If the numbers -3, M, N, and 5 are in arithmetic progression and –3

Question

If the numbers -3, M, N, and 5 are in arithmetic progression and –3 < M < N < 5, then M = ?

(A) –1/3
(B) 0
(C) 1/3
(D) 7/3
(E) 3

chetan2u · Answer

Since the numbers are in AP, -3+d=M, M+d=N and N+d=5.
Therefore, -3+3d=5......3d=8..... d=\frac{8}{3} 
-3+d = -3+\frac{8}{3}=\frac{-9+8}{3}=\frac{-1}{3}

A

randomavoidplease · Answer

Since numbers are in sequence, we can apply the formula a(n) = a(1) + (n-1)d to find the difference.

1st.  5 = -3 + 3(d) > d = \frac{8}{3} 
2nd.  M = -3 + \frac{8}{3} > m = \frac{-1}{3}

Ans . A

rocky620 · Answer

Let d be the common difference,

T1=-3
T2=M=>-3+d....eq1
T3=N=>-3+2d...eq2
T4=5=>-3+3d...eq3

From eq3, d=8/3

T2=M=>-3+8/3= -(9/3)+(8/3) = -(1/3)

Option A.

If the numbers -3, M, N, and 5 are in arithmetic progression and –3 <

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