Question

The sixth term of an A.P. is equal to 2, the value of the common difference of the A.P. which makes the product a₁ a₄ a₅ least is given by

Answer 1

Step-by-step explanation:

HEY MATE ............

Given that sixth term of an A.P is 2.

⇒a 1 +5d=2

Consider, p=a 1 a 4 a 5

⇒p=a 1 (a 1 +3d)(a 1 +4d)

Now substitute a 1 =2−5d in the above equation,

we get;

p=(2−5d)(2−2d)(2−d)

Now, solving the parentheses,

we get;

p=2[4−16d+17d^ 2 −5d^ 3 ]

Let,

S=−5d ^3 +17d ^2 −16d+4

Now, taking the derivative of S w.r.t d,

we get;

S′=−15d ^2 +34d−16

Notice that, for S′=0,

we get;d= 2/3 ,8/5

Now, taking the derivative of S′ w.r.t d,

we get;

S′′=−30d+34

At d= 2/3

, we get;

S′′=−20+34

So, S′′=14 which is positive.

Therefore, d= 2/3 gives minimum value.

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