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The 7th term of an A.P. is 20 and its 13th term is 32, then the A.P. is,

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8,10,12,14….

8,16,24,….

2,4,6,8….

1,2,3,4…

answer is A.

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Detailed Solution

Given that the 7th term of an A.P. is 20 and its 13th term is 32.
We know that the formula of

n^{th}

term in A.P. is,

T n = a + n − 1 d

.
The first term is a.
The common difference is d.
The number of terms are n.
For 7th term,

T 7 = a + 7 − 1 d ⇒ T 7 = a + 6 d

According to the question,

a + 6 d = 20...... (1)

Again,

T_{13} = a + (13 - 1) d

\Rightarrow 32 = a + 12 d . . . . . . (2)

On subtracting the equations (1),(2), we get,

a + 6 d - (a + 12 d) = 20 - 32

\Rightarrow a + 6 d - a - 12 d = 12

\Rightarrow - 6 d = - 12

\Rightarrow d = 2

Replace d=2 in equation (1),

a + 6 d = 20 ⇒ a + 6 × 2 = 20 ⇒ a + 12 = 20 ⇒ a = 8

The arithmetic progression will be of form a,a+d,a+2d,a+3d,......
We substitute 8 for a and 2 for d in

n^{th}

term formula.
The second term is

a + d = 8 + 2 ⇒ a + d = 10

The third term is

a + 2 d = 8 + 2 × 2 ⇒ a + 2 d = 8 + 4 ⇒ a + 2 d = 12

The fourth term is,

a + 3 d = 8 + 3 × 2 ⇒ a + 3 d = 8 + 6 ⇒ a + 3 d = 14

The Arithmetic progression is 8,10,12,14,.......
Hence, option 1 is correct.

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