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S = [2a + (n - 1)d]

Therefore, the required sum of first 20 terms of the series = [2 ∙ 1 + (17 - 1) ∙ 7]

= [2 + 16 ∙ 7]

= [2 + 112]

= × 114

= 17 × 57

= 969

Example 3- If the 5th term and 12th term of an Arithmetic Progression are 30 and
65 respectively, find the sum of its 26 terms.

Solution- Let us assume that ‘a’ be the first term and ‘d’ be the common difference of
the given Arithmetic Progression.

According to the problem,

5th term of an Arithmetic Progression is 30

i.e., 5th term = 30

⇒ a + (5 - 1)d = 30

⇒ a + 4d = 30 ................... (i)

and 12th term of an Arithmetic Progression is 65

i.e., 12th term = 65

⇒ a + (12 - 1)d = 65

⇒ a + 11d = 65 .................... (ii)

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