Question

Please answer this question fast •‿•​

Answer 1

$\sf\blue{Question}$

$\sf{An \ AP \ has \ 27 \ terms. \ The \ sum \ of \ the}$

$\sf{middle \ term \ and \ the \ two \ terms \ adjacent \ on}$

$\sf{each \ sides \ of \ it \ is \ 177. \ The \ sum \ of \ the}$

$\sf{last \ three \ terms \ is \ 321. \ Find \ the \ first}$

$\sf{terms \ of \ AP.}$

__________________________________

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{First \ three \ terms \ of \ an \ AP \ are}$

$\sf{7, \ 11 \ and \ 15 \ respectively.}$

$\sf\orange{\underline{\underline{Given:}}}$

$\sf{\implies{Sum \ of \ middle \ term \ and \ the}}$

$\sf{two \ terms \ adjacent \ on \ each \ side \ of \ it \ is \ 177.}$

$\sf{\implies{The \ sum \ of \ the \ last \ three \ terms}}$

$\sf{is \ 321.}$

$\sf\pink{To \ find:}$

$\sf{The \ first \ three \ terms \ of \ an \ AP.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{Let \ the \ 27 \ terms \ of \ an \ AP \ be}$

${(a-13d), \ (a-12d), \ (a-11d), \ (a-10d), \ (a-9d),}$

${(a-8d), \ (a-7d), \ (a-6d), \ (a-5d), \ (a-4d), \ (a-3d),}$

${(a-2d), \ (a-d), \ a, \ (a+d), \ (a+2d), \ (a+3d), \ (a+4d),}$

${(a+5d), \ (a+6d), \ (a+7d), \ (a+8d), \ (a+9d), \ (a+10d),}$

${(a+11d), \ (a+12d) \ and \ (a+13d)}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{(a-d)+a+(a+d)=177}$

$\sf{3a=177}$

$\sf{a=\frac{177}{3}}$

$\boxed{\sf{a=59...(1)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{(a+11d)+(a+12d)+(a+13d)=321}$

$\sf{3a+36d=321}$

$\sf{But, \ a=59 \ from \ (1)}$

$\sf{3(59)+36d=321}$

$\sf{177+36d=321}$

$\sf{36d=321-177}$

$\sf{36d=144}$

$\sf{d=\frac{144}{36}}$

$\boxed{\sf{d=4}}$

$\sf{\therefore{First \ three \ terms \ of \ an \ AP \ are}}$

$\sf{(a-13d)=59-13(4)=59-52=7,}$

$\sf{(a-12d)=59-12(4)=59-48=11,}$

$\sf{(a-11d)=59-11(4)=59-44=15}$

$\sf\purple{\tt{\therefore{First \ three \ terms \ of \ an \ AP \ are}}}$

$\sf\purple{\tt{7, \ 11 \ and \ 15 \ respectively.}}$

Answer 2

ɢɪᴠᴇɴ :-

An AP has 27 terms. The sum of the middle term and the Two adjacent on each sides of it is 177.If the sum of the last three terms Ais 321.

ᴛᴏ ғɪɴᴅ :-

• First three terms

sᴏʟᴜᴛɪᴏɴ :-

➥ According to 1st condition :-

• Middle term = a14
• Adjacent terms = a13 and a15

➱ a13 + a14 + a15 = 177. ----(given)

➱ (a + 12d) + (a + 13d) + (a + 14d) = 177

➱ 3a + 39d = 177

➱ 3(a + 13d) = 177

➱ (a + 13d) = 177/3

➱ (a + 13d) = 59

➱ a = (59 - 13d) -----(1)

➥ According to 2nd condition :-

• Last three terms are a25 , a26 , a27

➱ a25 + a26 + a27 = 321 ---(given)

➱ (a + 24d) + (a + 25d) + (a + 26d) = 321

➱ 3a + 75d = 321

➱ 3(a + 25d) = 321

➱ (a + 25d) = 321/3

➱ (a + 25d) = 107

➱ a = (107 - 25d) --(2)

From (1) and (2) , we get,

➱ 59 - 13d = 107 - 25d

➱ 25d - 13d = 107 - 59

➱ 12d = 48

➱ d = 48/12

➱ d = 4

Put d = 4 in (1) , we get,

➱ a = (59 - 13d)

➱ a = (59 - 13×4)

➱ a = (59 - 52)

➱ a = 7

Hence,

• First term (a1) = 7
• Second term (a2) = a + d = 7 + 4 = 11
• Third term (a3) = a + 2d = 7 + 2×4 = 15

Hence,

• Last three terms are 7 , 11 , 15