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Determine the AP whose third term is 16 and 7th term exceeds the 5th term by 12.

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2, 4, 6, 8, \dots

3, 9, 15, 21, \dots \dots

4, 10, 16, 22 \dots \dots

5, 10, 15, 20, \dots \dots

answer is C.

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Detailed Solution

We know that

n^{th}

terms of arithmetic progression =

a_{n} = a + (n - 1) d

a_{n}

n^{th}

term

a_{}

= first term

d

= common difference

n

= number of terms
Assume that
1st term of given AP =

a_{}

3rd term =

a_{3}

5th term =

a_{5}

7th term =

a_{7}

As per given condition

\Rightarrow a_{3} = 16

\Rightarrow a + (3 - 1) d = 16

\Rightarrow a + (3 - 1) d = 16

\Rightarrow a + 2 d = 16 \dots \dots

equation (1)
Now,

a_{7} - a_{5} = 12

\Rightarrow [a + (7 - 1) d] - [a + (5 - 1) d] = 12

\Rightarrow 2 d = 12

\Rightarrow d = 6

Put the value of d in the equation (1), we get

\Rightarrow a + 2 d = 16

a + 2 \times 6 = 16

a + 12 = 16

a = 4

Therefore,
The arithmetic progression will be

4, 4 + 6, 4 + 2 \times 6, 4 + 3 \times 6, \dots

Hence the Sequence is

4, 10, 16, 22, \dots

Correct option is 3.

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