Thr-6th-term-of-an-AP-is-equal-to-2-the-value-of-the-common-difference-of-the-AP-Which-makes-the-product-a-1-a-4-a-5-least-is-given-by-

Question Number 46922 by 786786AM last updated on 02/Nov/18

Thr 6 th term of an AP is equal to 2, the

value of the common difference of the AP .

Which makes the product a_{1} a_{4} a_{5} least

is given by

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Nov/18

a + 5 d = 2

p = a (a + 3 d) (a + 4 d)

p = (2 - 5 d) (2 - 5 d + 3 d) (2 - 5 d + 4 d)

p = (2 - 5 d) (2 - 2 d) (2 - d)

l n p = l n (2 - 5 d) + l n (2 - 2 d) + l n (2 - d)

\frac{1}{p} \times \frac{d p}{d (d)} = \frac{1}{2 - 5 d} \times (- 5) + \frac{1}{2 - 2 d} \times (- 2) + \frac{1}{2 - d} \times (- 1)

D p = p \times [\frac{- 5}{2 - 5 d} + \frac{- 2}{2 - 2 d} + \frac{- 1}{2 - d}]

= - 5 (2 - 2 d) (2 - d) - 2 (2 - 5 d) (2 - d) - (2 - 5 d) (2 - 2 d)

= - 5 (4 - 6 d + 2 d^{2}) - 2 (4 - 12 d + 5 d^{2}) - 1 (4 - 14 d + 10 d^{2})

f o r m a x / m i n D p = 0

20 - 30 d + 10 d^{2} + 8 - 24 d + 10 d^{2} + 4 - 14 d + 10 d^{2}

= 30 d^{2} - 68 d + 32

15 d^{2} - 34 d + 16 = 0

d = \frac{34 \pm \sqrt{34^{2} - 4 \times 15 \times 16}}{2 \times 15}

d = \frac{34 \pm \sqrt{1156 - 960}}{30}

d = \frac{34 \pm 14}{30} (\frac{48}{30}, \frac{20}{30}) \to (\frac{8}{5}, \frac{2}{3})

D p = 30 d^{2} - 68 d + 16

D^{2} p = 60 d - 68

v a l u e o f D^{2} p a t d = \frac{8}{5} i s

60 \times \frac{8}{5} - 68 = 96 - 68 = 28 + v e

v a l u e a t \frac{2}{3} i s

60 \times \frac{2}{3} - 68 = 40 - 68 = - 28 - v e

s o a t d = \frac{2}{3} t h e p r o d u c t i s m a x i m u m a n d d = \frac{8}{5}

i t i s m i n i m u m \dots

n o w a + 5 d = 2

a + 5 \times \frac{8}{5} = 2 a = - 6

a_{1} \times a_{4} \times a_{5}

(- 6) \times (- 6 + 3 \times \frac{8}{5}) (- 6 + 4 \times \frac{8}{5})

= (- 6) \times (\frac{- 30 + 24}{5}) (\frac{- 30 + 32}{5})

= (- 6) \times (\frac{- 6}{5}) (\frac{2}{5}) = \frac{72}{25} p l s c h e c k \dots .