Question

5 terms are in A.P.Sum of middle 3 terms is 24.Product of the first and the 5th is 48.find the terms.?

Answer 1

`12+10+8+6+4`

`4+6+8+10+12`

Explanation:

The nth term of an AP is given by:

`a+(n-1)d`

And the sum is given by:

`S_n=n/2(2a+(n-1)d)`

Where `a` is the first term, `d` is the common difference and `n` is the nth term.

If we find the sum of the first 4 terms and subtract the 1st term from this, we will have the sum of the 3 middle terms.

`S_4-S_1=n/2(2a+(n-1)d)-(a+(n-1)d)=24`

`S_4-S_1=4/2(2a+3d)-a=24`

`3a+6d=24color(white)(888)[1]`

Product of 1st and 5th terms:

`a*(a+4d)=48`

`a^2+4ad=48color(white)(888)[2]`

We now need to solve our 2 equations simultaneously:

From [1]

`d=(24-3a)/6`

Substituting in [2]

`a^2+4a((24-3a)/6)=48`

`a^2-16a-48=0`

`(a-12)(a-4)=0=> a=12 , a=4`

Using in [1]:

`3(12)+6d=24=>d=-2`

`3(4)+6d=24=>d=2`

So we have 2 sets:

`a=12 , d= -2`

`a=4 . d=2`

Test:

1st term x 5 term = 48

`12 *4=48`

`4*12=48`

Sum of middle 3 terms:

`S_4-S_1=4/2[2(12)+3(-2)]-(12)=28`

`S_4-S_1=4/2[(2(4)+3(12)]-(4)=28`

So there are 2 AP's that meet the conditions:

`12+10+8+6+4`

`4+6+8+10+12`