Question

find the sum of the first 11 terms of AP:2,6,10,14​

Answer 1

Answer:

a+(n-1)d

2+11-1×4

2+10×4

42

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Answer 2

QUESTION:

find the sum of the first 11 terms of AP:2,6,10,14

ANSWER:

We use the sum formula here;

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

where.

a = first term

d = common difference

now come to main question;

AP = 2, 6 , 10 , 14

here,

first term = 2

common difference = 6- 2 = 4

so,

$s(11) = \frac{11}{2} [2 \times 2 + (11 - 1)4]$

$s(11) = \frac{11}{2} (4 + 10 \times 4) \\ s(11) = \frac{11}{2} \times 44 \\ s(11) = 11 \times 22 \\ s(11) = 242$

hence,

$\huge\red{s(11) = 242}$

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