Divide 28 into four parts in A.P. so that ratio of the product of first and third with the product of second and fourth is 8:15.Now find the largest term in those four terms ?

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Solution

Here $4 a = 28$ So $a = 7.$
Also $\frac{(a - 3 d) (a + d)}{(a - d) (a + 3 d)} = \frac{8}{15}$
$15 [a^{2} - 3 d^{2} - 2 a d] = 8 [a^{2} - 3 d^{2} + 2 a d]$
$7 [a^{2} - 3 d^{2}] = 46 a d$
$7 (49 - 3 d^{2}) = 46 \times 7. d$
$49 - 3 d^{2} = 46 d$
$3 d^{2} + 46 d - 49 = 0$
$(d - 1) (3 d + 49) = 0$
$d = 1$
Required numbers are $4, 6, 8, 10$

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Divide 28 into four parts in A.P. so that ratio of the product of first and third with the product of second and fourth is 8:15.Now find the largest term in those four terms ?

Here 4a=28 So a=7.Also (a−3d)(a+d)(a−d)(a+3d)=81515[a2−3d2−2ad]=8[a2−3d2+2ad]7[a2−3d2]=46ad7(49−3d2)=46×7.d49−3d2=46d3d2+46d−49=0(d−1)(3d+49)=0 d=1 Required numbers are 4,6,8,10