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ravikumar

Find S_nand t_n of the series 9, 16, 29, 54, 103,............. please give solution very soon

umang

Hey !
subtract 1st term from 2nd and u get the series as 7,13,25,49,...
again , subtracting , u get 6,12,24... which r in GP with r=2 .

joyfrancis

nth term is = n+2+6(2^n-1)

aditi_g

yep umang is rite........but do u know wat to do after that.......
ill tell u in case u dunno how to proceed...
c whn diff of diff is in gp thn the general term is of the form
A r^ n + bx n + c
u know r = 2
for 1 st term 2A + b + c = 9
for 2nd
4A + 2b + c=16
for 3rd
8 A+ 3b + 3=29
frm these equations u vcan find a , b ,c
and thn find the sum.....i hope ull be able to solve it....in case u hv any prob ur free to nudge me.........
rate me if it was useful

priyesh

aditi_g is correct when second difference is in G.P

nth term = a r^n + bn + c

when second difference is in A.P

nth term = cubic in n

also when first diff is in A.P

nth term = quadratic in n

when first diff is in G.P

then nth term is ar^n + b

spideyunlimited

nth term of the progression = n + 2 + [ 6 .2^(n-1)]

Sum of n terms = n²+ 2n + 6._{[1 ]}

^{[ n ]} 2^(n-1)

(**328**)

Here's and easier way:

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