Answer:
Yes it is.
Explanation:
Let the AP be a, a+d, a+2d, a+3d and so on, then
4a + 6d = 44 which gives
a = (22–3d)/2 = 11 -1.5d
Assume that the AP has n terms, then nth term is given as :
a + (n-1)d = 11 - 1.5d + (n-1)d = 11 - (n-2.5)d
The second equation then becomes:
(11 - 1.5d)(n-2.5d) = 140
which simplifies to
42n -110d - 6nd + 15d^2 = 560 or put simply
6n = 15d -5 - (5 * 7 * 17)/(d-7)
As n is an integer, so must be 6n and therefore so must be (5*7*17)/(d-7) which means d = 12, 14 or 24
which gives 6n = 56, 120 and 320 respectively.
Again as n is an integer, the only possible answer is n = 20, which occurs when d = 14 and a is -10 which gives us the series -10, 4, 18, 32 .. 256
The terms be a-3d,a-d, a+d,a+3d
Sum = 4*a = 44.
a = 11;
Product
( a-3d)(a+3d) = 140
11*11- 9*d*d = 140
d*d = - 19/9.
Negative. No real common difference.
The meaning of ‘ extreme' is important.
The AP has more than four terms.
a,a+d,a+2d,a+3d,….,a+(n-1)d.
Sum condition
4a + 6d = 44
2a + 3d = 11 eqn-1