum of first 15 term of an A.P is 255 and that of 20 is 440 ,find the sum of n term.​

Answer 1

Answer:

acha laga to like and follow or else reject

Answer 2

Answer:

• N terms of given A.P be 2n + n².

Step-by-step explanation:

${\underline{\bf Given:}}\\ \\ \\ \longrightarrow \sf Sum\;of\;first\;15\;term\;(S_{15})=255\\ \\ \\ \longrightarrow \sf Sum\;of\;first\;20\;terms\;(S_{20})=440\\ \\ \\ {\underline{\bf To\;Find:}}\\ \\ \\ \longrightarrow \sf Sum\;of\;n\;terms.\\ \\ \rule{200}{1}$

$\longrightarrow \sf S_{n}=\dfrac{n}{2}[2a+(n-1)d]\\ \\ \\ \longrightarrow \sf S_{15}=\dfrac{15}{2}[2a+(15-1)d]\\ \\ \\ \longrightarrow \sf 255= \dfrac{15}{2}[2a+14d]\\ \\ \\ \longrightarrow \sf 255 = \dfrac{15}{2}\times 2[a+7d]\\ \\ \\ \longrightarrow \sf 255=15[a+7d]\\ \\ \\ \longrightarrow \sf \dfrac{255}{15}=a+7d\\ \\ \\ \longrightarrow \sf 17=a+7d\;\;\;\;\;\;..........(1)\\ \\ \rule{200}{1}$

$\longrightarrow \sf S_{n}=\dfrac{n}{2}[2a+(n-1)d]\\ \\ \\ \longrightarrow \sf S_{20}=\dfrac{20}{2}[2a+(20-1)d]\\ \\ \\ \longrightarrow \sf 440= 10[2a+19d]\\ \\ \\ \longrightarrow \sf \dfrac{440}{10}=2a+19d\\ \\ \\ \longrightarrow \sf 44=2a+19d\;\;\;\;\;\;..........(2)\\ \\ \rule{200}{1}$

${\underline{\bf Now,\;we\;will\;solve\;eq.\;(1)\;and\;eq.\;(2),\;by\;substitution\;method.}}\\ \\ \\ \longrightarrow \sf a+7d=17\;\;\;\;\;\;........(1)\\ \\ \\ \longrightarrow \sf 2a+19d=44\;\;\;\;\;\;..........(2)\\ \\ \rule{200}{1}$

$\longrightarrow \sf a+7d=17\\ \\ \\ \longrightarrow \sf a=17-7d\\ \\ \\ {\underline{\bf Put\;the\;value\;of\;a\;in\;eq.\;(2),\;we\;get}}\\ \\ \\ \longrightarrow \sf 2a + 19d=44\\ \\ \\ \longrightarrow \sf 2[17-7d]+19d=44\\ \\ \\ \longrightarrow \sf 34 - 14d+19d=44\\ \\ \\ \longrightarrow \sf 5d=10\\ \\ \\ \longrightarrow \sf d=\dfrac{10}{5}\\ \\ \\ \longrightarrow {\boxed{\sf d=2}}$

${\underline{\bf Now, put\;value\;of\;d\;in\;eq.\;(1),\;we\;get,}} \\ \\ \\ \longrightarrow \sf a+7d=17\\ \\ \\ \longrightarrow \sf a+7(2)=17\\ \\ \\ \longrightarrow \sf a+14 = 17\\ \\ \\ \longrightarrow \sf a=17-14\\ \\ \\ \longrightarrow {\boxed{\sf a=3}}\\ \\ \rule{200}{1}$

${\underline{\bf Now,\;we\;will\;find\;sum\;of\;n\;terms,}}\\ \\ \\ \longrightarrow \sf S_{n}=\dfrac{n}{2}[2a+(n-1)d]\\ \\ \\ \longrightarrow \sf S_{n}=\dfrac{n}{2}[6+(n-1)2]\\ \\ \\ \longrightarrow \sf S_{n}=\dfrac{n}{2}[6+2n-2]\\ \\ \\ \longrightarrow \sf S_{n}=\dfrac{n}{2}[4+2n]\\ \\ \\ \longrightarrow \sf S_{n}=\dfrac{n}{2}\times 2[2+n]\\ \\ \\ \longrightarrow {\boxed{\bf S_{n}=2n+n^{2}}}$

${\boxed{\bf Hence,\;n\;terms\;of\;given\;A.P.=2n+n^{2}}}$

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