The first term of an AP is $148$ and the common difference is $- 2$ . If the AM of first $n$ terms of the AP is $125$ , then the value of $n$ is

18

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24

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30

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Solution

The correct option is D $24$
Given, $a = 148, d = - 2$
AM of $n$ terms $= 125$
$\Rightarrow \frac{a_{1} + a_{2} + \dots + a_{n}}{n} = 125$
$\Rightarrow \frac{\frac{n}{2} [2 a + (n - 1) d]}{n} = 125$
$\Rightarrow 2 a + (n - 1) d = 250$
$\Rightarrow 2 \times 148 + (n - 1) (- 2) = 250$
$\Rightarrow 296 - 2 n + 2 = 250$
$\Rightarrow 298 - 2 n = 250$
$\Rightarrow n = 24$

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The first term of an AP is 148 and the common difference is −2. If the AM of first n terms of the AP is 125, then the value of n is

The correct option is D 24Given, a=148,d=−2AM of n terms =125⇒a1+a2+⋯+ann=125⇒n2[2a+(n−1)d]n=125⇒2a+(n−1)d=250⇒2×148+(n−1)(−2)=250⇒296−2n+2=250⇒298−2n=250⇒n=24

The first term of an AP is $148$ and the common difference is $- 2$ . If the AM of first $n$ terms of the AP is $125$ , then the value of $n$ is