Determine the common difference of AP whose third term is $16$ and the $7 t h$ term exceeds the $5 t h$ term by $12$

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Solution

The correct option is D 6
As the $7 t h$ term exceeds the $5 t h$ term by $12$ , so the $5 t h$ term will exceed the $3 r d$ term by $12$ as well
So, $n_{3} = 16$
$n_{5} = 28$
$n_{7} = 40$
$n_{4}$ or $n_{6}$ can be calculated by taking average of the preceding and next term
So, $n_{4} = (28 + 16) / 2 = 22$
This gives the $d = 6$
$A P : 4, 10, 16, 22, 28, 34, 40, 46, . . . . . . .$

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Similar questions

16

12

Determine the A.P. whose third term is $16$ and the $7^{t h}$ term exceeds the $5^{t h}$ term by $12.$

7^{t h}

5^{t h}

5^{t h}

7^{t h}

63, 65, 67

3, 10, 17

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