A sequence of three real numbers form an A.P. and the first term is 25. If 20 is subtracted from second term and 4 subtracted from third term, the resulting numbers form a G.P., then maximum value of second term of G.P. will be

Question

A sequence of three real numbers form an A.P. and the first term is 25. If 20 is subtracted from second term and 4 subtracted from third term, the resulting numbers form a G.P., then maximum value of second term of G.P. will be (A) 125 (B) 45 (C) 55 (D) 75

Tardigrade Admin · Accepted Answer

text Let term be 25,25+ d , 25+2 d ∴ 25,5+ d , 21+2 d ⇒ G . P . ∴(5+ d )2=25(21+2 d ) ⇒ d 2-40 d -500=0 ⇒( d -50)( d +10)=0 ⇒ d =50 text or d =-10 ∴ text maximum value of second term of G.P. =5+50=55

Q. A sequence of three real numbers form an A.P. and the first term is 25. If 20 is subtracted from second term and 4 subtracted from third term, the resulting numbers form a G.P., then maximum value of second term of G.P. will be

125

45

55

75

$Let term be 25, 25 + d, 25 + 2 d$
$∴ 25, 5 + d, 21 + 2 d \Rightarrow G . P .$
$∴ (5 + d)^{2} = 25 (21 + 2 d)$
$\Rightarrow d^{2} - 40 d - 500 = 0 \Rightarrow (d - 50) (d + 10) = 0$
$\Rightarrow d = 50 or d = - 10$
$∴ maximum value of second term of G.P. = 5 + 50 = 55$

Q. A sequence of three real numbers form an A.P. and the first term is 25. If 20 is subtracted from second term and 4 subtracted from third term, the resulting numbers form a G.P., then maximum value of second term of G.P. will be

125

45

55

75

Let term be 25,25+d,25+2d ∴25,5+d,21+2d⇒G.P. ∴(5+d)2=25(21+2d) ⇒d2−40d−500=0⇒(d−50)(d+10)=0 ⇒d=50 or d=−10 ∴ maximum value of second term of G.P. =5+50=55

$Let term be 25, 25 + d, 25 + 2 d$
$∴ 25, 5 + d, 21 + 2 d \Rightarrow G . P .$
$∴ (5 + d)^{2} = 25 (21 + 2 d)$
$\Rightarrow d^{2} - 40 d - 500 = 0 \Rightarrow (d - 50) (d + 10) = 0$
$\Rightarrow d = 50 or d = - 10$
$∴ maximum value of second term of G.P. = 5 + 50 = 55$