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Is it possible to have more than one arithmetic series with four terms whose sum is 44? Explain.​

User Dkruchok
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2 votes

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

User Tom Tresansky
by
8.2k points
2 votes

Answer:

Yes, look at the picture I attached for explanation

Explanation:

Is it possible to have more than one arithmetic series with four terms whose sum is-example-1
User Terrornado
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