If a1,a2,a3..............an are n postive numbers in arithmetic progression with common difference d≠0 and Sn = a1 + a2 + ... + an
a. Sn __ n - 2(√a1 + √a2 +...+√an) (>,<,=,≤,≥ or cannot say)
b. 2Sn2 > (np . a1 . an)
Find maximum value of p.
39good one.
1st one is > .
i am at present not able to find if any equality will occur.
1
by AM GM ineq.
1+a1≥2√a1
1+a2≥2√a2
.
.
.
1+an≥2√an
adding all
Sn≥-(n-2(√a1+√a2+...+√an))
so,
it will always be greater than RHS
second part is not clearly visible,
11@ dimensions....
your approach was correct (but there was a mistake in the conclusion)...
Using AM-GM,
1 + a1 > 2√a1
1 + a2 > 2√a2
.
.
.
1 + an > 2√an
Adding,
n + Sn > 2( √a1 + √a2 +...+ √an ) (Equality is not permissible bec. all ai's cannot be equal to 1 simultaneously as 'd≠0')
\Rightarrow Sn > 2( √a1 + √a2 +...+ √an ) - n
\Rightarrow Sn > - (n - 2( √a1 + √a2 +...+ √an ))
BUT Sn < or > n - 2( √a1 + √a2 +...+ √an ) cannot be said !!!!!
11@ Subash...
question is not 2Sn2 > np . √a1an
It is 2Sn2 > np . a1 . an
