Question

The first term 2 aps are equal and the ratio of the c.d is 1:2 if the 7 th term of first a.p and 21 th term of 2nd a.p are 23 and 125 respectively. Find 2 aps

Answer 1

$\blue{Question}$

The first term of two AP are equal and the ratio of the common difference is 1:2 if the 7th term of first AP and 21th term of 2nd AP are 23 and 125 respectively. Find two AP.

_____________________________________

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

$\sf{The \ required \ two \ AP \ are \ 5, \ 8, \ 11,... }$

$\sf{and \ 5, \ 11, \ 17,... \ respectively. }$

$\sf\orange{Given:}$

$\sf{\implies{First \ term(a) \ of \ two \ AP \ are \ equal.}}$

$\sf{\implies{Ratio \ of \ common \ difference(d)=1:2}}$

$\sf{\implies{7^{th} \ term \ of \ first \ AP=23}}$

$\sf{\implies{21^{rd} \ term \ of \ second \ AP=125}}$

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

$\sf{The \ two \ AP.}$

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

$\sf{Let, Common \ difference \ of \ first \ AP \ be \ d}$

$\sf{\therefore{Common \ difference \ of \ second \ AP=125}}$

$\boxed{\sf{tn=a+(n-1)d}}$

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

$\sf{t7=a+(7-1)d}$

$\sf{\therefore{a+6d=23...(1)}}$

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

$\sf{t21=a+(21-1)(2d)}$

$\sf{\therefore{a+20(2d)=125}}$

$\sf{\therefore{a+40d=125...(2)}}$

$\sf{Subtract \ equation(2) \ from \ equation \ (1)}$

$\sf{a+40d=125}$

$\sf{-}$

$\sf{a+6d=23}$

___________________

$\sf{\therefore{34d=102}}$

$\sf{\therefore{d=\frac{102}{34}}}$

$\boxed{\sf{\therefore{d=3}}}$

$\sf{Substitute \ d=3 \ in equation(1)}$

$\sf{a+6\times3=23}$

$\sf{\therefore{a=23-18}}$

$\boxed{\sf{\therefore{a=5}}}$

$\sf{First \ AP:}$

$\sf{t1=a=5,}$

$\sf{t2=a+d=5+3=8,}$

$\sf{t3=a+2d=5+6=11.}$

$\sf{Second \ AP:}$

$\sf{t1=a=5,}$

$\sf{t2=a+2d=5+6=11,}$

$\sf{t3=a+2(2d)=5+12=17.}$

$\sf\purple{\tt{\therefore{The \ required \ two \ AP \ are \ 5, \ 8, \ 11,... }}}$

$\sf\purple{\tt{and \ 5, \ 11, \ 17,... \ respectively. }}$

Answer 2

$\large\bf\underline{Given:-}$

• First term of two AP 's are equal .
• the ratio of the common difference = 1:2
• 7th term of first AP = 23
• 21st term of 2nd AP = 125

$\large\bf\underline {To \: find:-}$

• Both AP's

$\huge\bf\underline{Solution:-}$

Let the common difference of 1st AP be 1d

Let the common difference of 2nd AP be 2d

$\bf \blacktriangleright \:T_n = a+(n-1)d \\\\\longrightarrow \rm \: T_7 = a + 6d \\ \\ \longrightarrow \rm \:a + 6d = 23......(i) \\ \\ \longrightarrow \rm \: T_{21} = a + 20 \times 2d \\ \\ \longrightarrow \rm\:a+40d = 125.........(ii)$

Solving eq (i) and (ii).

$\rm \: a + 6d = 23 \\ \rm \: a + 40d = 125 \\ \rm \: - \: \: \: \: \: - \: \: \: \: \: \: \: \: - \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \rm \: \: \: \: \: \: - 34d \: = - 102 \\ \\ \rm \: \: \: \: \: \: \: \: \: \: \: \: \: d = \cancel\frac{ - 102}{ - 34} \\ \\ \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: d = 3$

$\dag \rm \: putting \: value \: of \: d = 3 \: in \: eq.(i) \\ \\ \rm \dashrightarrow \: a + 6d = 23 \\ \\ \rm \dashrightarrow \: a + 6 \times 3 = 23 \\ \\ \rm \dashrightarrow \: a + 18 = 23 \\ \\ \rm \dashrightarrow \: a = 23 - 18 \\ \\ \bf\dashrightarrow \: a = 5$

So, common difference of 1st AP 1d = 3

Common difference of 2nd AP 2d = 6

First term of both AP's are same = 5

So, The AP's are :-

1st term of 1st AP = 5

2nd term of 1st AP = a+ d

⠀⠀⠀⠀⠀➝ 5 + 3

⠀⠀⠀⠀⠀➝ 8

3rd term of 1st AP = a+ 2d

⠀⠀⠀⠀⠀➝ 5 + 2× 3

⠀⠀⠀⠀⠀➝ 5 + 6

⠀⠀⠀⠀⠀➝ 11

• So, 1st AP = 5, 8 ,11 .....

1st term of 2nd AP = 5

2nd term of 2nd AP = a+ d

⠀⠀⠀⠀⠀➝ 5 +6

⠀⠀⠀⠀⠀➝ 11

3rd term of 2nd AP = a + 2d

⠀⠀⠀⠀⠀➝ 5 +2 × 6

⠀⠀⠀⠀⠀➝ 5 + 12

⠀⠀⠀⠀⠀➝ 17

• So, 2nd AP = 5, 11,17....

Two AP's are as :-

• ★ 1st AP = 5 , 8 ,11
• ★ 2nd AP = 5, 11 ,17