Maths / Arithmetic Progression / General Term Of an AP

QUESTION

The fourth term of an AP is 0. Prove that its 25th term is thrice its 11th term.

EXPLANATION
Explain TypeExplanation Content
Text

$\dpi{120} \fn_jvn \large a_n=a+(n-1)d =0$

$\dpi{120} \fn_jvn \large a_4=a+(4-1)d = a+3d$

A.T.Q.

$\dpi{120} \fn_jvn \large a+3d =0$                                    (I)

$\dpi{120} \fn_jvn \large a_{25}=a+24d$

$\dpi{120} \fn_jvn \large a_{11}=a+10d$

A.T.Q.

$\dpi{120} \fn_jvn \large a_{25}= 3 Xa_{11}$

$\dpi{120} \fn_jvn \large a+ 24d= 3(a+10d)$

$\dpi{120} \fn_jvn \large a+ 24d= 3a+30d$

$\dpi{120} \fn_jvn \large a+ 24d - 3a-30d=0$

$\dpi{120} \fn_jvn \large -2a-6d=0$

$\dpi{120} \fn_jvn \large -2(a+3d)=0$

According to equation I

a+3d=0

LHS=RHS

Hence proved

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