THE SUM OF THE FIRST THREE TERMS OF AN ARITHMETIC SEQUENCE IS 15. THE SUM OF THEIR SQUARES IS 147. DETERMINE THE SEQUENCE.EXPLAIN YR REASONING AND JUSTIFY TR WORK.
tn=a+(3-1)d here i plugged in n=3 and Sn=15. what is next eqn?pl. try .thnx.
tn=a+2d
sn=n/2(a+tn) next what to do for sum of squares?
for Tn= a+(n-1)d but for square which formula applies?You have:
\(\displaystyle a + (a+d) + (a+2d) = 15\)
\(\displaystyle a^2+(a+d)^2 + (a+2d)^2 = 147 \)
Do you know how to solve simultaneous equations?
for Tn= a+(n-1)d but for square which formula applies?
Are you really that lazy that you need a formula to do every math problem? I don't know a formula for it, and I don't plan on deriving one or googling for one for you.
dear idn hw to solve simultaneous eqn of different degrees.thnx for yr @
Its the same process, only you may get more than one possible solution. Solve the first one for a or d, and plug into the second one. You will have a quadratic.
For example, if x+y=3 and x^2+(x+y)^2=10, I solve the first for y to get: y=3-x. Plug this into the second to get x^2+(x+[3-x])^2=10, or x^2+9=10. This gives two solutions of x=1 and x=-1 which give two solutions: x=1 and y=2, or x=-1 and y=4.
thnx. i belive i got -1, 5 nd 11 is rt ans.
-1+5+11= 15
and -1^2+ 5^2+11^2= 147
I think you mean (-1)^2. But yes, that is one solution. There is another.