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If fourth term of an A. P. is zero, show that its 25th term is thrice its 11th term.? ( A.P. = arithmetic progression )

Proof

1 Answer

Feb 19, 2016

(see below)

Explanation:

For an arithmetic progression with a difference of $d$ $d$ between terms and an initial value of $a_{0}$ $a_0$ :
$XXX a_{k} = a_{0} + k \cdot d$ $color(white)("XXX")a_k=a_0+k*d$

We are told that $a_{4} = a_{0} + 4 d = 0$ $a_4=a_0+4d=0$

$a_{11} = a_{0} + 11 d$ $a_11=a_0+11d$
$XXX = a_{0} + 4 d + 7 d$ $color(white)("XXX")=a_0+4d+7d$
$XXX = a_{4} + 7 d$ $color(white)("XXX")=a_4+7d$
$XXX = 7 d$ $color(white)("XXX")=7d$

$a_{25} = a_{0} + 25 d$ $a_25=a_0+25d$
$XXX = a_{0} + 4 d + 21 d$ $color(white)("XXX")=a_0+4d+21d$
$XXX = a_{4} + 21 d$ $color(white)("XXX")=a_4+21d$
$XXX = 21 d$ $color(white)("XXX")=21d$

$a_{25} = 3 \times a_{11}$ $a_25=3xxa_11$

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