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

1.	Find the sum of the first:11 terms of the A.P. : 2, 6, 10, 14, . . .
Answer» We know that the sum of terms for different arithmetic progressions is given by S_n = \(\frac{n}{2}\)[2a + (n − 1)d] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms Given A.P 2, 6, 10, 14,… to 11 terms. Common difference (d) = a₂ – a₁ = 10 – 6 = 4 Number of terms (n) = 11 First term for the given A.P. (a) = 2 So, S₁₁ = \(\frac{11}{2}\)[2(2) + (11 − 1)4] = \(\frac{11}{2}\)[2(2) + (10)4] = \(\frac{11}{2}\)[4 + 40] = 11 × 22 = 242 Hence, the sum of first 11 terms for the given A.P. is 242

Answer»

We know that the sum of terms for different arithmetic progressions is given by

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

Given A.P 2, 6, 10, 14,… to 11 terms.

Common difference (d) = a₂ – a₁ = 10 – 6 = 4

Number of terms (n) = 11

First term for the given A.P. (a) = 2

So,

S₁₁ = \(\frac{11}{2}\)[2(2) + (11 − 1)4]

= \(\frac{11}{2}\)[2(2) + (10)4]

= \(\frac{11}{2}\)[4 + 40]

= 11 × 22

= 242

Hence, the sum of first 11 terms for the given A.P. is 242

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