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In an AP, the first term is -4, the last term is 29 and the sum of all its terms is 150. Find its common difference.

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#### Solution

Suppose there are n terms in the AP.

Here , a = 4 , l = 29 and s_{n} = 150

s_{n} = 150

`⇒ n/2 (-4 +29 ) = 150 [ s_n = n/2 (a+l) ]`

` ⇒ n = (150xx 2) / 25 = 12`

Thus, the AP contains 12 terms.

Let d be the common difference of the AP.

∴ a_{12} = 29

⇒ - 4 + (12-1) × d = 29 [ a_{n} = a+(n-1)d]

⇒ 11d = 29 +4 = 33

⇒ d= 3

Hence, the common difference of the AP is 3.

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