[Hint : Find n for an< 0]
Given A.P. is \(121, 117, 113 ……\)
\(a = 121\) and \(d = 117 − 121 = −4\)
\(a_n = a + (n − 1) d\)
\(a_n = 121 + (n − 1) (−4)\)
\(a_n= 121 − 4n + 4\)
\(a_n= 125 − 4n \)
We have to find the first negative term of this A.P.
Therefore, \(an < 0\)
\(125 - 4n < 0\)
\(125 < 4n\)
\(n > \frac {125}{4}\)
\(n > 31.25\)
Therefore, 32nd term will be the first negative term of this A.P.
The common difference of the A.P.: $3,\,3+\sqrt{2},\,3+2\sqrt{2},\,3+3\sqrt{2},\,\ldots$ will be:
Let $a_1, a_2, a_3, \ldots$ be an AP If $a_7=3$, the product $a_1 a_4$ is minimum and the sum of its first $n$ terms is zero, then $n !-4 a_{n(n+2)}$ is equal to :
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$
निम्नलिखित विषय पर संकेत बिंदुओं के आधार पर लगभग 120 शब्दों में एक अनुच्छेद लिखिए |
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