Math, asked by neelammahadik702, 2 months ago

Nth term for 11,18,37,74,135

Answers

Answered by prashantraj0929
0

Answer:

nth term= 11+ (n-1)7

nth term = 11+ 7n- 7

nth term = 4+ 7n

Answered by ravenklaw33exe
0

Answer:

ans=n^2+10

Step-by-step explanation:

so we have a cubic equation as the answer here and we have the 4 formulas for this specific cubic sequence. In the end we get the cubic equation which is an^3+bn^2+cn+d and we substitute this with the values we get for the formulas we use to deduce the n th term of any cubic sequence

we first find the differences:

11,18,37,74,135
 7  19  37  61--first difference
   12   18  24---second difference
       6     6-------third difference

now that we got our 3 differences we procced by substituting them to our 4 formulas
note that these 4 formulas will apply to any cubic equation

for 3rd difference- 6a=(left most value in the 3rd diff which here is 6)
for 2nd difference- 12a+2b=(left most value in 2nd diff which here is 12)
for 1st difference- 7a+3b+c=(the left most value in the 1st diff which here is 7)

for d- a+b+c+d=(left most value in the original sequence which here is 11)

remember that we always use the left most values in all differences


now we substitute

6a=6
  a=6/6
  a=1                                    this is the value for an^3 we substitute in the                    

                                            final equation

12a+2b=12

       2b=12/12

          b=0                           this is bn^2 in the equation


7a+3b+c=7
            c=7-7
            c=0                          this is cn in the equation



a+b+c+d=11
            d=11-1
            d=10                          this is d in the equation



and now we substitute in the cubic equation to get our n th term

n=an^3+bn^2+cn+d
n=(1)n^3+(0)n^2+(0)n+10
n=n^3+10

there our final answer for the n th term of the sequence of 11,18,37,74,135

remembering the cubic formulas and the formulas used to find a,b,c,d is very important as those are the only things that can get you any where in these kind of questions.

Similar questions