Question

The sum of three consecutive numbers in A.P. is 12 and sum of their cubes is 408. Find the product of the numbers.

• A. 28
• B. 36
• C. 42
• D. 38
• E. 96

Answer 1

The correct option is A 28

Solution:
Let the three terms be p, q,r and let the common difference be d.
$p=q-d$
$r=q+d$
Sum $=12,q-d+q+q+d=12,q=4,p+r=8.$
Also $p^3+q^3+r^3=408$
$p^3+r^3=344$
$(p+r{)}^3=p^3+r^3+3pr(p+r)$
$8^3=p^3+r^3+3pr(p+r)$
$p+r=8,$ $p^3+r^3=344$
$pr=7$
So, $pqr=7\times 4=28.$

Alternate method: Let the three terms be (a – d), a and (a + d)
Sum of these three terms$=(a-d)+a+(a-d)=12\Rightarrow a=4.$
$(4-d{)}^3+4^3+(4+d{)}^3=408$
$4^3-d^3-3\times 4\times 4\times d^2+3\times 4\times d^2+4^3+4^3+d^3+3\times 4\times 4\times d+3\times 4\times d^2=408$
$\{4^3+4^3+4^3+12d^2+12d^2\}=408$
$24d^2=216$
$d=3.$
The three terns are 1, 4 and 7. Product$1\times 4\times 7=28.$