Question

sum of the first 15 terms of an ap is 405 and sum of first 21 terms is 756. find the 8th term and common difference

Answer 1

$\sf\large\underline\purple{Given:-}$

$\sf{\implies Sum\:_{(15th\: terms)}=405}$

$\sf{\implies Sum\:_{(21th\: terms)}=756}$

$\sf\large\underline\purple{To\: Find:-}$

$\sf{\implies Common\: difference\:and\:8th\: terms=?}$

$\sf\large\underline\purple{Solution:-}$

• To calculate the 8th terms of AP ,at first we have to find the value of first term and common difference by setting up equation as per given clue in the question:-]

$\sf\large\underline\purple{Formula\:used:-}$

$\tt{\implies S\:_{(n)}=\dfrac{n}{2}(2a+(n-1)d}$

$\tt{\implies S\:_{(15)}=\dfrac{15}{2}(2a+(15-1)d}$

$\tt{\implies 15(2a+14d)=405*2}$

$\tt{\implies 30a+210d=810------(i)}$

$\tt{\implies S\:_{(21)}=\dfrac{21}{2}(2a+(21-1)d}$

$\tt{\implies 21(2a+20d)=756*2}$

$\tt{\implies 42a+420d=1512-------(ii)}$

• Equation (i) multiplying by 42 and (ii) by 30 then subracting:-]

$\tt{\implies 1260a+8820d=34020}$

$\tt{\implies 1260a+12600d=45360}$

• By solving we get here:-]

$\tt{\implies -3780d=-11340\implies d=3}$

• Putting the value of d=3 in eq (I):-]

$\tt{\implies 30a+210d=810}$

$\tt{\implies 30a+210*3=810}$

$\tt{\implies 30a+630=810}$

$\tt{\implies 30a=810-630\implies a=6}$

• Now calculate 8th term of AP:-]

$\tt{\implies T\:_{(n)}=a+(n-1)d}$

$\tt{\implies T\:_{(8)}=6+(8-1)3}$

$\tt{\implies T\:_{(8)}=6+7*3}$

$\tt{\implies T\:_{(8)}=6+21}$

$\tt{\implies T\:_{(8)}=27}$

$\sf\large{Hence'}$

$\tt{\implies Term\:_{(8th)}=27}$

$\tt{\implies Common\:_{(difference)}=3}$

Answer 2

GIVEN :-

• Tʜᴇ sᴜᴍ ᴏғ ᴛʜᴇ ғɪʀsᴛ 15 ᴛᴇʀᴍs ᴏғ ᴀ.ᴘ ɪs 405 .

• Aɴᴅ ᴛʜᴇ sᴜᴍ ᴏғ ғɪʀsᴛ 21 ᴛᴇʀᴍs ᴏғ ᴛʜᴀᴛ ᴀ.ᴘ ɪs 756 .

TO FIND :-

1. Tʜᴇ 8ᴛʜ ᴛᴇʀᴍ .
2. ᴄᴏᴍᴍᴏɴ ᴅɪғғᴇʀᴇɴᴄᴇ .

SOLUTION :-

Wᴇ ʜᴀᴠᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

$\huge\orange\star$ $\bf\purple{S_n\:=\:\dfrac{n}{2}\:[2a\:+\:(n\:-\:1)\:d]\:}$

Wʜᴇʀᴇ,

• $\bf\red{a}$ = Fɪʀsᴛ ᴛᴇʀᴍ

• $\bf\red{n}$ = ɴᴏ. ᴏғ ᴛᴇʀᴍs

• $\bf\red{d}$ = ᴄᴏᴍᴍᴏɴ ᴅɪғғᴇʀᴇɴᴄᴇ

Aᴄᴄᴏʀᴅɪɴɢ ᴛᴏ ᴛʜᴇ ǫᴜᴇsᴛɪᴏɴ,

ᴄᴀsᴇ - 1 :-

$\purple\checkmark\:\bf{S_{15}\:=\:\dfrac{15}{2}\:[2a\:+\:(15\:-\:1)\:d]}$

➟ $\rm{15\:(2a\:+\:14d)\:=\:405\times{2}\:}$

➟ $\bf\orange{30a\:+\:210d\:=\:810\:}$----(1)

ᴄᴀsᴇ - 2 :-

$\purple\checkmark\:\bf{S_{21}\:=\:\dfrac{21}{2}\:[2a\:+\:(21\:-\:1)\:d]\:}$

➟ $\rm{21\:(2a\:+\:20d)\:=\:756\times{2}\:}$

➟ $\bf\pink{42a\:+\:420d\:=\:1512\:}$----(2)

☯︎ ᴍᴜʟᴛɪғʟʏ 2 ɪɴ ᴛʜᴇ ᴇǫᴜ.(1),

➟ $\bf\red{60a\:+\:420d\:=\:1620\:}$----(3)

➪ Nᴏᴡ sᴜʙsᴛʀᴀᴄᴛɪɴɢ ᴇǫᴜ.(2) ғʀᴏᴍ ᴇǫᴜ.(3),

➳ 60a+420d-(42a+420d) = 1620-1512

➳ 60a + 420d - 42a - 420d = 108

➳ 18a = 108

➳ a = 108/18

➳ a = 6

ᴘᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ 'a' ɪɴ ᴛʜᴇ ᴇǫᴜ.(1), ᴡᴇ ɢᴇᴛ

➳ 30 × 6 + 210d = 810

➳ 180 + 210d = 810

➳ 210d = 810 - 180

➳ 210d = 630

➳ d = 630/210

➳ d = 3

__________________________

$\bf\blue{t_{8}\:=\:a\:+\:(8\:-\:1)\:d\:}$

$\rm{\implies\:t_{8}\:=\:6\:+\:7\times{3}\:}$

$\rm{\implies\:t_{8}\:=\:(6\:+\:21)\:}$

$\bf\green{\implies\:t_{8}\:=\:27\:}$

___________________________

$\huge\red\therefore$ [1] Tʜᴇ 8ᴛʜ ᴛᴇʀᴍ ᴏғ ᴛʜᴇ ᴀ.ᴘ ɪs "27" .

$\huge\red\therefore$ [2] Cᴏᴍᴍᴏɴ ᴅɪғғᴇʀᴇɴᴄᴇ ɪs "3" .