Question

the sum of three consecutive terms of an ap is 9 and the sum of their squares is 30 find the sum of n terms of the serirs

Answer 1

Let the numbers are, (a-d),a & (a+d)(where d is the common diff). Now as per condition:(a-d)+a+(a+d)=9,or, a-d+a+a+d=9. So we get, 3a=9 & a=9/3=3

Again as per 2nd condition:(a-d)^2+a^2+(a+d)^2=35,or,a^2-2ad+d^2+a^2+a^2+2ad+d^2=35, or, 3a^2+2d^2=35, as a=3 so we get, 3×3^2+2d^2=35.or, 2d^2=35-27=8, or, d^2=4(taking sqr.root of bothe sides we get the value of d as 2). The consequitive numers are 1, 3 & 5. So the sum of the numbers in the series upto n th term is n^2. In order to find the sum of infinite series of this A.P let us go as under:Let T=1+2+3+4+5+…….or,T=1+(2+3+4)+(5+6+7)+………or,T=1+9+18+27+………………… or,T=1+9(1+2+3+4+………

or,T=1+9T,or, 9T-T=-1,or,8T=-1,or, T=—(1/8). Similarly, let S =1+3+5+7+……..let us add S with S while shifting the number one step forward. Then it becomes,

S=1+3+5+7+…….

S= +1+3+5+……..

So, 2S=1+4+8+12+……

or, 2S=1+4(1+2+3+4+….

or,2S=1+4T=1-4×1/8( as we calculated vale of T earlier).

or,2S=1-1/2=1/2

or,S=1/4.