Question

1. Find the sum of the following AP:

2.7.12.....to 10 terms
06.17.28....to 100 terms​
please answer

Answer 1

245 and 55050

Explanation:

The sum to n terms of an AP is

`S_(n)` =
`(n)/(2)` [ 2a₁ + (n - 1)d ]

where a₁ is the first term and d the common difference

In the first progression a₁ = 2 and d = 7 - 2 = 5, thus

`S_(10)` =
`(10)/(2)` [ (2 × 2) + (9 × 5) ] = 5 × ( 4 + 45 ) = 5 × 49 = 245

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In the second progression a₁ = 6 and d = 17 - 6 = 11, thus

`S_(100)` =
`(100)/(2)` [ (2 × 6) + (99 × 11) ]

= 50(12 + 1089) = 50 × 1101 = 55050