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Maths maths help not sure?
The n th term of geometric progression is 7X3^(2-n). Find the sum to infinity of the progression.
how to solve...
3 Answers
- ?Lv 710 years agoFavorite Answer
Tn = 7X3^(2-n)
Put n= 1,2,3,4, ..............
T1 = 7X3^(1) = 21 .........(i)
T2 = 7X3^0 = 7 ............(ii)
T3 = 7X3^-1 = 7/3 .........(iii)
T4 = 7x3^-2 = 7/9 .......(iv)
Therefore T1=a= 21 and common ratio =r = T2/T1 = 1/3
S= Sum to infinity = a/(1-r) = 21/(1-1/3) = 21/(2/3) = 63/2 =31.5 ...........Ans
- Anonymous10 years ago
Tn = 7 x 3^(2 -- n) = 63/3^n
T1 = 63/3 = 21
T2 = 63/3^2 = 7
Sum of infinite terms = (T1) / [ 1 -- T2/T1] = 21 / [1 -- 7/21] = 21*21/(21 -- 7) = 31 1/2 ANSWER
- ?Lv 45 years ago
f ' (x)=3 [(x^2+3)^2]. (2x) f '' (x)= 6 [x.2(x^2+3).2x +(x^2+3)^2 ] (employing product rule) =6[4x^2(x^2+3) + (x^2+3)^2] =6[4x^4+12x^2+x^4+9+6x^2] =30x^4+108x^2+fifty 4. you may get ideas for any sort of maths problems in few clicks interior a count of seconds.
