the sum of the three numbers in an ap is12 their product is 48 find those numbers
Answers
Answer:
Step-by-step explanation:
For odd number of AP terms, I use symbol for median (central) element.
So, here I take central number as c and common difference as d .
So, the terms are c−d , c and c+d .
Sum of the terms is 3c which is 21 , value of c is 7 .
At this stage, we do not know value of d .
The terms are 7−d , 7 and 7+d .
The product of these numbers is 7(49−d2) which is equal to 315.
7(49−d2)=315
=>49−d2=45
=>d2=4
=>d=±2 .
So, the sequence becomes 5 , 7 , 9 or 9 , 7 , 5 .
p.s.:
Informal method:
Product is 315=3×3×5×7 .
If we select middle factor, the remaining factors should be one on either side of the middle factor.
315 is between 63 and 73 and is multiple of 7 . So, I feel 7 should be middle term. Let us try 7 first.
For 7 , remaining factor 45 . Pairs of factors on either side of 7 are (1,45),(3,15) and (5,9) .
Wow, we got pair (5,9) which is equidistant from 7 .
The terms are 5,7,9 or 9,7,5 .
Let us check other possibilities.
We cannot select 1 as others will be on same side.
For 3 , remaining multiplier is 105 . Only possible factors of 105 (one each on either side are 1 and 105 .
For 5 , remaining 63 pairs are ( 1,63) and (3,21) . Not equidistant from 5 ; not acceptable.
For 9 , remaining 35 , one larger and one smaller than 9 divisors of 35 are 1 and 35 . Not equidistant from 9 . Not acceptable.
For 15 , remaining 21 . Factors one on either side of 15 are 1 and 21 . Not equidistant from 15 . Not acceptable.
For 21 , remaining 15 . No factors on both sides of 21 . No further checking required.
We got one answer of 5,7,9 or 9,7,5 .
Answer:
let the three numbers be :
a+d , a , a-d
given the sum is 12 so add the numbers
3a = 12
a = 4 ( therefore the first number of the ap is 4)
(a+d)(a)(a-d) = 48 ( we take a to the left hand side it becomes division)
(a2 - d2) = 48/4
(42 - d2) = 12
16 - d2 = 12
-d2 = -4
d= 2
therefore the numbers are 2,4,6.
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