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Answer 1 · 2016-08-31T17:22:50

$The Reqd. Sum = \frac{580}{9} \approx 64.44$ $" The Reqd. Sum"=580/9~~64.44$ .

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.!