The sum of $40$ terms of an A.P. whose first term is $2$ and common difference $4$ , will be

3200

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1600

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200

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2800

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Solution

The correct option is A $3200$
We know the formula:
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
Here, n=number of terms=40
a=first term=2
d=common difference=4
So, putting the above values in formula we get,
$S_{40} = \frac{40}{2} [2 (2) + (40 - 1) 4] = 20 [4 + 39 \times 4] = 20 [4 + 156] = 20 \times 160 = 3200$
$∴$ Sum of 40 terms=3200

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The sum of 40 terms of an A.P. whose first term is 2 and common difference 4, will be

The correct option is A 3200We know the formula:Sn=n2[2a+(n−1)d]Here, n=number of terms=40 a=first term=2 d=common difference=4So, putting the above values in formula we get,S40=402[2(2)+(40−1)4]=20[4+39×4]=20[4+156]=20×160=3200∴ Sum of 40 terms=3200

The sum of $40$ terms of an A.P. whose first term is $2$ and common difference $4$ , will be