Question

Find the mth term of an Arithmetic progression whose 12th term exceeds the 5th term by
14 and the sum of both terms is 36.​

Answer 1

Answer:

$\sf{The \ m^{th} \ term \ is \ 1+2m.}$

Given:

$\sf{In \ an \ AP,}$

$\sf{\leadsto{12^{th} \ term \ exceeds \ the \ 5^{th}}}$

$\sf{term \ by \ 14.}$

$\sf{\leadsto{The \ sum \ of \ both \ terms \ is \ 36.}}$

To find:

$\sf{The \ m^{th} \ term \ of \ an \ AP.}$

Solution:

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

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

$\sf{t_{12}-t_{5}=14}$

$\sf{\therefore{(a+11d)-(a+4d)=14}}$

$\sf{\therefore{7d=14}}$

$\sf{\therefore{d=\dfrac{14}{7}}}$

$\boxed{\sf{\therefore{d=2}}}$

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

$\sf{t_{12}+t_{5}=36}$

$\sf{\therefore{(a+11d)+(a+4d)=36}}$

$\sf{\therefore{2a+15d=36}}$

$\sf{But \ d=2,}$

$\sf{\therefore{2a+15(2)=36}}$

$\sf{\therefore{2a+30=36}}$

$\sf{\therefore{2a=36-30}}$

$\sf{\therefore{2a=6}}$

$\sf{\therefore{a=\dfrac{6}{2}}}$

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

____________________________

$\sf{Here, \ a=3 \ and \ d=2}$

$\sf{t_{n}=a+(n-1)d... formula}$

$\sf{\therefore{t_{m}=3+(m-1)\times2}}$

$\sf{\therefore{t_{m}=3+2m-2}}$

$\sf{\therefore{t_{m}=1+2m}}$

$\sf\purple{\tt{\therefore{The \ m^{th} \ term \ is \ 1+2m.}}}$

Answer 2

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• 12th term exceeds the 5th term by 14
• the sum of both terms is 36.

$\sf{\bold{\green{\underline{\underline{To\: Find}}}}}$

• The m th term of AP

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

$\sf{\large{\bold{\pink{\bigstar{\boxed{\boxed{t_n = a + ( n - 1 ) d }}}}}}}$

$\sf\implies t_5 = a + ( 5 - 1 ) d$

$\sf\implies t_5 = a + 4 d$

$\sf{\bold{\blue{\underline{\underline{t_5 = a + 4d }}}}}$

$\sf\implies t_12 = a + ( 12 - 1 ) d$

$\sf\implies t_5 = a + 11 d$

$\sf{\bold{\blue{\underline{\underline{t_{12}= a + 12d }}}}}$

• 12th term exceeds the 5th term by 14

$\sf ( a + 11d ) - ( a + 4d ) = 14$

$\sf a + 11d - a - 4d = 14$

$\sf 7d = 14$

$\sf d = \dfrac{14}{7}$

$\sf d = 2$

$\sf{\large{\bold{\orange{\bigstar{\boxed{\boxed{d = 2 }}}}}}}$

• the sum of both terms is 36

$\sf\implies ( a + 11d ) + ( a + 4d ) = 36$

$\sf\implies a + 11d + a + 4d = 36$

$\sf\implies 2a + 15d = 36$

$\sf\implies 2a + 15\times 2 = 36 ( d = 2 )$

$\sf\implies 2a + 30 = 36$

$\sf\implies 2a = 36 - 30$

$\sf\implies 2a = 6$

$\sf\implies a = \dfrac{6}{2}$

$\sf\implies a = 3$

$\sf{\large{\bold{\orange{\bigstar{\boxed{\boxed{a = 3 }}}}}}}$

__________________________

$\sf{\large{\bold{\pink{\bigstar{\boxed{\boxed{t_n = a + ( n - 1 ) d }}}}}}}$

$\sf\implies t_m = 3 + ( m - 1 )2$

$\sf\implies t_m = 3 + 2m - 2$

$\sf\implies t_m = 1 + 2m$

$\sf{\large{\bold{\orange{\bigstar{\boxed{\boxed{t_m = 1 + 2m}}}}}}}$