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The sum of the first 3 terms of an AP is 6 and that of the last 3 terms is 16. If the AP has 13 terms, what is the Arithmetic mean of the three terms that are exactly at the middle of the progression?

Solution

✅ Correct Option: 2

In an AP, the sum of three consecutive terms =3×= 3 \times the middle term.

First three terms: 3×T2=63 \times T_2 = 6, so T2=2T_2 = 2

Last three terms (T11,T12,T13T_{11}, T_{12}, T_{13}): 3×T12=163 \times T_{12} = 16, so T12=163T_{12} = \dfrac{16}{3}


With 1313 terms, the three middle terms are T6,T7,T8T_6, T_7, T_8.

Their arithmetic mean is the middle one, T7T_7.


T7T_7 is exactly 55 terms after T2T_2 and 55 terms before T12T_{12}, so it is their average:

T7=T2+T122T_7 = \dfrac{T_2 + T_{12}}{2}

=2+1632= \dfrac{2 + \frac{16}{3}}{2}

=223×12= \dfrac{22}{3} \times \dfrac{1}{2}

=113= \dfrac{11}{3}

Therefore, the required arithmetic mean is 113\dfrac{11}{3}.

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