Find the sum of the first $11$ terms of the A.P: $2, 6, 10, 14, . . . .$

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Solution

There are two formulae to find the sum of the terms of an $A P$ .
In this case we know the following:
$a = 2, d = 4, n = 11$
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{11} = \frac{11}{2} [2 (2) + (10) 4]$
$S_{11} = 5.5 [4 + 40]$
$S_{11} = 242.$

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Q. The sum of first 11 terms of the A.P 2, 6, 10, 14,... is .

Q. STATEMENT -

1

: The sum of first

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242.

STATEMENT -

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: The sum of first

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

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Find the sum of the first 11 terms of the A.P: 2,6,10,14,....

There are two formulae to find the sum of the terms of an AP.In this case we know the following:a=2,d=4,n=11Sn=n2[2a+(n−1)d]S11=112[2(2)+(10)4]S11=5.5[4+40]S11=242.

Find the sum of the first $11$ terms of the A.P: $2, 6, 10, 14, . . . .$

There are two formulae to find the sum of the terms of an $A P$ .
In this case we know the following:
$a = 2, d = 4, n = 11$
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{11} = \frac{11}{2} [2 (2) + (10) 4]$
$S_{11} = 5.5 [4 + 40]$
$S_{11} = 242.$