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Sum

Find the 12^{th} term from the end of the AP: –2, –4, –6,..., –100.

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#### Solution

Given AP : –2, –4, –6,…., –100

Here, first term (a) = –2

Common difference (d) = –4 – (–2) = –2

And the last term (l) = –100

We know that, the n^{th} term of an AP

From the end a_{n} = l – (n – 1 )d

Where l is the last term

And d is the common difference.

∴ 12^{th} term from the end

i.e., a_{12} = –100 – (12 – 1)(–2)

= –100 + (11)(2)

= – 100 + 22

= –78

Hence the 12^{th} term from the end is –78.

Concept: Arithmetic Progression

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