1 If the AM ,GM and HM of the first and last terms of the series 25,26,27, - Maths - Sequences and Series

Question

1 If the AM ,GM and HM of the first and last terms of the series
25,26,27, N-1,N are the terms of the series,find the value of
N?
2 The sum of all possible products of the first 'N' natural
numbers taken two bu two is:
a)1/24n(n+1)(n-1)(3n+2)
b)n(n+1)(n-1)(2n+3)/24
c)n(n+1)(n-1)(2n+1)/6
d) none of these

Gursheen kaur. · Accepted Answer

Solution 2)

As we know that

${(x_{1} + x_{2} + x_{3} + .......... + x_{n})}^{2} = ({x_{1}}^{2} + {x_{2}}^{2} + {x_{3}}^{2} + .... + {x_{n}}^{2}) + 2 (s u m o f p o s s i b l e p d t s t a k e n t w o a t a t i m e)$

${(\sum_{i = 1}^{n} x_{i})}^{2} = \sum_{i = 1}^{n} {x_{i}}^{2} + 2 [\sum_{i = 1}^{n} \sum_{j = 1}^{n} x_{i} x_{j}] \Rightarrow \sum_{i = 1}^{n} \sum_{j = 1}^{n} x_{i} x_{j} = \frac{1}{2} [{(\sum_{i = 1}^{n} x_{i})}^{2} - \sum_{i = 1}^{n} {x_{i}}^{2}] = \frac{1}{2} {[\frac{n (n + 1)}{2}]}^{2} - \frac{1}{2} . \frac{n (n + 1) (2 n + 1)}{6} = \frac{1}{4} n (n + 1) [\frac{n (n + 1)}{2} - \frac{2 n + 1}{3}] = \frac{1}{4} n (n + 1) [\frac{3 n (n + 1) - 2 (2 n + 1)}{6}] = \frac{n (n + 1) (n - 1) (3 n + 2)}{24}$

Hence, the sum of all possible products of the first n natural numbers taken two by two is

$\frac{n (n + 1) (n - 1) (3 n + 2)}{24}$

Solution 1)

$A M = \frac{25 + N}{2}, G M = \sqrt{25 N} = 5 \sqrt{N} a n d H M = \frac{50 N}{25 + N}$

$N o w, A M \geq G M \geq H M$

therefore, according to the statement ,

$\frac{25 + N}{2} - 2 = 5 \sqrt{N} - 1 = \frac{50 N}{25 + N}$