[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 :
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD.