# Math probs??

Could anyone here help me in finding the solutions for the problems below???? Thanks😀
1. 12, 13, 17, 26, X, 67, 103. find X. the answer here is 42, but i can't formulate an equation.
2. evaluate (0.7)[SUP]20[/SUP] + (0.7)[SUP]19[/SUP](0.3)(20)....+ (0.7)[SUP]4[/SUP](0.3)[SUP]16[SUP][SUB]20[/SUB]C[SUB]16[/SUB]
3. the arch of an underpass is a semi-ellipse 60ft wide and 20ft high. find the clearance at the edge of a lane if the edge is 20ft from the center.

Thanks again!!😀

## Replies

• KALYANpwn
hello ans for this 12, 13, 17, 26, X, 67, 103. find X. the answer here is 42, but i can't formulate an equation?

ans: (13-12 =1 i.e, 1^2) ,(17-13=4 i.e., 2^2), (26-17=9 i.e., 3^2), (26+ 4^2 =42), (67-42 =25 i.e., 5^2=25),
• maria flor
KALYANpwn
hello ans for this 12, 13, 17, 26, X, 67, 103. find X. the answer here is 42, but i can't formulate an equation?

ans: (13-12 =1 i.e, 1^2) ,(17-13=4 i.e., 2^2), (26-17=9 i.e., 3^2), (26+ 4^2 =42), (67-42 =25 i.e., 5^2=25),
thanks!!!!
• Ramani Aswath
[SUP][/SUP]
maria flor
Could anyone here help me in finding the solutions for the problems below???? Thanks😀
1. 12, 13, 17, 26, X, 67, 103. find X. the answer here is 42, but i can't formulate an equation.
The series is given by:
nth term S[SUB]n[/SUB] = 12 + Sum[SUB]1[/SUB][SUP]n[/SUP]((n-1)2)

e.g.: 5 th term S[SUB]5[/SUB] = 12 +(0[SUP]2[/SUP] + 1[SUP]2[/SUP] +2[SUP]2 [/SUP]+ 3[SUP]2[/SUP] + 4[SUP]2[/SUP]) = 12 + (0+1+4+9+16) = 12 + 30 = 42
• Ramani Aswath
S[SUB]n[/SUB] = 12 + Sum[SUB]1[/SUB][SUP]n[/SUP]((n-1)2) can also be written as:
S[SUB]n[/SUB] = 12 + (2n[SUP]3[/SUP] - 3n[SUP]2 [/SUP]+ n)/6

S[SUB]5[/SUB] = 12 + (250-75+5)/6 = 12 + 180/6 = 12 + 30 = 42 ​which checks.
• Sagar07
i think answer for second question is 1^20=1....
because given problem is the expansion of the (0.7+0.3)^20........
• Ramani Aswath
maria flor
Could anyone help finding the solutions for the problems below? 3. the arch of an underpass is a semi-ellipse 60ft wide and 20ft high. find the clearance at the edge of a lane if the edge is 20ft from the center.
Semi major axis= 30 and Semi minor axis= 20.
Ellipse is x[SUP]2[/SUP]/900 + y[SUP]2[/SUP] /400 = 1, or
4x2 + 9y2 = 3600 with origin at the road center.
For x = 20 (road edge from center) the clearance above is given by the value of y at x = 20
Solving the equation gives y = 14.907 ft. Seems correct.

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