Question #ff2fc

Question

1605 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-31T17:22:50

#" The Reqd. Sum"=580/9~~64.44#.

Let the A.P. be,

#a,a+d,a+2d,a+3d,..., "where", a,d, in RR, and, d!=0#.

Let #t_n, and, S_n# be the #n^(th)# term, &, the sum of first #n#

terms of A.P. resp.

We know that, #t_n=a+(n-1)d, and, S_n=n/2[2a+(n-1)d]#.

#t_3:t_6=9:4 ...."[Given]"rArr (a+2d)/(a+5d)=9/4...........(1)#

Further, # "Given "S_5=60 rArr 5/2(2a+4d)=60, or, (a+2d)=12#.

By #(1)," then, "a+5d=(4*12)/9=16/3#

Therefore, #(a+2d)-(a+5d)=12-16/3 rArr -3d=20/3 rArr d=-20/9#.

Since, #a+2d=12#, we have,

#a=12-2(-20/9)=12+40/9=148/9#.

Finally, the Reqd. Sum #=S_10=10/2(2a+9d)#

#=5[2*148/9+9(-20/9)]#

#=5[296/9-180/9]#

#=(5*116)/9#

#=580/9~~64.44#.

Enjoy Maths.!