If the $6^{th}$ term of an A.S is 64, find the sum of the first 11 terms.

Given :

The $6^{th}$ term of an A.P is 64.

To do :

We have to find the sum of the first 11 terms of the A.P.

Solution :

We know that $n^{th}$ term of an A.P is

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

$a_{6} = a + (6-1)d$

$64 = a + 5d$........................................(i)

Sum of n terms of an A.P is, 

$$S_{n} = \frac{n}{2} [2a + (n-1)d]$$

$S_{11} = \frac{11}{2} [2a + (11-1)d]$

$S_{11} = \frac{11}{2} [2a + 10d]$

Take 2 as common from $2a + 10d$,

$S_{11} = \frac{11}{2} \times 2 [a + 5d]$

$S_{11} = 11 \times [a+5d]$................(ii)

Substitue (i) in (ii),

$S_{11} = 11 \times 64$

$S_{11} = 704$

Therefore, the sum of the first 11 terms of an A.P is 704.

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Updated on: 10-Oct-2022

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