Find the sum of each of the following APs:

(i) 2, 7, 12, 17, ... to 19 terms
(ii)
(iii)
(iv)
(v)

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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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Q. Find the sum of each of the following APs:

(i) 2, 7, 12, 17, ... to 19 terms
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Find the sum of each of the following APs:
(i) 2, 7, 12, 17,... to 19 terms.
(ii) 9, 7, 5, 3, ... to 14 terms.
(iii) -37, -33, -29, ... to 12 terms.
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Q. Find the sum of first 19 terms of the AP 2, 7, 12, 17, ..... .

Q. Find the nth term of each of the following APs:
(i) 5, 11, 17, 23, ....
(ii) 16, 9, 2, –5, ....

Find the number of terms in each of the following APs:

(i) $7, 13, 19...., 205$

(ii) $18, \frac{31}{2}, 13, . . . - 47$

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Find the sum of each of the following APs: (i) 2, 7, 12, 17, ... to 19 terms (ii) (iii) (iv) (v)

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.

Find the sum of each of the following APs:

(i) 2, 7, 12, 17, ... to 19 terms
(ii)
(iii)
(iv)
(v)

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.