The first term of an AP is 5, last term is 45 and the sum is 400. Find the number of terms and the common difference.

n = 16 a n d d = 8 / 3

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n = 16 a n d d = 16 / 3

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n = 8 a n d d = 16 / 3

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n = 8 a n d d = 8 / 3

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Solution

The correct option is A $(n = 16 a n d d = \frac{8}{3})$
Given, first term $a = 5$ , last term $l = 45$ and $S_{n}$ = 400
Since, $S_{n} = \frac{n}{2} (a + l)$
$\Rightarrow 400 = \frac{n}{2} (5 + 45)$
$\Rightarrow 400 = \frac{n}{2} (50)$
$\Rightarrow 400 = n (25)$
$\Rightarrow n = 16$
Since, $l = a + (n - 1) d$
$\Rightarrow 45 = 5 + (16 - 1) d$
$\Rightarrow 45 - 5 = (15) d$
$\Rightarrow 40 = 15 d$
$\Rightarrow d = \frac{40}{15}$
$\Rightarrow d = \frac{8}{3}$

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The first term of an AP is 5, last term is 45 and the sum is 400. Find the number of terms and the common difference.

The correct option is A (n=16 and d=83)Given, first term a=5, last term l=45 and Sn = 400Since, Sn=n2(a+l)⇒400=n2(5+45)⇒400=n2(50)⇒400=n(25)⇒n=16Since, l=a+(n−1)d⇒45=5+(16−1)d⇒45−5=(15)d⇒40=15d⇒d=4015⇒d=83