IIT JEE , AIEEE Forum - Mathematics , Algebra

cute_ria

Q,.a >0 b>0 nd a+b =1 show that {a +1/a}²+{b + 1/b}²

25/2??

Q.find a three digit number such that its digits are in GP nd the digits of

the number obtained from it by subtrating 400 form an AP??

Q.. let a₁ a₂ ......a_n a_n+1 ......be an AP

S₁ =a₁ +a₂+ a₃+.....+a_n

S₂ =a_n+1 + a_n+2+......+a_2n

S₃= a_2n+1+a_2n+2+ ....a_3n

..............

prove that the sequences S₁ ,S₂ ,S₃ ...... is an arithmetic progression whose

common difference is n² times the common difference of the given progression.

srujana

Q2)Let us start from the A.P,

let the digits be

a-d , a , a+d
the value of the number is 100(a-d)+10(a)+a+d

When 400 is added to it, we will get a 3 digit no. which is in G.P
The value of the number is

100(4+a-d)+10(a) +a+d

Hence the digits are
4+a-d a a+d

(a+d)(4+a-d)=a^2

simpifying u get

4(a+d)=d^2
i.e, 4 times a number is a perfect square, as 4 is also a perfect square, a+d=1

=>d=+ or-2
d cannot be 2 as a becomes negative, hence d= -2

Hence the number is

531

the required number is 531 +400= 931

Avinash_Bhat

3rd question:

a₁, a₂, a₃, a₄, ...................... are in A.P. with common difference d

S₁ = n/2 (2a₁ + (n - 1)d) = a₁n + n²d/2 - dn/2

S₂ = (S₁ + S₂) - S₁ = n (2a₁ + (2n - 1)d) - S₁ = a₁n + 3n²d/2 - dn/2

S₃ = (S₁ + S₂ + S₃) - (S₁ + S₂) = a₁n + 5n²d/2 - dn/2

Now note that : 2S₂ = S₁ + S₃
S₁, S₂, S₃, .......................................... are in A.P. whose difference is S₂ - S₁= S₃ - S₂ = n²d

nadeemoidu

An easier way to do the third question.

S₁=   a₁ + a₂ + a₃.....     a_n
S₂=   a_n+1 + a_n+2 + a_n+3.....     a_2n           = a₁ +nd + a₂ +nd + a₃ +nd...
                                                                =S₁ + n*nd
S₃=   a_2n+1 + a_2n+2 + a_2n+3.....     a_3n      = S₁ + 2 n*nd
S₄=   a_3n+1 + a_3n+2 + a_3n+3.....     a4n      = S₁ + 3 n*nd

...
...
...
So obviously it is an AP of CD=n*nd

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