Let $s_1,s_2,\ldots,s_{10}$ be the sums of 12-term APs with first terms $1,2,\ldots,10$ and common differences $1,3,5,\ldots,19$. Then $\sum_{i=1}^{10}s_i$ is equal to
Let \(n \in N, n > 25\). Let \(A, G, H\) denote the arithmetic mean, geometric mean, and harmonic mean of 25 and \(n\). The least value of \(n\) for which \(A, G, H \in \{25, 26, \ldots, n\}\) is
The number of common terms in the progressions $4,9,14,19,\ldots$ up to $25^{\text{th}}$ term and $3,6,9,12,\ldots$ up to $37^{\text{th}}$ term is:
166. If \(a+c,\ a+b,\ b+c\) are in G.P. and \(a, c, b\) are in H.P. where \(a, b, c > 0\), then the value of \(\dfrac{a+b}{c}\) is:
In a non constant arithmetic progression having odd number of terms, having positive integral common difference, the ratio of the sum of the 1st, 3rd, 5th, 7th, ... terms to the sum of remaining terms is 13 : 12, then the number of terms in the arithmetic progression, is:
If sum of the series $1+\dfrac{\sqrt5-\sqrt3}{2\sqrt5}+\dfrac{8-2\sqrt{15}}{30}+\dfrac{14\sqrt5-18\sqrt3}{60\sqrt5}+\cdots=2+\dfrac{a+\sqrt{15}}{b}\log_b\!\left(\dfrac{a}{c}\right)$; $a,b,c\in\mathbb{N}$, $\gcd(a,b,c)=1$, then $2(a^2+b^2+c^2)$ is
Let \(\dfrac{1}{x_1}, \dfrac{1}{x_2}, \ldots, \dfrac{1}{x_n}\) (\(x_i \neq 0\) for \(i = 1, 2, \ldots, n\)) be in AP such that \(x_1 = 4\) and \(x_{21} = 20\). If n is the least positive integer for which \(x_n > 50\), then \(\displaystyle\sum_{i=1}^{n}\left(\dfrac{1}{x_i}\right)\) is equal to
Find the number of common terms to the two sequences 17, 21, 25, ..., 417 and 16, 21, 26, ..., 466.
If sum of the series $1+\dfrac{\sqrt5-\sqrt3}{2\sqrt5}+\dfrac{8-2\sqrt{15}}{30}+\dfrac{14\sqrt5-18\sqrt3}{60\sqrt5}+\cdots=2+\dfrac{a+\sqrt{15}}{b}\log_b\!\left(\dfrac{a}{c}\right)$; $a,b,c\in\mathbb{N}$, $\gcd(a,b,c)=1$, then $2(a^2+b^2+c^2)$ is