Find the sum of first 19 terms of the AP 2, 7, 12, 17, ..... .

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Solution

Here, a = 2, d = (7 - 2) = 5 and n = 19
Using the formula, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ , we get:
$S_{19} = \frac{19}{2} [2 \times 2 + (19 - 1) 5] [∵ a = 2, d = 5 and n = 19] = \frac{19}{2} [4 + 90] = 19 \times 47 = 893$
Hence, the sum of the first 19 terms of the given AP is 893.

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Find the sum of first 19 terms of the AP 2, 7, 12, 17, ..... .

Here, a = 2, d = (7 - 2) = 5 and n = 19 Using the formula, Sn = n22a + n-1d, we get: S19=1922×2+19-15 [∵ a = 2, d = 5 and n= 19] = 1924+90=19×47=893 Hence, the sum of the first 19 terms of the given AP is 893.

Here, a = 2, d = (7 - 2) = 5 and n = 19
Using the formula, $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ , we get:
$S_{19} = \frac{19}{2} [2 \times 2 + (19 - 1) 5] [∵ a = 2, d = 5 and n = 19] = \frac{19}{2} [4 + 90] = 19 \times 47 = 893$
Hence, the sum of the first 19 terms of the given AP is 893.