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Given that, $4^{\text{th}}$ term of an arithmetic progression is $8.$

$\Rightarrow \boxed{T_{4} = 8}$

We know that, $n^{\text{th}}$ term of an arithmetic progression $: T_{n} = a + (n-1)d$

Where $a =$ the first term, $d =$ common difference.

Now, $T_{4} = a + (4-1)d$

$ \Rightarrow 8 = a + 3d$

$ \Rightarrow \boxed{a + 3d = 8}$

The sum of an arithmetic progression $: S_{n} = \frac{n}{2} \left[ 2a + (n-1) d \right]$

$\Rightarrow S_{7} = \frac{7}{2} \left[ 2a + (7-1) d \right]$

$\Rightarrow S_{7} = \frac{7}{2} (2a + 6d)$

$\Rightarrow S_{7} = \frac{7}{2} \times 2 (a + 3d)$

$\Rightarrow S_{7} = 7 \times 8 \quad [\because a + 3d = 8]$

$\Rightarrow \boxed{S_{7} = 56}$

$\therefore$ The sum of the first $7$ terms of the arithmetic progression is $56.$

Correct Answer $: \text{C}$
edited by
1 votes
1 votes

Another Approach:
It is given 4th term=8 so consider an AP with 7 terms as following
a-3d , a-2d , a-d , a , a+d , a+2d , a+3d

Here 4th term = a = 8
Sum of the above series= 7*8=56

Answer= 56

Answer:

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