The sum of all natural numbers such that and H.C.F. is...

(a): We have, and H.C.F. So, can be multiple of 7 or 13 or both Let Sum of all natural numbers which are divisible by 7 and lie between 100 and 200 Sum of all natural numbers which are divisible by 13 and lie between 100 and 200 Sum of all natural numbers which are divisible by both 7 and 13 and lie...

The sum of all natural numbers quotesingle{} n quotesingle{} such that 100 < n < 200 and H.C.F. (91,n) > 1 is: Sequence & Series (Mathematics)

The sum of all natural numbers quotesingle{} n quotesingle{} such that 100 < n < 200 and H.C.F. (91,n) > 1 is: Sequence & Series (Mathematics) | DivineJEE

If a_{n} = \frac{3}{4} - ( \frac{3}{4} )^{2} + ( \frac{3}{4} )^{3} + ( - 1)^{n - 1}( \frac{3}{4} )^{n} and b_{n} = 1 - a_{n}, then find the least natural number n_{0} such that b_{n} > a_{n}\forall n ? n_{0}

Different A.P. are constructed with the first term 100 , the last term 199, and integral common differences. The sum of the common differences of all such A.P. having at least 3 terms and at most 33 terms is

Number of 4 -digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to

Let a,b be two non-zero real numbers. If p and r are the roots of the equation x^{2} - 8ax + 2a = 0 and q and s are the roots of the equation \end{enumerate} x^{2} + 12bx + 6b = 0, such that \frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{s} are in A.P., then a^{- 1} - b^{- 1} is equal to

Let A_{1},G_{1} and H_{1} denote the arithmetic, geometric and harmonic means . respectively of two distinct positive numbers. For n ? 2 let A_{n - 1} and H_{n - 1} have . arithmetic, geometric and harmonic means as A_{n},G_{n},H_{n} respectively. Which of the following statements is correct?

For any odd integer n ? 1,n^{3} - (n - 1)^{3} + + ( - 1)^{n - 1}1^{3} =

If a_1, a_2, a_n are in A.P. where a_i>0 for all i show that &\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}&=\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}

The sum of the first n terms of the series (1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2 . 6^{2} + ..) is (\frac{n(n + 1)^{2}}{2}, when n is even. when n is odd, the sum is

If a_{1},a_{2},a_{3}, are in A.P. such that a_{1} + a_{7} + a_{16} = 40, then the sum of the first 15 terms of this A.P. is

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