Question

the 8th term of an ap is zero prove that 38 term is triple it's 18th term

Answer 1

Heya !!!




8th term = 0



A + 7D = 0



A = -7D -------(1)



To Prove:-


38th term = 3 ( 18th term )




A + 37D = 3 ( A + 17D)


-7D + 37D = 3 ( -7D + 17D)




30D = 3( 10D)



30D = 30D ......... PROVED.......



HOPE IT WILL HELP YOU...... :-)

Answer 2

GIVEN:-

$\bf \: •8th \: term \: of \: AP \: is \: zero$

TO PROVE:-

$\bf \: 38th \: term \: of \: AP \: is \: triple \: its \: 18th \: term$

SOLUTION:-

$\bf In \: the \: given \: AP \: let \: a \: and \: d \: be \: the \: first \: and \: last \: term.$

$\bf Then,T_n=a+(n-1)d$

$\bf \implies T_8=a+(8-1)d,T_{38}=a+(38-1)d,\\\bf T_{18}=a+(18-1)d$

$\bf \implies T_8=(a+7d),T_{38}=(a+37d),\\\bf T_{18}=(a+17d)$

$\bf \: Now ,T_8=0\implies a+7d=0\implies a=-7d \: \: \: \: ------(1)$

$\bf \therefore \: T_{38}=(a+37d)=(-7d+37d)=30d \: \: \: [using (1)]$

$\bf And ,T_{18}=(a+17d)=(-7d+17d)=10d \: \: \: \: [using \: (1)]$

$\bf \therefore T_{38}=30d=3×(10d)=3×T_{18}$

$\bf \: Hence \: ,38th \: term \: is \: triple \: its \: 18th \: term$